diff options
Diffstat (limited to '3204')
201 files changed, 3738 insertions, 0 deletions
diff --git a/3204/CH10/EX10.1/Ex10_1.sce b/3204/CH10/EX10.1/Ex10_1.sce new file mode 100644 index 000000000..182c1c3ea --- /dev/null +++ b/3204/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,29 @@ +// Initilization of variables
+W1=400 // N // vertical load at pt C
+W2=600 // N // vertical load at pt D
+W3=400 // N // vertical load at pt E
+l=2 // m // l= Lac=Lcd=Lde=Leb
+h=2.25 // m // distance of the cable from top
+L=2 // m // dist of A from top
+// Calculations
+// Solving eqn's 1&2 using MATRIX for Xb & Yb
+A=[-L 4*l;-h 2*l]
+B=[((W1*l)+(W2*2*l)+(W1*3*l));(W1*l)]
+C=inv(A)*B
+// Now consider the F.B.D of BE, Take moment at E
+y_e=(C(2)*l)/C(1) // m / here y_e is the distance between E and the top
+theta_1=atand(y_e/l) // degree // where theta_1 is the angle between BE and the horizontal
+T_BE=C(1)/cosd(theta_1) // N (T_BE=T_max)
+// Now consider the F.B.D of portion BEDC
+// Take moment at C
+y_c=((C(2)*6)-(W3*4)-(W2*2))/(C(1)) // m
+theta_4=atand(((y_c)-(l))/(l)) // degree
+T_CA=C(1)/cosd(theta_4) // N // Tension in CA
+// Results
+clc
+printf('(i) The horizontal reaction at B (Xb) is %f N \n',C(1))
+printf('(i) The vertical reaction at B (Yb) is %f N \n',C(2))
+printf('(ii) The sag at point E (y_e) is %f m \n',y_e)
+printf('(iii) The tension in portion CA (T_CA) is %f N \n',T_CA)
+printf('(iv) The max tension in the cable (T_max) is %f N \n',T_BE)
+printf('(iv) The max slope (theta_1) in the cable is %f degree \n',theta_1)
diff --git a/3204/CH10/EX10.2/Ex10_2.sce b/3204/CH10/EX10.2/Ex10_2.sce new file mode 100644 index 000000000..2cc70570e --- /dev/null +++ b/3204/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,31 @@ +// Initiization of variables
+W1=100 // N // Pt load at C
+W2=150 // N // Pt load at D
+W3=200 // N // Pt load at E
+l=1 // m // l=Lac=Lcd=Lde=Leb
+h=2 // m // dist between Rb & top
+Xa=200 // N
+Xb=200 // N
+// Calculations
+// consider the F.B.D of entire cable
+// Take moment at A
+Yb=((W1*l)+(W2*2*l)+(W3*3*l)-(Xb*h))/(4*l) // N
+Ya=W1+W2+W3-Yb // N // sum Fy=0
+// Now consider the F.B.D of AC
+// Take moment at C,
+y_c=(Ya*l)/Xa // m
+theta_1=atand(y_c/l) // degree
+T_AC=Xa/cosd(theta_1) // N // T_AC*cosd(theta_1)=horizontal component of tension in the cable
+// here, T_AC=T_max
+T_max=T_AC // N
+// Now consider the F.B.D of portion ACD
+y_d=((Ya*2*l)-(W1*l))/(Xa) // m // taking moment at D
+theta_2=atand(((y_d)-(y_c))/(l)) // degree
+T_CD=Xa/(cosd(theta_2)) // N
+// Results
+clc
+printf('(i) The component of support reaction at A (Ya) is %f N \n',Ya)
+printf('(i) The component of support reaction at B (Yb) is %f N \n',Yb)
+printf('(ii) The tension in portion AC (T_AC) of the cable is %f N \n',T_AC)
+printf('(ii) The tension in portion CD (T_CD) of the cable is %f N \n',T_CD)
+printf('(iii) The max tension in the cable is %f N \n',T_max)
diff --git a/3204/CH10/EX10.3/Ex10_3.sce b/3204/CH10/EX10.3/Ex10_3.sce new file mode 100644 index 000000000..88c56787e --- /dev/null +++ b/3204/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,23 @@ +// Initilization of variables
+w=75 // kg/m // mass per unit length of thw pipe
+l=20 // m // dist between A & B
+g=9.81 // m/s^2 // acc due to gravity
+y=2 // m // position of C below B
+// Calculations
+// Let x_b be the distance of point C from B
+// In eq'n x_b^2+32*x_b-320=0
+a=1
+b=32
+c=-320
+x_b=(-b+sqrt(b^2-(4*a*c)))/(2*a) // m // we get x_b by equating eqn's 1&2
+// Now tension T_0
+T_0=((w*g*x_b^2)/(2*y))*(10^-3) //kN // from eq'n 1
+// Now the max tension occurs at point A,hence x is given as,
+x=20-x_b // m
+w_x=w*g*x*10^(-3) // kN
+T_max=sqrt((T_0)^2+(w_x)^2) // kN // Maximum Tension
+// Results
+clc
+printf('The lowest point C which is situated at a distance (x_b) from support B is %f m \n',x_b)
+printf('The maximum tension (T_max) in the cable is %f kN \n',T_max)
+printf('The minimum tension (T_0) in the cable is %f kN \n',T_0)
diff --git a/3204/CH10/EX10.4/Ex10_4.sce b/3204/CH10/EX10.4/Ex10_4.sce new file mode 100644 index 000000000..ce4b95727 --- /dev/null +++ b/3204/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+m=0.5 // kg/m // mass of the cable per unit length
+g=9.81 // m/s^2
+x=30 // m // length AB
+y=0.5 // m // dist between C & the horizontal
+x_b=15 // m // dist of horizontal from C to B
+// Calculations
+w=m*g // N/m // weight of the cable per unit length
+T_0=(w*x_b^2)/(2*y) // N // From eq'n 1
+T_B=sqrt((T_0)^2+(w*x/2)^2) // N // Tension in the cable at point B
+W=T_B // N // As pulley is frictionless the tension in the pulley on each side is same,so W=T_B
+// Slope of the cable at B,
+theta=acosd(T_0/T_B) // degree
+// Now length of the cable between C & B is,
+S_cb=x_b(1+((2/3)*(y/x_b)^2)) // m
+// Now total length of the cable AB is,
+S_ab=2*S_cb // m
+// Results
+clc
+printf('(i) The magnitude of load W is %f N \n',W)
+printf('(ii) The angle of the cable with the horizontal at B is %f degree \n',theta)
+printf('(iii) The total length of the cable AB is %f m \n',S_ab)
diff --git a/3204/CH10/EX10.5/Ex10_5.sce b/3204/CH10/EX10.5/Ex10_5.sce new file mode 100644 index 000000000..97afc0868 --- /dev/null +++ b/3204/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+x=30 // m // distance between two electric poles
+Tmax=400 // N // Max Pull or tension
+w=3 // N/m // weight per unit length of the cable
+// Calculations
+// The cable is assumed to be parabolic in shape, its eq'n is y=w*x^2/2*T_0.....(eq'n 1). Substuting the co-ordinates of point B (l/2,h), where h is the sag in the cable.This gives, T_0=(w*(l/2)^2)/(2*h)=wl^2/8*h
+// Now the maximum pull or tension occurs at B,
+T_B=Tmax // N
+// Hence T_B=Tmax=sqrt(T_0^2+(w*l/2)^2). On simplyfyingthis eq'n we get,
+h=sqrt(x^2/(16*(((Tmax*2)/(w*x))^2-(1)))) // m
+// Results
+clc
+printf('The smallest value of the sag in the cable is %f m \n',h)
diff --git a/3204/CH10/EX10.6/Ex10_6.sce b/3204/CH10/EX10.6/Ex10_6.sce new file mode 100644 index 000000000..1f53c9701 --- /dev/null +++ b/3204/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,19 @@ +// Initilization of variables
+l=200 // m // length of the cable
+m=1000 // kg // mass of the cable
+S=50 // m // sag in the cable
+s=l/2 // m
+g=9.81 // m/s^2
+// Calculations
+w=(m*g)/l // N/m // mass per unit length of the cable
+// Substuting the values s=l/2 & y=c+S in eq'n 1 to get the value of c,
+c=7500/100 // m
+Tmax=sqrt((w*c)^2+(w*s)^2) // N // Maximum Tension
+// To determine the span (2*x) let us use the eq'n of catenary, y=c*cosh(x/c), where y=c+50. On simplyfying we get y/c=cosh(x/c), here let y/c=A
+y=c+50
+A=y/c
+x=c*(acosh(A)) // m
+L=2*x // m // where L= span
+// Results
+clc
+printf('The horizontal distance between the supports and the max Tension (L) is %f m \n',L)
diff --git a/3204/CH12/EX12.10/Ex12_10.sce b/3204/CH12/EX12.10/Ex12_10.sce new file mode 100644 index 000000000..8ee58271f --- /dev/null +++ b/3204/CH12/EX12.10/Ex12_10.sce @@ -0,0 +1,49 @@ +// Initilization of variables
+b1=120 // mm // width of the flange pate of L-section
+d1=20 // mm // depth of the flange plate
+b2=20 // mm // width/thickness of the web
+d2=130 // mm // depth of the web
+// Calculations
+// (a) Location of the centroid of the composite area
+A_1=b1*d1 // mm^2 // area of the flange plate
+A_2=b2*d2 // mm^2 // area of the web
+y_1=d2+(d1/2) // mm // y-coordinate of the centroid
+y_2=d2/2 // mm // y-coordinate of the centroid
+x_1=60 // mm // x-coordinate of the centroid
+x_2=110 // mm // x-coordinate of the centroid
+y_c=((A_1*y_1)+(A_2*y_2))/(A_1+A_2) // mm // from the bottom edge
+x_c=((A_1*x_1)+(A_2*x_2))/(A_1+A_2) // mm // from the bottom edge
+// (b) Moment of Inertia of the composite area about the centroidal x-axis
+// Area (A_1) M.I of area A_1 about x-axis
+I_x1=(b1*(d1^3))/12 // mm^4
+// M.I of the area A_1 about the centroidal x-axis of the composite area (By parallel-axis theorem)
+OC_1=d2+(d1/2) // mm // from the bottom edge
+OC_2=d2/2 // mm // from the bottom edge
+OC=y_c // mm // from the bottom edge
+d_1=(d2-y_c)+(d1/2) // mm
+d_2=y_c-OC_2 // mm
+I_X1=(I_x1)+(A_1*d_1^2) // mm^4
+// Area(A_2) M.I of area A_2 about x-axis
+I_x2=(b2*d2^3)/12 // mm^4
+// M.I of the area A_2 about the centroidal x-axis of the composite area (By parallel-axis theorem)
+I_X2=(I_x2)+(A_2*d_2^2) // mm^4
+// COMPOSITE AREA:M.O.I of the composite area about the centroidal x-axis
+I_x=(I_X1)+(I_X2) // mm^4
+// (c) Moment of Inertia of the composite area about the centroidal y-axis
+// Area (A_1) M.I of area A_1 about y-axis
+I_y1=(d1*(b1^3))/12 // mm^4
+// M.I of the area A_1 about the centroidal y-axis of the composite area (By parallel-axis theorem)
+d_3=x_c-(b1/2) // mm // distance between c &c1 along x axis
+I_Y1=(I_y1)+(A_1*d_3^2) // mm^4
+// Area(A_2) M.I of area A_2 about y-axis
+I_y2=(d2*b2^3)/12 // mm^4
+// M.I of the area A_2 about the centroidal y-axis of the composite area (By parallel-axis theorem)
+d_4=b1-x_c-(b2/2) // mm // distance between c &c2 along x axis
+I_Y2=(I_y2)+(A_2*d_4^2) // mm^4
+// COMPOSITE AREA:M.O.I of the composite area about the centroidal y-axis
+I_y=(I_Y1)+(I_Y2) // mm^4
+// Results
+clc
+printf('The M.O.I of the composite area about the centroidal x-axis is %f mm^4 \n',I_x)
+printf('The M.O.I of the composite area about the centroidal Y-axis is %f mm^4 \n',I_y)
+// NOTE: The answer for I_x given in text book is 0.76*10^6 insted of 10.76*10^6
diff --git a/3204/CH12/EX12.14/Ex12_14.sce b/3204/CH12/EX12.14/Ex12_14.sce new file mode 100644 index 000000000..c615359db --- /dev/null +++ b/3204/CH12/EX12.14/Ex12_14.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+b=1 // cm // smaller side of the L-section
+h=4 // cm // larger side of the L-section
+// Calculations
+// (A) RECTANGLE A_1: Using the paralel axis theorem
+Ixy=0
+I_xy1=(Ixy)+((h*b)*(b/2)*(h/2)) // cm^4
+// (B) RECTANGLE A_2: Using the paralel axis theorem
+I_xy2=(Ixy)+((b*(h-1))*(1+(3/2))*(b/2)) // cm^4
+// Product of inertia of the total area
+I_xy=I_xy1+I_xy2 // cm^4
+// Calculations
+clc
+printf('The Product of inertia of the L-section is %f cm^4 \n',I_xy)
diff --git a/3204/CH12/EX12.15/Ex12_15.sce b/3204/CH12/EX12.15/Ex12_15.sce new file mode 100644 index 000000000..3f391ca12 --- /dev/null +++ b/3204/CH12/EX12.15/Ex12_15.sce @@ -0,0 +1,29 @@ +// Initilization of variables
+I_x=1548 // cm^4 // M.O.I of the Z-section about X-axis
+I_y=2668 // cm^4 // M.O.I of the Z-section about Y-axis
+b=12 // cm // width of flange of the Z-section
+d=3 // cm // depth of flange of the Z-section
+t=2 // cm // thickness of the web of the Z-section
+h=6 // cm // depth of the web of the Z-section
+//Calculations
+A_1=b*d // cm^2 // area of top flange
+x_1=-5 // cm // distance of the centroid from X-axis for top flange
+y_1=4.5 // cm // distance of the centroid from Y-axis for top flange
+A_2=t*h // cm^2 // area of web
+x_2=0 // cm // distance of the centroid from X-axis for the web
+y_2=0 // cm // distance of the centroid from Y-axis for the web
+A_3=b*d // cm^2 // area of bottom flange
+x_3=5 // cm // distance of the centroid from X-axis for top flange
+y_3=-4.5 // cm // distance of the centroid from Y-axis for top flange
+// Product of Inertia of the total area is,
+I_xy=((A_1*x_1*y_1)+(A_3*x_3*y_3)) // cm^4
+// The direction of the principal axes is,
+theta_m=(atand((2*I_xy)/(I_y-I_x)))/2 // degree
+// Principa M.O.I
+I_max=((I_x+I_y)/2)+(sqrt(((I_x-I_y)/2)^2+(I_xy)^2)) // cm^4
+I_mini=((I_x+I_y)/2)-(sqrt(((I_x-I_y)/2)^2+(I_xy)^2)) // cm^4
+// Results
+clc
+printf('The principal axes of the section about O is %f degree \n',theta_m)
+printf('The Maximum value of principal M.O.I is %f cm^4 \n',I_max)
+printf('The Minimum value of principal M.O.I is %f cm^4 \n',I_mini)
diff --git a/3204/CH12/EX12.7/Ex12_7.sce b/3204/CH12/EX12.7/Ex12_7.sce new file mode 100644 index 000000000..13059d1b9 --- /dev/null +++ b/3204/CH12/EX12.7/Ex12_7.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+A= 50 // cm^2 // area of the shaded portion
+J_A=22.5*10^2 // cm^4 // polar moment of inertia of the shaded portion
+d=6 // cm
+// Calculations
+J_c=J_A-(A*d^2)
+// substuting the value of I_x from eq'n 2 in eq'n 1 we get,
+I_y=J_c/3 // cm^4 // M.O.I about Y-axis
+// Now from eq'n 2,
+I_x=2*I_y // cm^4 // M.O.I about X-axis
+// Results
+clc
+printf('The centroidal moment of inertia about X-axis (I_x) is %f cm^4 \n',I_x)
+printf('The centroidal moment of inertia about Y-axis (I_y) is %f cm^4 \n',I_y)
diff --git a/3204/CH12/EX12.8/Ex12_8.sce b/3204/CH12/EX12.8/Ex12_8.sce new file mode 100644 index 000000000..1d174e15a --- /dev/null +++ b/3204/CH12/EX12.8/Ex12_8.sce @@ -0,0 +1,32 @@ +// Initilization of variables
+b=20 // cm // width of the pate
+d=30 // cm // depth of the plate
+r=15 // cm // radius of the circular hole
+h=20 // cm // distance between the centre of the circle & the x-axis
+// Calculations
+// (a) Location of the centroid of the composite area
+A_1=b*d // cm^2 // area of the plate
+y_1=d/2 // cm // y-coordinate of the centroid
+A_2=(%pi*r^2)/4 // cm^2 // area of the circle removed (negative)
+y_2=h // cm // y-coordinate of the centroid
+y_c=((A_1*y_1)-(A_2*y_2))/(A_1-A_2) // cm // from the bottom edge
+// (b) Moment of Inertia of the composite area about the centroidal x-axis
+// Area (A_1) M.I of area A_1 about x-axis
+I_x1=(b*(d^3))/12 // cm^4
+// M.I of the area A_1 about the centroidal x-axis of the composite area (By parallel-axis theorem)
+OC_1=15 // cm // from the bottom edge
+OC_2=20 // cm
+OC=12.9 // cm // from the bottom edge
+d_1=OC_1-OC // cm
+d_2=OC_2-OC // cm
+I_X1=(I_x1)+(A_1*d_1^2) // cm^4
+// Area(A_2) M.I of area A_2 about x-axis
+I_x2=(%pi*r^4)/64 // cm^2
+// M.I of the area A_2 about the centroidal x-axis of the composite area (By parallel-axis theorem)
+I_X2=(I_x2)+(A_2*d_2^2) // cm^4
+// COMPOSITE AREA:M.O.I of the composite area about the centroidal x-axis
+I_x=(I_X1)-(I_X2) // cm^4
+// Results
+clc
+printf('The M.O.I of the composite area about the centroidal x-axis is %f cm^4 \n',I_x)
+// There may be a small error in the answer due to decimal point discrepancy
diff --git a/3204/CH12/EX12.9/Ex12_9.sce b/3204/CH12/EX12.9/Ex12_9.sce new file mode 100644 index 000000000..b20336816 --- /dev/null +++ b/3204/CH12/EX12.9/Ex12_9.sce @@ -0,0 +1,32 @@ +// Initilization of variables
+b1=80 // mm // width of the flange pate
+d1=20 // mm // depth of the flange plate
+b2=40 // mm // width/thickness of the web
+d2=60 // mm // depth of the web
+// Calculations
+// (a) Location of the centroid of the composite area
+A_1=b1*d1 // mm^2 // area of the flange plate
+y_1=d2+(d1/2) // mm // y-coordinate of the centroid
+A_2=b2*d2 // mm^2 // area of the web
+y_2=d2/2 // mm // y-coordinate of the centroid
+y_c=((A_1*y_1)+(A_2*y_2))/(A_1+A_2) // mm // from the bottom edge
+// (b) Moment of Inertia of the composite area about the centroidal x-axis
+// Area (A_1) M.I of area A_1 about x-axis
+I_x1=(b1*(d1^3))/12 // mm^4
+// M.I of the area A_1 about the centroidal x-axis of the composite area (By parallel-axis theorem)
+OC_1=70 // mm // from the bottom edge
+OC_2=30 // mm // from the bottom edge
+OC=y_c // mm // from the bottom edge
+d_1=(d2-y_c)+(d1/2) // mm
+d_2=y_c-OC_2 // mm
+I_X1=(I_x1)+(A_1*d_1^2) // mm^4
+// Area(A_2) M.I of area A_2 about x-axis
+I_x2=(b2*d2^3)/12 // mm^4
+// M.I of the area A_2 about the centroidal x-axis of the composite area (By parallel-axis theorem)
+I_X2=(I_x2)+(A_2*d_2^2) // mm^4
+// COMPOSITE AREA:M.O.I of the composite area about the centroidal x-axis
+I_x=(I_X1)+(I_X2) // mm^4
+// Results
+clc
+printf('The M.O.I of the composite area about the centroidal x-axis is %f mm^4 \n',I_x)
+// NOTE: The answer given in the text book is 2.31*10^3 insted of 2.31*10^6.
diff --git a/3204/CH13/EX13.1/Ex13_1.sce b/3204/CH13/EX13.1/Ex13_1.sce new file mode 100644 index 000000000..1683b56a6 --- /dev/null +++ b/3204/CH13/EX13.1/Ex13_1.sce @@ -0,0 +1,8 @@ +// Initilization of variables
+W=1000 // N // weight to be raised
+// Calculations
+// From the Principle of virtual work,
+P=W/2 // N
+// Results
+clc
+printf('The value of force (i.e P) that can hold the system in equilibrium is %f N \n',P)
diff --git a/3204/CH13/EX13.7/Ex13_7.sce b/3204/CH13/EX13.7/Ex13_7.sce new file mode 100644 index 000000000..a493cc1ea --- /dev/null +++ b/3204/CH13/EX13.7/Ex13_7.sce @@ -0,0 +1,20 @@ +// Initilization of variables
+P=1000 // N // Force acting at the hinge of the 1st square
+Q=1000 // N // Force acting at the hinge of the 2nd square
+// Calculations
+// Chosing the co-ordinate system with originat A, we can write,
+theta=45 // degree
+// Forces that do work are P,Q & X_B. Applying the principle of virtual work & Simplyfying and solving for X_B,
+X_B=((2*P)/6)*(cosd(theta)/sind(theta)) // N
+// Now give a virtual angular displacement to the whole frame about end A such that line AB turns by an angle delta_phi.
+// The force doing work are P,Q&Y_B.Applying the principle of virtual work & Simplyfying this eq'n and solving for Y_B,
+Y_B=((3*Q)+P)/6 // N
+// Simply by removing the support at A & replacing it by the reactions X_A & Y_A we can obtain,
+X_A=X_B // N
+Y_A=P+Q-Y_B // N
+// Results
+clc
+printf('The Horizontal component of reaction at A (X_A) is %f N \n',X_A)
+printf('The Vertical component of reaction at A (Y_A) is %f N \n',Y_A)
+printf('The Horizontal component of reaction at B (X_B) is %f N \n',X_B)
+printf('The Vertical component of reaction at B (Y_B) is %f N \n',Y_B)
diff --git a/3204/CH14/EX14.10/Ex14_10.sce b/3204/CH14/EX14.10/Ex14_10.sce new file mode 100644 index 000000000..b8dd45d23 --- /dev/null +++ b/3204/CH14/EX14.10/Ex14_10.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+a=10 // m/s^2 // acceleration of the particle
+S_5th=50 // m // distance travelled by the particle during the 5th second
+t=5 // seconds
+// Calculations
+// The distance travelled by the particle in time t is given by, S=(u*t)+(1/2)*a*t^2.....(consider this as eq'n 1)
+// Here, The distance travelled by the particle in the 5th second=The distance travelled in 5 seconds - The distance travelled in 4 seconds..... (consider eq'n 2)
+// Using eq'n 1: S_(0-5)=(5*u)+(1/2)*10*5^2 = 5*u+125.....(consider eq'n 3)
+// again, S_(0-4)=(4*u)+(1/2)*10*4^2 = 4*u+80....(consider eq'n 4)
+// Now,put eq'n 3&4 in eq'n 2 and solve for u. We get, 50=[(5*u+125)-(4*u+80)] i.e 50=u+45
+u=(S_5th)-45 // m/s
+// Calculations
+clc
+printf('The initial velocity of the particle is %f m/s \n',u)
diff --git a/3204/CH14/EX14.11/Ex14_11.sce b/3204/CH14/EX14.11/Ex14_11.sce new file mode 100644 index 000000000..38ff8a24c --- /dev/null +++ b/3204/CH14/EX14.11/Ex14_11.sce @@ -0,0 +1,17 @@ +// Initilization of variables
+// Conditions given are
+t=1 // s
+x=14.75 // m
+v=6.33 // m/s
+// Calculations
+// We use expression 1,2 & 3 to find distance,velocity & acceleration of the particle after 2 sec
+T=2 // sec
+X=(T^4/12)-(T^3/3)+(T^2)+(5*T)+9 // m // eq'n 3
+V=(T^3/3)-(T^2)+(2*T)+5 // m/s
+a=(T^2)-(2*T)+2 // m/s^2
+// Results
+clc
+printf('The distance travelled by the particle is %f m \n',X)
+printf('The velocity of the particle is %f m/s \n',V)
+printf('The acceleration of the particle is %f m/s^2 \n',a)
+// The answer may vary due to decimal point error
diff --git a/3204/CH14/EX14.12/Ex14_12.sce b/3204/CH14/EX14.12/Ex14_12.sce new file mode 100644 index 000000000..fdf68968f --- /dev/null +++ b/3204/CH14/EX14.12/Ex14_12.sce @@ -0,0 +1,13 @@ +// Calculations
+// From eq'n 2 it is clear that velocity of the particle becomes zero at t=3 sec
+t=3 // sec .. from eq'n 2
+// Position of particle at t=3 sec
+x=(t^3)-(3*t^2)-(9*t)+12 // m // from eq'n 1
+// Acc of particle at t=3 sec
+a=6*(t-1) // m/s^2 // from eq'n 3
+// Results
+clc
+printf('The time at which the velocity of the particle becomes zero is %f sec \n',t)
+printf('The position of the partice at t=3 sec is %f m \n',x)
+printf('The acceleration of the particle is %f m/s^2 \n',a)
+// Ref textbook for the graphs
diff --git a/3204/CH14/EX14.15/Ex14_15.sce b/3204/CH14/EX14.15/Ex14_15.sce new file mode 100644 index 000000000..42e8aac3a --- /dev/null +++ b/3204/CH14/EX14.15/Ex14_15.sce @@ -0,0 +1,9 @@ +// Initilization of variables
+F=250 // N // Force acting on a body
+m=100 // kg // mass of the body
+// Calculations
+// Using the eq'n of motion
+a=F/m // m/s^2
+// Results
+clc
+printf('The acceleration of the body is %f m/s^2 \n',a)
diff --git a/3204/CH14/EX14.16/Ex14_16.sce b/3204/CH14/EX14.16/Ex14_16.sce new file mode 100644 index 000000000..c3024a6c0 --- /dev/null +++ b/3204/CH14/EX14.16/Ex14_16.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+a=1 // m/s^2 // downward/upward acceleration of the elevator
+W=500 // N // Weight of man
+g=9.81 // m/s^2 // acceleration due to gravity
+// Calculations
+// (a) Downward Motion
+R_1=W*(1-(a/g)) // N // (Assume pressure as R_1)
+// (b) Upward Motion
+R_2=W*(1+(a/g)) // N // (Assume pressure as R_2)
+// Results
+clc
+printf('(a) The pressure transmitted to the floor by the man for Downward motion of the elevator is %f N \n',R_1)
+printf('(b) The pressure transmitted to the floor by the man for Upward motion of the elevator is %f N \n',R_2)
diff --git a/3204/CH14/EX14.17/Ex14_17.sce b/3204/CH14/EX14.17/Ex14_17.sce new file mode 100644 index 000000000..f6e47e71e --- /dev/null +++ b/3204/CH14/EX14.17/Ex14_17.sce @@ -0,0 +1,20 @@ +// Initilization of variables
+W=5000 // N // Total weight of the elevator
+u=0 // m/s
+v=2 // m/s // velocity of the elevator
+s=2 // m // distance traveled by the elevator
+t=2 // seconds // time to stop the lift
+w=600 // N // weight of the man
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Acceleration acquired by the elevator after travelling 2 m is given by,
+a=sqrt((v^2-u^2)/(2*s)) // m/s^2
+// (a) Let T be the the tension in the cable which is given by eq'n,
+T=W*(1+(a/g)) // N
+// (b) Motion of man
+// Let R be the pressure experinced by the man.Then from the Eq'n of motion of man pressure is given as,
+R=w*(1-(a/g)) // N
+// Results
+clc
+printf('(a) The Tensile force in the cable is %f N \n',T)
+printf('(b) The pressure transmitted to the floor by the man is %f N \n',R)
diff --git a/3204/CH14/EX14.18/Ex14_18.sce b/3204/CH14/EX14.18/Ex14_18.sce new file mode 100644 index 000000000..40bbf8e27 --- /dev/null +++ b/3204/CH14/EX14.18/Ex14_18.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+M_1=10 // kg // mass of the 1st block
+M_2=5 // kg // mass of the 2nd block
+mu=0.25 // coefficient of friction between the blocks and the surface
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+a=g*(M_2-(mu*M_1))/(M_1+M_2) // m/s^2 // from eq'n 5
+T=M_1*M_2*g*(1+mu)/(M_1+M_2) // N // from eq'n 6
+// Results
+clc
+printf('The acceleration of the masses is %f m/s^2 \n',a)
+printf('The tension in the string is %f N \n',T)
diff --git a/3204/CH14/EX14.19/Ex14_19.sce b/3204/CH14/EX14.19/Ex14_19.sce new file mode 100644 index 000000000..9218ea01d --- /dev/null +++ b/3204/CH14/EX14.19/Ex14_19.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+M_1=150 // kg // mass of the 1st block
+M_2=100 // kg // mass of the 2nd block
+mu=0.2 // coefficient of friction between the blocks and the inclined plane
+g=9.81 // m/s^2 // acc due to gravity
+theta=45 // degree // inclination of the surface
+// Calculations
+// substuting the value of eq'n 3 in eq'n 1 & solving for T,we get value of T as,
+T=((M_1*M_2*g)*(sind(theta)+2-(mu*cosd(theta))))/((4*M_1)+(M_2)) // N
+// Results
+clc
+printf('The tension in the string during the motion of the system is %f N \n',T)
diff --git a/3204/CH14/EX14.20/Ex14_20.sce b/3204/CH14/EX14.20/Ex14_20.sce new file mode 100644 index 000000000..abb0e4025 --- /dev/null +++ b/3204/CH14/EX14.20/Ex14_20.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+M_1=5 // kg // mass of the 1st block
+theta_1=30 // degree // inclination of the 1st plane
+M_2=10 // kg // mass of the 2nd block
+theta_2=60 // degree // inclination of the 2nd plane
+mu=0.33 // coefficient of friction between the blocks and the inclined plane
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// solving eq'n 1 & 2 for a we get,
+a=((((M_2*(sind(theta_2)-(mu*cosd(theta_2))))-(M_1*(sind(theta_1)+(mu*cosd(theta_1))))))*g)/(M_1+M_2) // m/s^2
+// Results
+clc
+printf('The acceleration of the masses is %f m/s^2 \n',a)
diff --git a/3204/CH14/EX14.21/Ex14_21.sce b/3204/CH14/EX14.21/Ex14_21.sce new file mode 100644 index 000000000..f6aa0ef95 --- /dev/null +++ b/3204/CH14/EX14.21/Ex14_21.sce @@ -0,0 +1,21 @@ +// Initilization of variables
+S=5 // m // distance between block A&B
+mu_A=0.2 // coefficient of friction between the block A and the inclined plane
+mu_B=0.1 // coefficient of friction between the block B and the inclined plane
+theta=20 // degree // inclination of the pane
+g=9.81 // m/s^2 // acc due to gravity
+// Calculatio//
+// EQUATION OF MOTION OF BLOCK A:
+// Let a_A & a_B be the acceleration of block A & B.
+a_A=(g*sind(theta))-(mu_A*g*cosd(theta)) // m/s^2 // from eq'n 1 & eq'n 2
+// EQUATION OF MOTION OF BLOCK B:
+a_B=g*((sind(theta))-(mu_B*cosd(theta))) // m/s^2 // from eq'n 3 & Rb
+// Now the eq'n for time of collision of the blocks is given as,
+t=sqrt((S*2)/(a_B-a_A)) // seconds
+S_A=(1/2)*a_A*t^2 // m // distance travelled by block A
+S_B=(1/2)*a_B*t^2 // m // distance travelled by block B
+// Results
+clc
+printf('The time before collision is %f seconds \n',t)
+printf('The distance travelled by block A before collision is %f m \n',S_A)
+printf('The distance travelled by block B before collision is %f m \n',S_B)
diff --git a/3204/CH14/EX14.22/Ex14_22.sce b/3204/CH14/EX14.22/Ex14_22.sce new file mode 100644 index 000000000..8aad11eba --- /dev/null +++ b/3204/CH14/EX14.22/Ex14_22.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+P=50 // N // Weight of the car
+Q=100 // N // Weight of the rectangular block
+g=9.81 // m/s^2 // acc due to gravity
+b=25 // cm // width of the rectangular block
+d=50 // cm // depth of the block
+// Calculations
+a=(Q*g)/(4*P+2*Q) // m/s^2 // from eq'n 4
+W=(Q*(P+Q))/(4*P+Q) // N // from eq'n 6
+// Resuts
+clc
+printf('The maximum value of weight (W) by which the car can be accelerated is %f N \n',W)
+printf('The acceleration is %f m/s^2 \n',a)
diff --git a/3204/CH14/EX14.23/Ex14_23.sce b/3204/CH14/EX14.23/Ex14_23.sce new file mode 100644 index 000000000..95fcf24f1 --- /dev/null +++ b/3204/CH14/EX14.23/Ex14_23.sce @@ -0,0 +1,10 @@ +// Initilization of variables
+P=40 // N // weight on puley r_1
+Q=60 // N // weight on pulley r_2
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// The eq'n for acceleration of pulley Pi.e a_p is,
+a_p=(((2*P)-(Q))/((4*P)+(Q)))*2*g // m/s^2
+// Results
+clc
+printf('The downward acceleration of P is %f m/s^2 \n',a_p)
diff --git a/3204/CH14/EX14.24/Ex14_24.sce b/3204/CH14/EX14.24/Ex14_24.sce new file mode 100644 index 000000000..a14a9e70a --- /dev/null +++ b/3204/CH14/EX14.24/Ex14_24.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+M=15 // kg // mass of the wedge
+m=6 // kg // mass of the block
+theta=30 // degree // angle of the wedge
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+a_A=((m*g*cosd(theta)*sind(theta))/((M)+(m*(sind(theta))^2)))/(g) // g // By eliminating R_1 from eq'n 1&3.
+// Here, assume a_r is the acceleration of block B relative to wedge A which is given by substuting a_A in eq'n 2
+a_r=(((g*sind(theta))*(m+M))/((M)+(m*(sind(theta))^2)))/(g) // g
+// Results
+clc
+printf('(a) The acceleration of the wedge is %f g \n',a_A)
+printf('(b) The acceleration of the bock relative to the wedge is %f g \n',a_r)
diff --git a/3204/CH14/EX14.25/Ex14_25.sce b/3204/CH14/EX14.25/Ex14_25.sce new file mode 100644 index 000000000..a3404c0bd --- /dev/null +++ b/3204/CH14/EX14.25/Ex14_25.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+P=30 // N // weight on pulley A
+Q=20 // N // weight on pulley B
+R=10 // N // weight on puey B
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Solving eqn's 6 & 7 using matrix for a & a_1, we get
+A=[70 -40;-10 30]
+B=[10;-10]
+C=inv(A)*B
+// Acceleration of P is given as,
+P=C(1) // m/s^2
+// Acceleration of Q is given as,
+Q=C(2)-C(1) // m/s^2
+// Acceleration of R is given as,
+R=-(C(2)+C(1)) // m/s^2 // as R is taken to be +ve
+// Results
+clc
+printf('The acceleration of P is %f g \n',P)
+printf('The acceleration of Q is %f g \n',Q)
+printf('The acceleration of R is %f g \n',R)
+// Here the -ve sign indicates deceleration or backward/downward acceleation.
diff --git a/3204/CH14/EX14.3/Ex14_3.sce b/3204/CH14/EX14.3/Ex14_3.sce new file mode 100644 index 000000000..9db304e2c --- /dev/null +++ b/3204/CH14/EX14.3/Ex14_3.sce @@ -0,0 +1,15 @@ +// Initilization of variables
+a_T=0.18 // m/s^2 // acc of trolley
+// Calculations
+a_B=-a_T/3 // m/s^2 // from eq'n 4
+t=4 // seconds
+v_T=a_T*t // m/s // velocity of trolley after 4 seconds
+v_B=-v_T/3 // m/s // from eq'n 3
+S_T=(1/2)*a_T*t^2 // m // distance moved by trolley in 4 sec
+S_B=-S_T/3 // m // from eq'n 2
+// Results
+clc
+printf('The acceleration of block B is %f m/s^2 \n',a_B)
+printf('The velocity of trolley & the block after 4 sec is %f m/s & %f m/s \n',v_T,v_B)
+printf('The distance moved by the trolley & the block is %f m & %f m \n',S_T,S_B)
+// The -ve sign indicates that the velocity or the distance travelled is in opposite direction.
diff --git a/3204/CH14/EX14.30/Ex14_30.sce b/3204/CH14/EX14.30/Ex14_30.sce new file mode 100644 index 000000000..1453cbdb0 --- /dev/null +++ b/3204/CH14/EX14.30/Ex14_30.sce @@ -0,0 +1,30 @@ +// Initilization of variables
+W=1 // kg/m // weight of the bar
+L_AB=0.6 // m // length of segment AB
+L_BC=0.30 // m // length of segment BC
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D.
+theta_1=atand(5/12) // slope of bar AB // here theta_1= atan(theta)
+theta_2=asind(5/13) // theta_2=asin(theta)
+theta_3=acosd(12/13) // theta_3=acos(theta)
+M_AB=L_AB*W // kg acting at D // Mass of segment AB
+M_BC=L_BC*W // kg acting at E // Mass of segment BC
+// The various forces acting on the bar are:
+// Writing the eqn's of dynamic equilibrium
+Y_A=(L_AB*g)+(L_BC*g) // N // sum F_y=0
+// Using moment eq'n Sum M_A=0:Here,in this eq'n the values are as follows,
+AF=L_BC*cosd(theta_3)
+DF=L_BC*sind(theta_2)
+AH=(L_AB*cosd(theta_3))+((L_BC/2)*sind(theta_2))
+IG=(L_AB*sind(theta_2))-((L_BC/2)*cosd(theta_3))
+// On simplifying and solving moment eq'n we get a as,
+a=((2*L_AB*L_BC*g*sind(theta_2))-(L_BC*g*(L_BC/2)*cosd(theta_3)))/((2*L_AB*L_BC*cosd(theta_3))+(L_BC*(L_BC/2)*sind(theta_2))) // m/s^2
+X_A=0.9*a //N // from eq'n of dynamic equilibrium
+R_A=sqrt(X_A^2+Y_A^2) // N // Resultant of R_A
+alpha=atand(Y_A/X_A) // degree
+// Results
+clc
+printf('The acceleration is %f m/s^2 \n',a)
+printf('The reaction at A (R_A) is %f N \n',R_A)
+printf('The angle made by the resultant is %f degree \n',alpha)
diff --git a/3204/CH14/EX14.4/Ex14_4.sce b/3204/CH14/EX14.4/Ex14_4.sce new file mode 100644 index 000000000..602f17a8f --- /dev/null +++ b/3204/CH14/EX14.4/Ex14_4.sce @@ -0,0 +1,16 @@ +// Initiliztion of variables
+v_B=12 // cm/s // velocity of block B
+u=0
+s=24 // cm // distance travelled by bock B
+t=5 // seconds
+// Calculations
+a_B=v_B^2/(2*s) // cm/s^2 // using eq'n v^2-u^2=28*a*s for block B. Here u=0
+a_A=(3/2)*a_B // cm/s^2 // from eq'n 4 // Here a_A is negative which means acceleration is in opposite direction. However we consider +ve values for further calculations
+v_A=u+(a_A*t) // m/s // using eq'n v=u+(a*t)
+S_A=(u*t)+((1/2)*a_A*t^2) // m // using eq'n S=(u*t)+((1/2)*a*t^2)
+// Results
+clc
+printf('The acceleration of block A (a_A) is %f cm/s^2 \n',a_A)
+printf('The acceleration of block B (a_B) is %f cm/s^2 \n',a_B)
+printf('The velocity of block A (v_A) after 5 seconds is %f m/s \n',v_A)
+printf('The position of block A (S_A) after 5 seconds is %f m \n',S_A)
diff --git a/3204/CH14/EX14.5/Ex14_5.sce b/3204/CH14/EX14.5/Ex14_5.sce new file mode 100644 index 000000000..90aedae9d --- /dev/null +++ b/3204/CH14/EX14.5/Ex14_5.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+u=72*(1000/(60*60)) // km/hr // speed of the vehicle
+s=300 // m // distance where the light is turning is red
+t=20 // s // traffic light timed to remain red
+// Calculations
+// Now to find the acceleration we use the eq'n s=u*t+(1/2)*a*t^2
+a=(((s)-(u*t))*2)/t^2 // m/s^2 (Deceleration)
+v=(u+(a*t))*((60*60)/1000) // km/hr // here we multiply with (60*60)/1000 to convert m/s to km/hr
+// Results
+clc
+printf('(a) The required uniform acceleration of the car is %f m/s^2 \n',a)
+printf('(b) The speed at which the motorist crosses the traffic light is %f km/hr \n',v)
diff --git a/3204/CH14/EX14.6/Ex14_6.sce b/3204/CH14/EX14.6/Ex14_6.sce new file mode 100644 index 000000000..7bc6b29c7 --- /dev/null +++ b/3204/CH14/EX14.6/Ex14_6.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+S=50 // m // height of the tower
+v=25 // m/s // velocity at which the stone is thrown up from the foot of the tower
+g=9.81 // m/s^2 // acc due to graity
+// Calculations
+// The equation of time for the two stones to cross each other is given as,
+t=S/v // seconds
+S_1=(1/2)*g*t^2 // m // from the top
+// Results
+clc
+printf('The time (t) at which the two stones cross each other is %f seconds \n',t)
+printf('The two stones cross each other (from top) at a distance of %f m \n',S_1)
diff --git a/3204/CH14/EX14.7/Ex14_7.sce b/3204/CH14/EX14.7/Ex14_7.sce new file mode 100644 index 000000000..5092b6be2 --- /dev/null +++ b/3204/CH14/EX14.7/Ex14_7.sce @@ -0,0 +1,22 @@ +// Intilization of variables
+acc=0.5 // m/s^2 // acceleration of the elevator
+s=25 // m // distance travelled by the elevator from the top
+u=0 // m/s
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Using eq'n the eq'n v^2-u^2=2*a*s, solving for v we get,
+v=sqrt((2*acc*s)+(u^2)) // m/s
+// Now solving eq'n 1 & 2 for t we get, (4.655*t^2)-(5*t)+(25)=0
+// Find the roots of the eq'n using the eq'n,t=(-b+sqrt(b^2-(4*a*c)))/(2*a).In this eq'n the values of a,b & c are,
+a=4.655
+b=-5
+c=-25
+t=(-b+sqrt((b^2)-(4*a*c)))/(2*a) // seconds
+// Let S_1 be the distance travelled by the elevator after it travels 25 m from top when the stone hits the elevator,This disance S_1 is given as,
+S_1=(v*t)+((1/2)*acc*t^2) // m
+// Let S be the total dist from top when the stone hits the elevator,
+S=S_1+s // m
+// Results
+clc
+printf('The time taken by the stone to hit the elevator is %f seconds \n',t)
+printf('The distance (S)travelled by the elevator at the time of impact is %f m \n',S)
diff --git a/3204/CH14/EX14.9/Ex14_9.sce b/3204/CH14/EX14.9/Ex14_9.sce new file mode 100644 index 000000000..5fc24cb7e --- /dev/null +++ b/3204/CH14/EX14.9/Ex14_9.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+v=60 // km/hr // velocity of the train
+d1=15 // km // Distance travelled by the local train from the velocity-time graph (here d1= Area OED)
+d2=12 // km // from the velocity-time graph (here d2= Area OABB')
+d3=3 // km // from the velocity-time graph (here d3= Area BB'C)
+// Calculations
+t_1=d2/v // hr // time of travel for first 12 kms
+t_2=(2*d3)/v // hr // time of for next 3 kms
+// Total time of travel for passenger train is given by eq'n
+t=t_1+t_2 // hr
+// Now time of travel of the local train (let it be T) is given as,
+T=2*t // hr
+V_max=(2*d1)/T // km/hr
+// Results
+clc
+printf('The maximum speed of the local train is %f km/hr \n',V_max)
diff --git a/3204/CH15/EX15.10/Ex15_10.sce b/3204/CH15/EX15.10/Ex15_10.sce new file mode 100644 index 000000000..9030cd721 --- /dev/null +++ b/3204/CH15/EX15.10/Ex15_10.sce @@ -0,0 +1,21 @@ +// Initilization of variables
+mu_a=0.40 // coefficient of friction under block A
+n=40 // r.p.m // speed of rotation of frame
+W_A=120 // N // weight of block A
+W_B=80 // N // weight of block B
+r_1=1.2 // m // distance between W_A & axis of rotation
+r_2=1.6 // m // distance between W_B & axis of rotation
+g=9.81 // m/s^2
+// Calculations
+// Consider the F.B.D of block A
+N=W_A // N // sum F_y=0
+omega=(2*%pi*n)/60 // rad/sec
+a_n=omega^2*r_1 // m/s^2
+T=((W_A/g)*a_n)-(mu_a*W_A) // N
+// Now consider the F.B.D of block B
+A_n=omega^2*r_2 // m/s^2
+N_1=(W_B/g)*A_n // N // sum F_x=0
+mu=(T-W_B)/N_1 // sum F_x=0
+// Results
+clc
+printf('The coefficient of friction of block B is %f \n',mu)
diff --git a/3204/CH15/EX15.12/Ex15_12.sce b/3204/CH15/EX15.12/Ex15_12.sce new file mode 100644 index 000000000..c20e17a1b --- /dev/null +++ b/3204/CH15/EX15.12/Ex15_12.sce @@ -0,0 +1,9 @@ +// Initilization of variables
+W=10000 // N // Weight of the locomotive
+// Calculations
+// Consider the various derivations given in the textbook
+R_max=W/20 // N // eq'n for max reaction
+// The position of occurence of maximum thrust cannot be defined here. Refer textbook for the answer
+// Results
+clc
+printf('The maximum lateral thrust is %f N \n',R_max)
diff --git a/3204/CH15/EX15.13/Ex15_13.sce b/3204/CH15/EX15.13/Ex15_13.sce new file mode 100644 index 000000000..10bd065d4 --- /dev/null +++ b/3204/CH15/EX15.13/Ex15_13.sce @@ -0,0 +1,8 @@ +// Initilization of variables
+W=10 // N // Weight of the ball
+// Calculations
+// consider the eq'n derived to find the reaction, given as
+R=W*(1+((2*%pi^2)/9)) // N
+// Results
+clc
+printf('The value of the reaction is %f N \n',R)
diff --git a/3204/CH15/EX15.15/Ex15_15.sce b/3204/CH15/EX15.15/Ex15_15.sce new file mode 100644 index 000000000..a5f6df981 --- /dev/null +++ b/3204/CH15/EX15.15/Ex15_15.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+P=50 // N // Weight of ball P
+Q=50 // N // Weight of ball Q
+R=100 // N // Weight of the governing device
+l=0.3 // m // length of each side
+theta=30 // degree
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D
+r=l*sind(theta) // m // Radius of circe
+// On solving eqn's 1,2 &3 we get the value of v as,
+v=sqrt(((Q+R)*g*r)/((sqrt(3))*Q)) // m/s
+// But the eq'n v=omega*r we get the value of N as,
+N=(60*v)/(2*%pi*r) // r.p.m
+// Results
+clc
+printf('The speed of rotation is %f r.p.m \n',N)
+// NOTE: In the text book (A.K. Tayal) this sum is numbered as 'EXAMPLE 15.14' which is incorrect.
diff --git a/3204/CH15/EX15.16/Ex15_16.sce b/3204/CH15/EX15.16/Ex15_16.sce new file mode 100644 index 000000000..56176f4a1 --- /dev/null +++ b/3204/CH15/EX15.16/Ex15_16.sce @@ -0,0 +1,17 @@ +// Initilization of variables
+Q=20 // N // Weight of the governor device
+W=10 // N // Weight of the fly balls
+theta=30 // degree // angle between the vertical shaft and the axis AB
+l=0.2 // m // length of the shaft
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D
+// Radius of the circle is given as,
+r=Q*sind(theta)*(10^-2) // m
+// Solving eq'n 1 & 2 for v. The eq'n for v is given as,
+v=sqrt(((W*l*0.5)+(0.05*Q))/((W*0.2*sqrt(3))/(2*g*r))) // m/s
+// But, v=r*omega=2*pi*N*r/60. From this eq'n we get N as,
+N=(v*60)/(2*%pi*r) // r.p.m.
+// Results
+clc
+printf('The speed of the fly-balls is %f r.p.m \n',N)
diff --git a/3204/CH15/EX15.18/Ex15_18.sce b/3204/CH15/EX15.18/Ex15_18.sce new file mode 100644 index 000000000..ccb69ada8 --- /dev/null +++ b/3204/CH15/EX15.18/Ex15_18.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+r=50 // m // radius of the road
+mu=0.15 // coefficient of friction between the wheels and the road
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// The eq'n fo max speed of the vehicle without skidding is
+v=sqrt(mu*g*r) // m/s
+// The angle theta made with the vertical while negotiating the corner is
+theta=atand(v^2/(g*r)) // degree
+// Results
+clc
+printf('The maximum speed with which the vehicle can travel is %f m/s \n',v)
+printf('The angle made with the vertical is %f degree \n',theta)
diff --git a/3204/CH15/EX15.19/Ex15_19.sce b/3204/CH15/EX15.19/Ex15_19.sce new file mode 100644 index 000000000..df973d45c --- /dev/null +++ b/3204/CH15/EX15.19/Ex15_19.sce @@ -0,0 +1,10 @@ +// Initilization of variables
+v=100*(1000/3600) // m/s // or 100 km/hr
+r=250 // m // radius of the road
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// The angle of banking is given by eq'n,
+theta=atand((v^2)/(g*r)) // degree
+// Results
+clc
+printf('The angle of banking of the track is %f degree \n',theta)
diff --git a/3204/CH15/EX15.2/Ex15_2.sce b/3204/CH15/EX15.2/Ex15_2.sce new file mode 100644 index 000000000..3c179224b --- /dev/null +++ b/3204/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+r=200 // m // radius of the curved road
+v_1=72*(1000/3600) // m/s // initial speed of the car
+v_2=36*(1000/3600) // m/s // speed of the car after 10 seconds
+t=10 // seconds
+// Calculations
+A_n=v_1^2/r // m/s^2 // normal component of acceleration
+A_t=0 // since dv/dt=0 // tangential component of acceeration
+delv=v_1-v_2
+delt=t-0
+a_t=delv/delt // m/s^2 // tangential component of deceleration after the brakes are applied
+a_n=v_1^2/r // m/s^2 // normal component of deceleration after the brakes are applied
+// Results
+clc
+printf('The normal component of acceleration is %f m/s^2 \n',A_n)
+printf('The tangential component of acceleration is %f m/s^2 \n',A_t)
+printf('The normal component of deceleration is %f m/s^2 \n',a_n)
+printf('The tangential component of deceleration is %f m/s^2 \n',a_t)
diff --git a/3204/CH15/EX15.20/Ex15_20.sce b/3204/CH15/EX15.20/Ex15_20.sce new file mode 100644 index 000000000..6e18f9a78 --- /dev/null +++ b/3204/CH15/EX15.20/Ex15_20.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+W=10000 // N // Weight of the car
+r=100 // m // radius of the road
+v=10 // m/s // speed of the car
+h=1 // m // height of the C.G of the car above the ground
+b=1.5 // m // distance between the wheels
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// The reactions at the wheels are given by te eq'ns:
+R_A=(W/2)*(1-((v^2*h)/(g*r*b))) // N // Reaction at A
+R_B=(W/2)*(1+((v^2*h)/(g*r*b))) // N // Reaction at B
+// The eq'n for max speed to avoid overturning on level ground is,
+v_max=sqrt((g*r*(b/2))/(h)) // m/s
+// Results
+clc
+printf('The reaction at Wheel A (R_A) is %f N \n',R_A)
+printf('The reaction at Wheel B (R_B) is %f N \n',R_B)
+printf('The maximum speed at which the vehicle can travel without the fear of overturning is is %f m/s \n',v_max)
diff --git a/3204/CH15/EX15.21/Ex15_21.sce b/3204/CH15/EX15.21/Ex15_21.sce new file mode 100644 index 000000000..d31e57aaa --- /dev/null +++ b/3204/CH15/EX15.21/Ex15_21.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+W=1 // N // Weight of the bob
+theta=8 // degree // angle made by the bob with the vertical
+r=100 // m // radius of the curve
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// from eq'n 1 & 2 we get v as,
+v=(sqrt(g*r*tand(theta)))*(3600/1000) // km/hr
+T=W/cosd(theta) // N // from eq'n 2
+// Results
+clc
+printf('The speed of the cariage is %f km/hr \n',v)
+printf('The tension in the chord is %f N \n',T)
diff --git a/3204/CH15/EX15.3/Ex15_3.sce b/3204/CH15/EX15.3/Ex15_3.sce new file mode 100644 index 000000000..61d6e5141 --- /dev/null +++ b/3204/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,16 @@ +// Iintilization of variables
+r=250 // m // radius of the curved road
+a_t=0.6 // m/s^2 // tangential acceleration
+a=0.75 // m/s^2 // total acceleration attained by the car
+// Calculations
+a_n=sqrt(a^2-a_t^2) // m/s^2
+v=sqrt(a_n*r) // m/s
+// Using v=u+a*t
+u=0
+t=v/a_t // seconds
+// Now using v^2-u^2=2*a*s
+s=v^2/(2*a_t) // m
+// Results
+clc
+printf('The distance traveled by the car is %f m \n',s)
+printf('The time for which the car travels is %f seconds \n',t)
diff --git a/3204/CH15/EX15.5/Ex15_5.sce b/3204/CH15/EX15.5/Ex15_5.sce new file mode 100644 index 000000000..21e76a53c --- /dev/null +++ b/3204/CH15/EX15.5/Ex15_5.sce @@ -0,0 +1,31 @@ +// Initilization of variables
+v=10 // m/s // speed of the car
+r=200 // m // radius of the road
+t=15 // seconds
+// Calculations
+omega=(v/r) // radian/seconds // angular velocity of the car
+// Velocity in x & y direction is given by eq'n
+v_x=omega*r*sind(omega*(180/%pi)*t) // m/s // value of v_x is -ve but we consider it to be +ve for calculations
+v_y=omega*r*cosd(omega*(180/%pi)*t) // m/s
+// Acceleration in x & y direction is given by
+a_x=omega^2*r*cosd(omega*(180/%pi)*t) // m/s^2 // value of a_x is -ve but we consider it to be +ve for calculations
+a_y=omega^2*r*sind(omega*(180/3.14)*t) // m/s^2 // value of a_y is -ve but we consider it to be +ve for calculations
+a=sqrt(a_x^2+a_y^2) // m/s^2 // total acc
+phi=atand(a_y/a_x) // degrees // direction of acceleration
+// Components in tangential and normal directions
+// Velocity
+v_n=0 // m/s
+v_t=v // m/s
+// Acceleration
+a_n=v^2/r // m/s^2 // normal acc
+a_t=0 // tangential acc
+// angular position of the car after 15 sec
+theta=omega*(180/%pi)*t // degrees
+// Results
+clc
+printf('The component of velocity in X direction (v_x) is %f m/s \n',v_x)
+printf('The component of velocity in Y direction (v_y) is %f m/s \n',v_y)
+printf('The component of acceleration in X direction (a_x) is %f m/s^2 \n',a_x)
+printf('The component of acceleration in Y direction (a_y) is %f m/s^2 \n',a_y)
+printf('The total acceleration is %f m/s^2 and its direction is %f degrees \n',a,phi)
+printf('The normal acceleration is %f m/s^2 and tangential acceleration is %f m/s^2 \n',a_n,a_t)
diff --git a/3204/CH15/EX15.6/Ex15_6.sce b/3204/CH15/EX15.6/Ex15_6.sce new file mode 100644 index 000000000..e361b3705 --- /dev/null +++ b/3204/CH15/EX15.6/Ex15_6.sce @@ -0,0 +1,28 @@ +// Initilization of variables
+t=1 // seconds
+pi=3.14
+// Calculations
+// From the equations of r and theta given we find 1st & 2nd derative and substitute t=1sec Here we consider the 1st derative as r_1 & theta_1 and so on...
+r=(1.25*t^2)-(0.9*t^3) // m
+r_1=(1.25*(2*t))-(0.9*(3*t^2)) // m/s
+r_2=2.5-(0.9*3*(2*t)) // m/s^2
+theta=(pi/2)*(4*t-3*t^2) // radian
+theta_1=(pi/2)*(4-(6*t)) // rad/second
+theta_2=(pi/2)*(0-(6*t)) // rad/second^2
+// Velocity of collar P
+v_r=r_1 // m/s
+v_theta=r*theta_1 // m/s
+v=sqrt(v_r^2+v_theta^2) // m/s
+alpha=atand(v_theta/v_r) // degree
+// Acceleration of the collar P
+a_r=r_2-(r*theta_1^2) // m/s^2
+a_theta=(r*theta_2)+(2*r_1*theta_1) // m/s^2
+a=sqrt(a_r^2+a_theta^2) // m/s^2
+beta=atand(a_theta/a_r) // degree
+// Acceleration of collar P relative to the rod. Let it be a_relative
+a_relative=r_2 // m/s^2 // towards O
+// Calculations
+clc
+printf('The velocity of the collar is %f m/s \n',v)
+printf('The accelaration of the collar is %f m/s^2 \n',a)
+printf('The acceleration of the collar relative to the rod is %f m/s^2 \n',a_relative)
diff --git a/3204/CH15/EX15.7/Ex15_7.sce b/3204/CH15/EX15.7/Ex15_7.sce new file mode 100644 index 000000000..2f61528e6 --- /dev/null +++ b/3204/CH15/EX15.7/Ex15_7.sce @@ -0,0 +1,39 @@ +// Consider the eq'ns of motion from the book
+// The notations have been changed for the derivatives of r & theta
+// (1) At t=0 s
+theta_0=0
+theta_1=2*%pi // rad/s
+theta_2=0
+r_0=0
+r_1=10 // cm/s
+r_2=0
+// At t=0.3 s
+t=0.3 // sec
+theta=2*%pi*t // rad
+theta1=2*%pi // rad/s
+theta2=0
+r=10*t // cm
+r1=10 // cm/s
+r2=0
+// (i)
+//Velocity
+v_r=r_1 // cm/s
+v_theta=r_0*theta_1
+v=sqrt(v_r^2+v_theta^2) // cm/s
+// Acceleration
+a_r=r_2-(r_0*theta_1^2) // cm/s^2
+a_theta=(r_0*theta_2)+(2*r_1*theta_1) // cm/s^2
+a=sqrt(a_r^2+a_theta^2) // cm/s^2
+// (ii)
+// Velocity
+V_R=r1 // cm/s
+V_theta=r*theta1 // cm/s
+V=sqrt(V_R^2+V_theta^2) // cm/s
+// Acceleration
+A_r=r2-(r*theta1^2) // cm/s^2
+A_theta=(r*theta2)+(2*r1*theta1) // cm/s^2
+A=sqrt(A_r^2+A_theta^2) // cm/s^2
+// Results
+clc
+printf('The velocity and the acceleration of the partice at t=0 s is %f cm/s & %f cm/s^2 \n',v,a)
+printf('The velocity and the acceleration of the partice at t=0.3 s is %f cm/s & %f cm/s^2 \n',V,A)
diff --git a/3204/CH15/EX15.9/Ex15_9.sce b/3204/CH15/EX15.9/Ex15_9.sce new file mode 100644 index 000000000..5dcc297ca --- /dev/null +++ b/3204/CH15/EX15.9/Ex15_9.sce @@ -0,0 +1,7 @@ +// Calculations
+// Tension in the wire before it is cut
+T_ab=1/((2.747*0.643)+(0.766)) // From eqn's 1 & 2..Here T_ab is multiplied with W (i.e weight of small ball)
+T_AB=cosd(40) // Tension in the wire after the wire is cut. Again T_AB is multiplied with W.
+// Results
+clc
+printf('The tension in the wire before and after it is cut is respectively %f W & %f W \n',T_ab,T_AB)
diff --git a/3204/CH16/EX16.1/Ex16_1.sce b/3204/CH16/EX16.1/Ex16_1.sce new file mode 100644 index 000000000..edc53cac4 --- /dev/null +++ b/3204/CH16/EX16.1/Ex16_1.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+k=1000 // N/m // stiffness of spring
+x_1=0.1 // m // distance upto which the spring is stretched
+x_2=0.2 // m
+x_0=0 // initial position of spring
+// Calculations
+// Work required to stretch the spring by 10 cm from undeformed position is given as,
+U_10=-(k/2)*(x_1^2-x_0^2) // N-m
+// Work required to stretch from 10 cm to 20 cm is,
+U=-(1/2)*k*(x_2^2-x_1^2) // N-m
+// Results
+clc
+printf('The work of the spring force is %f N-m \n',U_10)
+printf('The work required to stretch the spring by 20 cm is %f N-m \n',U)
diff --git a/3204/CH16/EX16.10/Ex16_10.sce b/3204/CH16/EX16.10/Ex16_10.sce new file mode 100644 index 000000000..af64f5e8b --- /dev/null +++ b/3204/CH16/EX16.10/Ex16_10.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+m=5 // kg // mass of the block
+theta=30 // degree // inclination of the plane
+x=0.5 // m // distance travelled by the block
+k=1500 // N/m // stiffness of the spring
+mu=0.2 // coefficient of friction between the block and the surface
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the F.B.D of the block
+// Applying the principle of work and energy between the positions 1 & 2 and on further simplification we get the generic eq'n for delta as, 750*delta^2-16.03*delta-8.015=0. From this eq'n e have values of a.b & c as,
+a=750
+b=-16.03
+c=-8.015
+// Thus the roots of the eq'n are given as,
+delta=(-b+(sqrt(b^2-(4*a*c))))/(2*a) // m
+// Results
+clc
+printf('The maximum compression of the spring is %f m \n',delta)
diff --git a/3204/CH16/EX16.11/Ex16_11.sce b/3204/CH16/EX16.11/Ex16_11.sce new file mode 100644 index 000000000..ab81ec312 --- /dev/null +++ b/3204/CH16/EX16.11/Ex16_11.sce @@ -0,0 +1,10 @@ +// Initilization of variables
+M=10 // kg // Here M=M_1=M_2
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D
+// Applying the principle of conservation of energy and by equating the total energies at position 1 & position 2 we get v as,
+v=sqrt((M*g*4)/(25)) // m/s
+// Results
+clc
+printf('The velocity of mass M_2 is %f m/s \n',v)
diff --git a/3204/CH16/EX16.3/Ex16_3.sce b/3204/CH16/EX16.3/Ex16_3.sce new file mode 100644 index 000000000..0ab985d52 --- /dev/null +++ b/3204/CH16/EX16.3/Ex16_3.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+M_A=100 // kg // mass of block A
+M_B=150 // kg // mass of block B
+mu=0.2 // coefficient of friction between the blocks and the surface
+x=1 // m // distance by which block A moves
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D
+// Applying the principle of work and energy to the system of blocks A&B and on simplifying we get the value of v as,
+v=sqrt(((-mu*M_A*g)+(M_B*g))/(125)) // m/s
+// Results
+clc
+printf('The velocity of block A is %f m/s \n',v)
diff --git a/3204/CH16/EX16.4/Ex16_4.sce b/3204/CH16/EX16.4/Ex16_4.sce new file mode 100644 index 000000000..54c183024 --- /dev/null +++ b/3204/CH16/EX16.4/Ex16_4.sce @@ -0,0 +1,19 @@ +// Initilization of variables
+M=500*10^3 // kg // mass of the train
+u=0 // m/s // initial speed
+v=90*(1000/3600) // m/s // final speed
+t=50 // seconds
+F_r=15*10^3 // N // Frictioal resistance to motion
+// Calculations
+// Acceleration is given as,
+a=v/t // m/s^2
+// The total force required to accelerate the train is,
+F=M*a // N
+// The maximum power required is at, t=50s & v=25 m/s
+P=(F+F_r)*v*(10^-6) // MW
+// At any time after 50 seconds, the force required only to overcome the frictional resistance of 15*10^3 N is,
+P_req=F_r*v*(10^-3) // kW
+// Results
+clc
+printf('(a) The maximum power required is %f MW \n',P)
+printf('(b) The power required to maintain a speed of 90 km/hr is %f kW \n',P_req)
diff --git a/3204/CH16/EX16.5/Ex16_5.sce b/3204/CH16/EX16.5/Ex16_5.sce new file mode 100644 index 000000000..d5291a9aa --- /dev/null +++ b/3204/CH16/EX16.5/Ex16_5.sce @@ -0,0 +1,26 @@ +// Initilization of variables
+W=50 // N // Weight suspended on spring
+k=10 // N/cm // stiffness of the spring
+x_2=15 // cm // measured extensions
+h=10 // cm // height for position 2
+// Calculations
+// Consider the required F.B.D.
+// POSITION 1: The force exerted by the spring is,
+F_1=W // N
+// Extension of spring from undeformed position is x_1,
+x_1=F_1/k // cm
+// POSITION 2: When pulled by 10 cm to the floor. P.E of weight is,
+P.E_g=-W*h // N-cm // (P.E_g= P.E_gravity)
+// P.E of the spring with respect to position 1
+P.E_s=(1/2)*k*(x_2^2-x_1^2) // N-cm // (P.E_s= P.E_ spring)
+// Total P.E of the system with respect to position 1
+P.E_t=P.E_g+P.E_s // N-cm // (P.E_t= P.E_total)
+// Total energy of the system,
+E_2=P.E_t // N-cm
+// Total energy of the system in position 3 w.r.t position 1 is:
+x=-sqrt(100) // cm
+x=+sqrt(100) // cm
+// Results
+clc
+printf('The potential energy of the system is %f N-cm \n',E_2)
+printf('The maximum height above the floor that the weight W will attain after release is %f cm \n',x)
diff --git a/3204/CH16/EX16.6/Ex16_6.sce b/3204/CH16/EX16.6/Ex16_6.sce new file mode 100644 index 000000000..8a2c5a146 --- /dev/null +++ b/3204/CH16/EX16.6/Ex16_6.sce @@ -0,0 +1,15 @@ +// Initilization of variables
+m=5 // kg // mass of the ball
+k=500 // N/m // stiffness of the spring
+h=10 // cm // height of drop
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D.
+// In eq'n 1 substitute the respective values and simplify it further. In this eq'n of 2nd degree a=1 b=-0.1962 & c=-0.01962. Thus the roots of the eq'n is given as,
+a=1
+b=-0.1962
+c=-0.01962
+delta=((-b+(sqrt((b^2)-(4*a*c))))/(2*a))*(10^2) // cm // We consider the +ve value of delta
+// Results
+clc
+printf('The maximum deflection of the spring is %f cm \n',delta)
diff --git a/3204/CH16/EX16.7/Ex16_7.sce b/3204/CH16/EX16.7/Ex16_7.sce new file mode 100644 index 000000000..eb4b22357 --- /dev/null +++ b/3204/CH16/EX16.7/Ex16_7.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+m=5 // kg // mass of the ball
+k=500 // N/m // stiffness of the spring
+h=0.1 // m // height of vertical fall
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D
+// On equating the total energies at position 1 & 2 we get eq'n of delta as,
+delta=sqrt((2*m*g*h)/(k)) // m
+// Results
+clc
+printf('The maximum compression of the spring is %f m \n',delta)
diff --git a/3204/CH16/EX16.9/Ex16_9.sce b/3204/CH16/EX16.9/Ex16_9.sce new file mode 100644 index 000000000..5b5d660da --- /dev/null +++ b/3204/CH16/EX16.9/Ex16_9.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+m=5 // kg // mass of the collar
+k=500 // N/m // stiffness of the spring
+AB=0.15 // m // Refer the F.B.D for AB
+AC=0.2 // m // Refer the F.B.D for AC
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Consider the respective F.B.D
+// POSITION 1:
+P.E_1=m*g*(AB)+0
+K.E_1=0
+E_1=P.E_1+K.E_1 //
+// POSITION 2 : Length of the spring in position 2
+CB=sqrt(AB^2+AC^2) // m
+// x is the extension in the spring
+x=CB-AC // m
+// On substuting and Equating equations of total energies for position1 & position 2 we get the value of v as,
+v=sqrt(((E_1-((1/2)*k*x^2))*2)/m) // m/s
+// Results
+clc
+printf('The velocity of the collar will be %f m/s \n',v)
+// The answer given in the text book (v=16.4 m/s) is wrong.
diff --git a/3204/CH17/EX17.1/Ex17_1.sce b/3204/CH17/EX17.1/Ex17_1.sce new file mode 100644 index 000000000..23979a96f --- /dev/null +++ b/3204/CH17/EX17.1/Ex17_1.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+m=0.1 // kg // mass of ball
+// Calculations
+// Consider the respective F.B.D.
+// For component eq'n in x-direction
+delta_t=0.015 // seconds // time for which the ball &the bat are in contact
+v_x_1=-25 // m/s
+v_x_2=40*cosd(40) // m/s
+F_x_average=((m*(v_x_2))-(m*(v_x_1)))/(delta_t) // N
+// For component eq'n in y-direction
+delta_t=0.015 // sceonds
+v_y_1=0 // m/s
+v_y_2=40*sind(40) // m/s
+F_y_average=((m*v_y_2)-(m*(v_y_1)))/(delta_t) // N
+F_average=sqrt(F_x_average^2+F_y_average^2) // N
+// Results
+clc
+printf('The average impules force exerted by the bat on the ball is %f N \n',F_average)
diff --git a/3204/CH17/EX17.2/Ex17_2.sce b/3204/CH17/EX17.2/Ex17_2.sce new file mode 100644 index 000000000..ba58ea53f --- /dev/null +++ b/3204/CH17/EX17.2/Ex17_2.sce @@ -0,0 +1,20 @@ +// Initiliation of variables
+m_g=3000 // kg // mass of the gun
+m_s=50 // kg // mass of the shell
+v_s=300 // m/s // initial velocity of shell
+s=0.6 // m // distance at which the gun is brought to rest
+v=0 // m/s // initial velocity of gun
+// Calculations
+// On equating eq'n 1 & eq'n 2 we get v_g as,
+v_g=(m_s*v_s)/(-m_g) // m/s
+// Using v^2-u^2=2*a*s to find acceleration,
+a=(v^2-v_g^2)/(2*s) // m/s^2
+// Force required to stop the gun,
+F=m_g*(-a) // N // here we make a +ve to find the Force
+// Time required to stop the gun, using v=u+a*t:
+t=(-v_g)/(-a) // seconds // we take -a to consider +ve value of acceleration
+// Results
+clc
+printf('The recoil velocity of gun is %f m/s \n',v_g)
+printf('The Force required to stop the gun is %f N \n',F)
+printf('The time required to stop the gun is %f seconds \n',t)
diff --git a/3204/CH17/EX17.3/Ex17_3.sce b/3204/CH17/EX17.3/Ex17_3.sce new file mode 100644 index 000000000..1d670fe48 --- /dev/null +++ b/3204/CH17/EX17.3/Ex17_3.sce @@ -0,0 +1,17 @@ +// Initilization of variables
+m_m=50 // kg // mass of man
+m_b=250 // kg // mass of boat
+s=5 // m // length of the boat
+v_r=1 // m/s // here v_r=v_(m/b)= relative velocity of man with respect to boat
+// Calculations
+// Velocity of man is given by, v_m=(-v_r)+v_b
+// Final momentum of the man and the boat=m_m*v_m+m_b*v_b. From this eq'n v_b is given as
+v_b=(m_m*v_r)/(m_m+m_b) // m/s // this is the absolute velocity of the boat
+// Time taken by man to move to the other end of the boat is,
+t=s/v_r // seconds
+// The distance travelled by the boat in the same time is,
+s_b=v_b*t // m to right from O
+// Results
+clc
+printf('(a) The velocity of boat as observed from the ground is %f m/s \n',v_b)
+printf('(b) The distance by which the boat gets shifted is %f m \n',s_b)
diff --git a/3204/CH17/EX17.5/Ex17_5.sce b/3204/CH17/EX17.5/Ex17_5.sce new file mode 100644 index 000000000..7aa4c8f73 --- /dev/null +++ b/3204/CH17/EX17.5/Ex17_5.sce @@ -0,0 +1,28 @@ +// Initilization of variables
+M=250 // kg // mass of the boat
+M_1=50 // kg // mass of the man
+M_2=75 // kg // mass of the man
+v=4 // m/s // relative velocity of man w.r.t boat
+// Calculations
+// (a)
+// Let the increase in the velocity or the final velocity of the boat when TWO MEN DIVE SIMULTANEOUSLY is given by eq'n,
+deltaV_1=((M_1+M_2)*v)/(M+(M_1+M_2)) // m/s
+// (b) // The increase in the velocity or the final velocity of the boat when man of 75 kg dives 1st followed by man of 50 kg
+// Man of 75 kg dives first, So let the final velocity is given as
+deltaV_75=(M_2*v)/((M+M_1)+M_2) // m/s
+// Now let the man of 50 kg jumps next, Here
+deltaV_50=(M_1*v)/(M+M_1) // m/s
+// Let final velocity of boat is,
+deltaV_2=0+deltaV_75+deltaV_50 // m/s
+// (c)
+// The man of 50 kg jumps first,
+delV_50=(M_1*v)/((M+M_2)+(M_1)) // m/s
+// the man of 75 kg jumps next,
+delV_75=(M_2*v)/(M+M_2) // m/s
+// Final velocity of boat is,
+deltaV_3=0+delV_50+delV_75 // m/s
+// Results
+clc
+printf('(a) The Final velocity of boat when two men dive simultaneously is %f m/s \n',deltaV_1)
+printf('(b) The Final velocity of boat when the man of 75 kg dives first and 50 kg dives second is %f m/s \n',deltaV_2)
+printf('(3) The Final velocity of boat when the man of 50kg dives first followed by the man of 75 kg is %f m/s \n',deltaV_3)
diff --git a/3204/CH17/EX17.6/Ex17_6.sce b/3204/CH17/EX17.6/Ex17_6.sce new file mode 100644 index 000000000..7ebfe782d --- /dev/null +++ b/3204/CH17/EX17.6/Ex17_6.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+m_m=70 // kg // mass of man
+m_c=35 // kg // mass of canoe
+m=25/1000 // kg // mass of bullet
+m_wb=2.25 // kg // mass of wodden block
+V_b=5 // m/s // velocity of block
+// Calculations
+// Considering Initial Momentum of bullet=Final momentum of bullet & the block we have,Velocity of bullet (v) is given by eq'n,
+v=(V_b*(m_wb+m))/(m) // m/s
+// Considering, Momentum of the bullet=Momentum of the canoe & the man,the velocity on canoe is given by eq'n
+V=(m*v)/(m_m+m_c) // m/s
+// Results
+clc
+printf('The velocity of the canoe is %f m/s \n',V)
diff --git a/3204/CH17/EX17.8/Ex17_8.sce b/3204/CH17/EX17.8/Ex17_8.sce new file mode 100644 index 000000000..cac6cc4fd --- /dev/null +++ b/3204/CH17/EX17.8/Ex17_8.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+m=2 // kg // mass of the particle
+v_0=20 // m/s // speed of rotation of the mass attached to the string
+r_0=1 // m // radius of the circle along which the particle is rotated
+r_1=r_0/2 // m
+// Calculations
+// here, equating (H_0)_1=(H_0)_2 i.e (m*v_0)*r_0=(m*v_1)*r_1 (here, r_1=r_0/2). On solving we get v_1 as,
+v_1=2*v_0 // m/s
+// Tension is given by eq'n,
+T=(m*v_1^2)/r_1 // N
+// Results
+clc
+printf('The new speed of the particle is %f m/s \n',v_1)
+printf('The tension in the string is %f N \n',T)
diff --git a/3204/CH18/EX18.1/Ex18_1.sce b/3204/CH18/EX18.1/Ex18_1.sce new file mode 100644 index 000000000..3f8d861f8 --- /dev/null +++ b/3204/CH18/EX18.1/Ex18_1.sce @@ -0,0 +1,15 @@ +// Initilization of variables
+m_a=1 // kg // mass of the ball A
+v_a=2 // m/s // velocity of ball A
+m_b=2 // kg // mass of ball B
+v_b=0 // m/s // ball B at rest
+e=1/2 // coefficient of restitution
+// Calculations
+// Solving eqn's 1 & 2 using matrix for v'_a & v'_b,
+A=[1 2;-1 1]
+B=[2;1]
+C=inv(A)*B
+// Results
+clc
+printf('The velocity of ball A after impact is %f m/s \n',C(1))
+printf('The velocity of ball B after impact is %f m/s \n',C(2))
diff --git a/3204/CH18/EX18.10/Ex18_10.sce b/3204/CH18/EX18.10/Ex18_10.sce new file mode 100644 index 000000000..3145a54b5 --- /dev/null +++ b/3204/CH18/EX18.10/Ex18_10.sce @@ -0,0 +1,17 @@ +// Initilization of variables
+v_a=600 // m/s // velocity of the bullet before impact
+v_b=0 // m/s // velocity of the block before impact
+w_b=0.25 // N // weight of the bullet
+w_wb=50 // N // weight of wodden block
+mu=0.5 // coefficient of friction between the floor and the block
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+m_a=w_b/g // kg // mass of the bullet
+m_b=w_wb/g // kg // mass of the block
+// Let the common velocity be v_c which is given by eq'n (Principle of conservation of momentum)
+v_c=(w_b*v_a)/(w_wb+w_b) // m/s
+// Let the distance through which the block is displaced be s, Then s is given by eq'n
+s=v_c^2/(2*g*mu) // m
+// Results
+clc
+printf('The distance through which the block is displaced from its initial position is %f m \n',s)
diff --git a/3204/CH18/EX18.11/Ex18_11.sce b/3204/CH18/EX18.11/Ex18_11.sce new file mode 100644 index 000000000..b67388180 --- /dev/null +++ b/3204/CH18/EX18.11/Ex18_11.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+M=750 // kg // mass of hammer
+m=200 // kg // mass of the pile
+h=1.2 // m // height of fall of the hammer
+delta=0.1 // m // distance upto which the pile is driven into the ground
+g=9.81 // m/s^2 // acc due to gravity
+// Caculations
+// The resistance to penetration to the pile is given by eq'n,
+R=(((M+m)*g)+((M^2*g*h)/((M+m)*delta)))*(10^-3) // kN
+// Results
+clc
+printf('The resistance to penetration to the pile is %f kN \n',R)
diff --git a/3204/CH18/EX18.2/Ex18_2.sce b/3204/CH18/EX18.2/Ex18_2.sce new file mode 100644 index 000000000..0100f3a68 --- /dev/null +++ b/3204/CH18/EX18.2/Ex18_2.sce @@ -0,0 +1,26 @@ +// Initilization of variables
+m_a=2 // kg // mass of ball A
+m_b=6 // kg // mass of ball B
+m_c=12 // kg // mass of ball C
+v_a=12 // m/s // velocity of ball A
+v_b=4 // m/s // velocity of ball B
+v_c=2 // m/s // velocity of ball C
+e=1 // coefficient of restitution for perfectly elastic body
+// Calculations
+// (A)
+// Solving eq'n 1 & 2 using matrix for v'_a & v'_b,
+A=[2 6;-1 1]
+B=[48;8]
+C=inv(A)*B
+// Calculations
+// (B)
+// Solving eq'ns 3 & 4 simultaneously using matrix for v'_b & v'_c
+P=[1 2;-1 1]
+Q=[12;6]
+R=inv(P)*Q
+// Results (A&B)
+clc
+printf('The velocity of ball A after impact on ball B is %f m/s \n',C(1)) // here the ball of mass 2 kg is bought to rest
+printf('The velocity of ball B after getting impacted by ball A is %f m/s \n',C(2))
+printf('The final velocity of ball B is %f m/s \n',R(1)) // here the ball of mass 6 kg is bought to rest
+printf('The velocity of ball C after getting impacted by ball B is %f m/s \n',R(2))
diff --git a/3204/CH18/EX18.3/Ex18_3.sce b/3204/CH18/EX18.3/Ex18_3.sce new file mode 100644 index 000000000..34b3f2212 --- /dev/null +++ b/3204/CH18/EX18.3/Ex18_3.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+h_1=9 // m // height of first bounce
+h_2=6 // m // height of second bounce
+// Calculations
+// From eq'n (5) we have, Coefficient of restitution between the glass and the floor is,
+e=sqrt(h_2/h_1)
+// From eq'n 3 we get height of drop as,
+h=h_1/e^2 // m
+// Results
+clc
+printf('The ball was dropped from a height of %f m \n',h)
+printf('The coefficient of restitution between the glass and the floor is %f \n',e)
+// Here we use h`=h_1 & h``=h_2 because h` & h`` could not be defined in Scilab.
diff --git a/3204/CH18/EX18.4/Ex18_4.sce b/3204/CH18/EX18.4/Ex18_4.sce new file mode 100644 index 000000000..18ae94ad8 --- /dev/null +++ b/3204/CH18/EX18.4/Ex18_4.sce @@ -0,0 +1,33 @@ +// Initilization of variables
+e=0.90 // coefficient o restitution
+v_a=10 // m/s // velocity of ball A
+v_b=15 // m/s // velocity of ball B
+alpha_1=30 // degree // angle made by v_a with horizontal
+alpha_2=60 // degree // angle made by v_b with horizontal
+// Calculations
+// The components of initial velocity of ball A:
+v_a_x=v_a*cosd(alpha_1) // m/s
+v_a_y=v_a*sind(alpha_1) // m/s
+// The components of initial velocity of ball B:
+v_b_x=-v_b*cosd(alpha_2) // m/s
+v_b_y=v_b*sind(alpha_2) // m/s
+// From eq'n 1 & 2 we get,
+v_ay=v_a_y // m/s // Here, v_ay=(v'_a)_y
+v_by=v_b_y // m/s // Here, v_by=(v'_b)_y
+// On adding eq'n 3 & 4 we get,
+v_bx=((v_a_x+v_b_x)+(-e*(v_b_x-v_a_x)))/2 // m/s // Here. v_bx=(v'_b)_x
+// On substuting the value of v'_b_x in eq'n 3 we get,
+v_ax=(v_a_x+v_b_x)-(v_bx) // m/s // here, v_ax=(v'_a)_x
+// Now the eq'n for resultant velocities of balls A & B after impact are,
+v_A=sqrt(v_ax^2+v_ay^2) // m/s
+v_B=sqrt(v_bx^2+v_by^2) // m/s
+// The direction of the ball after Impact is,
+theta_1=atand(-(v_ay/v_ax)) // degree
+theta_2=atand(v_by/v_bx) // degree
+// Results
+clc
+printf('The velocity of ball A after impact is %f m/s \n',v_A)
+printf('The velocity of ball B after impact is %f m/s \n',v_B)
+printf('The direction of ball A after impact is %f degree \n',theta_1)
+printf('The direction of ball B after impact is %f degree \n',theta_2)
+// Her we use, (1) v'_a & v'_b as v_A & v_B.
diff --git a/3204/CH18/EX18.5/Ex18_5.sce b/3204/CH18/EX18.5/Ex18_5.sce new file mode 100644 index 000000000..a71f5d650 --- /dev/null +++ b/3204/CH18/EX18.5/Ex18_5.sce @@ -0,0 +1,19 @@ +// Initiization of variables
+theta=30 // degrees // ange made by the ball against the wall
+e=0.50
+// Calculations
+// The notations have been changed
+// Resolving the velocity v as,
+v_x=cosd(theta)
+v_y=sind(theta)
+V_y=v_y
+// from coefficient of restitution reation
+V_x=-e*v_x
+// Resultant velocity
+V=sqrt(V_x^2+V_y^2)
+theta=atand(V_y/(-V_x)) // taking +ve value for V_x
+// NOTE: Here all the terms are multiplied with velocity i.e (v).
+// Results
+clc
+printf('The velocity of the ball is %f v \n',V)
+printf('The direction of the ball is %f degrees \n',theta)
diff --git a/3204/CH18/EX18.6/Ex18_6.sce b/3204/CH18/EX18.6/Ex18_6.sce new file mode 100644 index 000000000..ada754797 --- /dev/null +++ b/3204/CH18/EX18.6/Ex18_6.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+e=0.8 // coefficient of restitution
+g=9.81 // m/s^2 // acc due to gravity
+// Calcuations
+// Squaring eqn's 1 &2 and Solving eqn's 1 & 2 using matrix for the value of h
+A=[-1 (2*g);-1 -(1.28*g)]
+B=[0.945^2;(-0.4*9.81)]
+C=inv(A)*B // m
+// Results
+clc
+printf('The height from which the ball A should be released is %f m \n',C(2))
+// The answer given in the book i.e 0.104 is wrong.
diff --git a/3204/CH18/EX18.7/Ex18_7.sce b/3204/CH18/EX18.7/Ex18_7.sce new file mode 100644 index 000000000..974032950 --- /dev/null +++ b/3204/CH18/EX18.7/Ex18_7.sce @@ -0,0 +1,9 @@ +// Initilization of variables
+theta_a=60 // degree // angle made by sphere A with the verticle
+e=1 // coefficient of restitution for elastic impact
+// Calculations
+// theta_b is given by the eq'n cosd*theta_b=0.875, hence theta_b is,
+theta_b=acosd(0.875) // degree
+// Results
+clc
+printf('The angle through which the sphere B will swing after the impact is %f degree \n',theta_b)
diff --git a/3204/CH18/EX18.8/Ex18_8.sce b/3204/CH18/EX18.8/Ex18_8.sce new file mode 100644 index 000000000..9edfd8a39 --- /dev/null +++ b/3204/CH18/EX18.8/Ex18_8.sce @@ -0,0 +1,39 @@ +// Initilization of variables
+m_a=0.01 // kg // mass of bullet A
+v_a=100 // m/s // velocity of bullet A
+m_b=1 // kg // mass of the bob
+v_b=0 // m/s // velocity of the bob
+l=1 // m // length of the pendulum
+v_r=-20 // m/s // velocity at which the bullet rebounds the surface of the bob // here the notation for v'_a is shown by v_r
+v_e=20 // m/s // velocity at which the bullet escapes through the surface of the bob // here the notation for v_a is shown by v_e
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Momentum of the bullet & the bob before impact is,
+M=(m_a*v_a)+(m_b*v_b) // kg.m/s......(eq'n 1)
+// The common velocity v_c ( we use v_c insted of v' for notation of common velocity) is given by equating eq'n 1 & eq'n 2 as,
+// (a) When the bullet gets embedded into the bob
+v_c=M/(m_a+m_b) // m/s
+// The height h to which the bob rises is given by eq'n 3 as,
+h_1=(1/2)*(v_c^2/g) // m
+// The angle (theta_1) by which the bob swings corresponding to the value of height h_1 is,
+theta_1=acosd((l-h_1)/l) // degree
+// (b) When the bullet rebounds from the surface of the bob
+// The velocity of the bob after the rebound of the bullet from its surface is given by equating eq'n 1 & eq'n 4 as,
+v_bob_rebound=M-(m_a*v_r) // m/s // here v_bob_rebound=v'_b
+// The equation for the height which the bob attains after impact is,
+h_2=(v_bob_rebound^2)/(2*g) // m
+// The corresponding angle of swing
+theta_2=acosd((l-h_2)/l) // degree
+// (c) When the bullet pierces and escapes through the bob
+// From eq'n 1 & 5 the velocity attained by the bob after impact is given as,
+v_b_escape=M-(m_a*v_e) // m/s // here we use, v_b_escape insted of v'_b
+// The equation for the height which the bob attains after impact is,
+h_3=(v_b_escape^2)/(2*g) // m
+// The corresponding angle of swing
+theta_3=acosd((l-h_3)/(l)) // degree
+// Results
+clc
+printf('(a) The maximum angle through which the pendulum swings when the bullet gets embeded into the bob is %f degree \n',theta_1)
+printf('(b) The maximum angle through which the pendulum swings when the bullet rebounds from the surface of the bob is %f degree \n',theta_2)
+printf('(c) The maximum angle through which the pendulum swings when the bullet escapes from other end of the bob the bob is %f degree \n',theta_3)
+// IN THIS SUM WE HAVE USED DIFFERENT NOTATIONS CONSIDERING DIFFERENT CASES BECAUSE IN THE TEXT BOOK WE HAD 3 VARIABLES WITH SAME NOTATION BUT WITH A DIFFERENT VALUE WHICH COULD NOT BE EXECUTED INTO SCILAB.
diff --git a/3204/CH18/EX18.9/Ex18_9.sce b/3204/CH18/EX18.9/Ex18_9.sce new file mode 100644 index 000000000..bc0f132fc --- /dev/null +++ b/3204/CH18/EX18.9/Ex18_9.sce @@ -0,0 +1,23 @@ +// Initilization of variables
+W_a=50 // N // falling weight
+W_b=50 // N // weight on which W_a falls
+g=9.81 // m/s^2 // acc due to gravity
+m_a=W_a/g // kg // mass of W_a
+m_b=W_b/g // kg // mass of W_b
+k=2*10^3 // N/m // stiffness of spring
+h=0.075 // m // height through which W_a falls
+// The velocity of weight W_a just before the impact and after falling from a height of h is given from the eq'n, ( Principle of conservation of energy)
+v_a=sqrt(2*g*h) // m/s
+// Let the mutual velocity after the impact be v_m (i.e v_m=v'), (by principle of conservation of momentum)
+v_m=(m_a*v_a)/(m_a+m_b) // m/s
+// Initial compression of the spring due to weight W_b is given by,
+delta_st=(W_b/k)*(10^2) // cm
+// Let the total compression of the spring be delta_t, Then delta_t is found by finding the roots from the eq'n........ delta_t^2-0.1*delta_t-0.000003=0. In this eq'n let,
+a=1
+b=-0.1
+c=-0.000003
+delta_t=((-b+(sqrt(b^2-(4*a*c))))/2*a)*(10^2) // cm // we consider the -ve value
+delta=delta_t-delta_st // cm
+// Results
+clc
+printf('The compression of the spring over and above caused by the static action of weight W_a is %f cm \n',delta)
diff --git a/3204/CH19/EX19.1/Ex19_1.sce b/3204/CH19/EX19.1/Ex19_1.sce new file mode 100644 index 000000000..d7006c7a4 --- /dev/null +++ b/3204/CH19/EX19.1/Ex19_1.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+v_t=10 // m/s // velocity of the train
+v_s=5 // m/s // velocity of the stone
+// Calculations
+// Let v_r be the relative velocity, which is given as, (from triangle law)
+v_r=sqrt(v_t^2+v_s^2) // m/s
+// The direction ofthe stone is,
+theta=atand(v_s/v_t) // degree
+// Results
+clc
+printf('The velocity at which the stone appears to hit the person travelling in the train is %f m/s \n',v_r)
+printf('The direction of the stone is %f degree \n',theta)
diff --git a/3204/CH19/EX19.2/Ex19_2.sce b/3204/CH19/EX19.2/Ex19_2.sce new file mode 100644 index 000000000..6e3b02830 --- /dev/null +++ b/3204/CH19/EX19.2/Ex19_2.sce @@ -0,0 +1,17 @@ +// Initilization of variables
+v_A=5 // m/s // speed of ship A
+v_B=2.5 // m/s // speed of ship B
+theta=135 // degree // angle between the two ships
+// Calculations
+// Here,
+OA=v_A // m/s
+OB=v_B // m/s
+// The magnitude of relative velocity is given by cosine law as,
+AB=sqrt((OA^2)+(OB^2)-(2*OA*OB*cosd(theta))) // m/s
+// where AB gives the relative velocity of ship B with respect to ship A
+// Applying sine law to find the direction, Let alpha be the direction of the reative velocity, then
+alpha=asind((OB*sind(theta))/(AB)) // degree
+// Results
+clc
+printf('The magnitude of relative velocity of ship B with respect to ship A is %f m/s \n',AB)
+printf('The direction of the relative velocity is %f degree \n',alpha)
diff --git a/3204/CH19/EX19.3/Ex19_3.sce b/3204/CH19/EX19.3/Ex19_3.sce new file mode 100644 index 000000000..41758417b --- /dev/null +++ b/3204/CH19/EX19.3/Ex19_3.sce @@ -0,0 +1,21 @@ +// Initilization of variables
+v_c=20 // km/hr // speed at which the cyclist is riding to west
+theta_1=45 // degree // angle made by rain with the cyclist when he rides at 20 km/hr
+V_c=12 // km/hr // changed speed
+theta_2=30 // degree // changed angle when the cyclist rides at 12 km/hr
+// Calculations
+// Solving eq'ns 1 & 2 simultaneously to get the values of components(v_R_x & v_R_y) of absolute velocity v_R. We use matrix to solve eqn's 1 & 2.
+A=[1 1;1 0.577]
+B=[20;12]
+C=inv(A)*B // km/hr
+// The X component of relative velocity (v_R_x) is C(1)
+// The Y component of relative velocity (v_R_y) is C(2)
+// Calculations
+// Relative velocity (v_R) is given as,
+v_R=sqrt((C(1))^2+(C(2))^2) // km/hr
+// And the direction of absolute velocity of rain is theta, is given as
+theta=atand(C(2)/C(1)) // degree
+// Results
+clc
+printf('The magnitude of absolute velocity is %f km/hr \n',v_R)
+printf('The direction of absolute velocity is %f degree \n',theta)
diff --git a/3204/CH19/EX19.4/Ex19_4.sce b/3204/CH19/EX19.4/Ex19_4.sce new file mode 100644 index 000000000..944597970 --- /dev/null +++ b/3204/CH19/EX19.4/Ex19_4.sce @@ -0,0 +1,38 @@ +// Initiization of variables
+a=1 // m/s^2 // acceleration of car A
+u_B=36*(1000/3600) // m/s // velocity of car B
+u=0 // m/s // initial velocity of car A
+d=32.5 // m // position of car A from north of crossing
+t=5 // seconds
+// Calculations
+// CAR A: Absolute motion using eq'n v=u+at we have,
+v=u+(a*t) // m/s
+// Now distance travelled by car A after 5 seconds is given by, s_A=u*t+(1/2)*a*t^2
+s_A=(u*t)+((1/2)*a*t^2)
+// Now, let the position of car A after 5 seconds be y_A
+y_A=d-s_A // m //
+// CAR B:
+// let a_B be the acceleration of car B
+a_B=0 // m/s
+// Now position of car B is s_B
+s_B=(u_B*t)+((1/2)*a_B*t^2) // m
+x_B=s_B // m
+// Let the Relative position of car A with respect to car B be BA & its direction be theta, then from fig. 19.9(b)
+OA=y_A
+OB=x_B
+BA=sqrt(OA^2+OB^2) // m
+theta=atand(OA/OB) // degree
+// Let the relative velocity of car A w.r.t. the car B be v_AB & the angle be phi. Then from fig 19.9(c). Consider small alphabets
+oa=v
+ob=u_B
+v_AB=sqrt(oa^2+ob^2) // m/s
+phi=atand(oa/ob) // degree
+// Let the relative acceleration of car A w.r.t. car B be a_A/B.Then,
+a_AB=a-a_B // m/s^2
+// Results
+clc
+printf('The relative position of car A relative to car B is %f m \n',BA)
+printf('The direction of car A w.r.t car B is %f degree \n',theta)
+printf('The velocity of car A relative to car B is %f m/s \n',v_AB)
+printf('The direction of car A w.r.t (for relative velocity)is %f degree \n',phi)
+printf('The acceleration of car A relative to car B is %f m/s^2 \n',a_AB)
diff --git a/3204/CH2/EX2.1/Ex_2_1.sce b/3204/CH2/EX2.1/Ex_2_1.sce new file mode 100644 index 000000000..49e1e44ef --- /dev/null +++ b/3204/CH2/EX2.1/Ex_2_1.sce @@ -0,0 +1,11 @@ +//Initilization of variables
+P=50 //N
+Q=100 //N
+beta=150 //degree // angle between P & the horizontal
+//Calculations
+R=sqrt(P^2+Q^2-(2*P*Q*cosd(beta))) // using the Trignometric solution
+Alpha=asind((sind(beta)*Q)/R)+15
+//Result
+clc
+printf('The magnitude of resultant is %f Newton (N) \n',R)
+printf('The direction of resultant is %f degree \n',Alpha)
diff --git a/3204/CH2/EX2.10/Ex2_10.sce b/3204/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..99fe93709 --- /dev/null +++ b/3204/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,13 @@ +//Initilization of variables
+F3=500 //N
+alpha=60 //degree //angle made by F3 with F2
+beta=40 //degree //angle made by F1 with F3
+theta=80 //degree //angle made by F1 with F2
+//Calculations
+// Solving by using law of sines
+F1=(F3*sind(alpha)/sind(theta)) //N //by law of sines
+F2=(F3*sind(beta)/sind(theta)) //N //by law of sines
+//Resuts
+clc
+printf('The force F1 is %f N \n',F1)
+printf('The force F2 is %f N \n',F2)
diff --git a/3204/CH2/EX2.11/Ex2_11.sce b/3204/CH2/EX2.11/Ex2_11.sce new file mode 100644 index 000000000..dd52a1bf0 --- /dev/null +++ b/3204/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,18 @@ +//Initilization of variables
+P=5000 //N
+lAB=5 //m
+lOB=1.443 // m
+alpha=30 //degree //angle made by force P with the beam
+//Calculations
+theta=atand(lOB/lAB) // degree // eq'n 1
+Xa=(P*cosd(alpha)) //N //using eq'n 4
+Ya=Xa*tand(theta) //N // from eq'n 3 & 4
+Rb=P*sind(alpha)-Ya // N from eq'n 5// substuting value of Ya in eq'n 5
+Ra=sqrt((Xa^2)+(Ya^2)) //N
+//Results
+clc
+printf('The X component of reaction at A is %f N \n',Xa)
+printf('The Y component of reaction at A is %f N \n',Ya)
+printf('The reaction at support A is %f N \n',Ra)
+printf('The reaction at support B is %f N \n',Rb)
+// The decimal point error might cause a small discrepancy in the answers
diff --git a/3204/CH2/EX2.12/Ex2_12.sce b/3204/CH2/EX2.12/Ex2_12.sce new file mode 100644 index 000000000..bbad225e3 --- /dev/null +++ b/3204/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,14 @@ +//Initilization of variables
+W=1000 //N
+OD=0.4 //m
+AD=0.3 //m
+AO=0.5 //m //AO=sqrt((0.4)^2+(0.3)^2)
+//Calculations
+Ra=W*AO/OD //N // The answer of Ra in the textbook is incorrect
+Rc=W*AD/OD //N
+alpha=atand(OD/AD) //degree
+//Results
+clc
+printf('The reaction at support A is %f N \n',Ra)
+printf('The reaction at support B is %f N \n',Rc)
+printf('The angle that Rc makes with horizontal %f degree \n',alpha)
diff --git a/3204/CH2/EX2.13/Ex2_13.sce b/3204/CH2/EX2.13/Ex2_13.sce new file mode 100644 index 000000000..24bf02c3e --- /dev/null +++ b/3204/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,12 @@ +//Initilization of variables
+W=2500 //N //This load acts at point B and C.
+alpha=30 //degree // angle made by T1 with +ve y-axis & T2 with +ve x-axis
+//Calculations
+T2=W-(((cosd(alpha))^2/(sind(alpha)))-(sind(alpha))) // N // substuting eq'n 1 in 2
+T1=(T2*cosd(30))/(sind(30)) //N // using eq'n 1
+T3=T2 //N // By equilibrium eq'n at point C(sumFx=0)
+//Results
+clc
+printf('Tension in portion AB is %f N \n',T1)
+printf('Tension in portion BC is %f N \n',T2)
+printf('Tension in portion CD is %f N \n',T3)
diff --git a/3204/CH2/EX2.15/Ex2_15.sce b/3204/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..99ec2a71b --- /dev/null +++ b/3204/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,11 @@ +//Initilization of variables
+d=0.6 //m //diameter of the wheel
+r=0.3 //m //radius of the wheel
+W=1000 //N //weight of the wheel
+h=0.15 //m //height of rectangular block
+//Calculations
+theta=atand((sqrt(h))/(sqrt(d-h)))
+P=(W*tand(theta)) //N // dividing eq'n 1 & 2
+//Resuts
+clc
+printf('The force P so that the wheel is just to roll over the block is %f N \n',P)
diff --git a/3204/CH2/EX2.16/Ex2_16.sce b/3204/CH2/EX2.16/Ex2_16.sce new file mode 100644 index 000000000..58c115e10 --- /dev/null +++ b/3204/CH2/EX2.16/Ex2_16.sce @@ -0,0 +1,12 @@ +//Initilization of variables
+Soa=1000 //N (tension)
+alpha=45 //degree //where alpha=(360/8)
+theta=67.5 //degree //angle made by bar AO with AB &AH
+//Calcultions
+Sab=Soa*(sind(theta)/sind(alpha)) // N // Using law of sines
+Sah=Sab //N
+Sob=(Sab*sind(180-2*(theta)))/sind(theta) //N
+//Results
+clc
+printf('The axial force in the bar AB is %f N \n',Sab) //Compression
+printf('The axial force in the bar OB is %f N \n',Sob) //Tension
diff --git a/3204/CH2/EX2.17/Ex2_17.sce b/3204/CH2/EX2.17/Ex2_17.sce new file mode 100644 index 000000000..3f43371ef --- /dev/null +++ b/3204/CH2/EX2.17/Ex2_17.sce @@ -0,0 +1,12 @@ +//Initilization of variables
+W=500 //N //weight of cylinder
+alpha=25 //degree //angle made by OA with horizontal
+beta=65 //degree //angle made by OB with horizontal
+theta=90 //degree // theta=(alpha+beta)
+//Calculations
+Ra=(W*sind(beta))/sind(theta) //N //from equilibrium eq'n
+Rb=(W*sind(alpha))/sind(theta) //N //from equilibrium eqn's
+//Results
+clc
+printf('The reaction at A is %f N \n',Ra)
+printf('The reaction at B is %f N \n',Rb)
diff --git a/3204/CH2/EX2.18/Ex2_18.sce b/3204/CH2/EX2.18/Ex2_18.sce new file mode 100644 index 000000000..c074ccddc --- /dev/null +++ b/3204/CH2/EX2.18/Ex2_18.sce @@ -0,0 +1,17 @@ +//Initilization of variables
+Wa=1000 //N //weight of sphere A
+Wb=400 //N //weight of sphere B
+Ra=0.09 //m //radius of sphere A
+Rb=0.05 //m //radius of sphere B
+theta=33.86 //degree //angle made by Rq with Wb
+alpha=60 //degree //angle made by Rl with horizontal
+//Calculations
+Rq=Wb/cosd(theta) //N //using sum Fy=0 for sphere B
+Rp=Rq*sind(theta) //N //using sum Fx=0 for sphere B
+Rl=(Rq*sind(theta))/sind(alpha) //N //using sum Fx=0 for sphere A
+Rn=((Wa)+(Rq*cosd(theta))-(Rl*cosd(alpha))) //N
+//Results
+clc
+printf('The reaction at point P is %f N \n',Rp)
+printf('The reaction at point L is %f N \n',Rl)
+printf('The reaction at point N is %f N \n',Rn)
diff --git a/3204/CH2/EX2.19/Ex2_19.sce b/3204/CH2/EX2.19/Ex2_19.sce new file mode 100644 index 000000000..a27f572d4 --- /dev/null +++ b/3204/CH2/EX2.19/Ex2_19.sce @@ -0,0 +1,11 @@ +//Initilization of variables
+P=50 //N
+Q=100 //N
+alpha=30 //degree //angle made by Rq with +ve Y-axis
+//Calculations
+theta=atand((P*cotd(alpha)-Q*tand(alpha))/(P+Q)) //degree
+T=Q/(cosd(theta)*cotd(alpha)-sind(theta)) //N
+//Results
+clc
+printf('The tension in the string is %f N \n',T)
+printf('The angle wich the string makes with the horizontal when the system is in equilibrium is %f N \n',theta)
diff --git a/3204/CH2/EX2.2/Ex2_2.sce b/3204/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..db84f1d78 --- /dev/null +++ b/3204/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,14 @@ +//Initilization of variables
+P=50 //N
+Q=100 //N
+beta=15 //degree // angle between P& the horizontal
+theta=45 //degree // angle between the resultant (R) & the horizontal
+//Calculations
+Rx=P*cosd(beta)+Q*cosd(theta) //N
+Ry=P*sind(beta)+Q*sind(theta) //N
+R=sqrt((Rx^2)+(Ry^2)) //N
+alpha=atand(Ry/Rx) //degree
+//Results
+clc
+printf('The magnitude of the resultant is %f N \n',R)
+printf('The ange of the resultant with x-axis is %f degree \n',alpha)
diff --git a/3204/CH2/EX2.20/Ex2_20.sce b/3204/CH2/EX2.20/Ex2_20.sce new file mode 100644 index 000000000..081286cac --- /dev/null +++ b/3204/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,14 @@ +//Initilization of variables
+theta1=50.5 //degree //is the angle made between BC & and BE
+theta2=36.87 //degree //is te angle ade between BA &BE
+g=9.81 //m/s^2
+Wa=15*g //N
+Wb=40*g //N
+Wc=20*g //N
+//Calculations
+R2=Wc/(sind(theta1)) //N //from F.B.D of cylinder C(sum Fy=0)
+R4=(Wb+R2*sind(theta1))/sind(theta2) //N //from F.B.D of cylinder B(sum Fy=0)
+R6=R4*cosd(theta2) //N //from F.B.D of cylinder A(sum Fx=0)
+//Results
+clc
+printf('The reaction between the cylinder A and the wall of the channel is %f N \n',R6)
diff --git a/3204/CH2/EX2.21/Ex2_21.sce b/3204/CH2/EX2.21/Ex2_21.sce new file mode 100644 index 000000000..f71e706ee --- /dev/null +++ b/3204/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,12 @@ +//Initilazation of variables
+F=1000 //N
+theta=30 //degree //angle made by the force with the beam AB
+Lab=3 //m
+Lae=2 //m
+Lce=1 //m
+//Calculations
+Re=(F*Lab*sind(theta))/Lae //N //Taking moment at A
+Rd=(Re*Lce)/(Lab*sind(theta)) //N //Taking moment about C
+//Results
+clc
+printf('The reaction at D due to force of 1000 N acting at B is %f N \n',Rd)
diff --git a/3204/CH2/EX2.23/Ex2_23.sce b/3204/CH2/EX2.23/Ex2_23.sce new file mode 100644 index 000000000..ec4c6aa1b --- /dev/null +++ b/3204/CH2/EX2.23/Ex2_23.sce @@ -0,0 +1,11 @@ +//Initilization of variables
+W=1000 //N
+r=0.30 //m //radius of the wheel
+h=0.15 //m //height of the obstacle
+//Calculations
+theta=asind(1) //degree //P is mini when sin(theta)=1 from eq'n of P
+Pmini=(W*sqrt((2*r*h)-(h^2)))/(r*sind(theta)) //N
+//Results
+clc
+printf('The least force required to just turn the wheel over the block is %f N \n',Pmini)
+printf('The angle wich should be made by Pmini with AC is %f degree \n',theta)
diff --git a/3204/CH2/EX2.4/Ex2_4.sce b/3204/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..283329c2f --- /dev/null +++ b/3204/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,12 @@ +//Initilization of variables
+Tac=3.5 //kN
+Tbc=3.5 //kN
+alpha=20 //degree //angle made by Tac with -ve X axis
+beta=50 //degree //angle made by Tbc with +ve X axis
+//Calculations
+theta=atand(((Tac*sind(alpha))+(Tbc*sind(beta)))/((Tac*cosd(alpha))-(Tbc*cosd(beta)))) //degree
+P=Tac*(cosd(alpha)-cosd(beta))/(cosd(theta)) //kN // from eq'n 1
+//Results
+clc
+printf('The maximum force that can be applied is %f kN \n',P)
+printf('The direction of applied force is %f degree \n',theta)
diff --git a/3204/CH2/EX2.8/Ex2_8.sce b/3204/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..b6b3964d3 --- /dev/null +++ b/3204/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,8 @@ +//Initilization of variables
+lAB=0.4 //m
+lBC=0.3 //m
+//Calculations
+alpha=atand(lAB/lBC) //degree
+//Results
+clc
+printf('The angle wich the force should make with the horizontal to keep the edge AB of the body vertical %f degree \n',alpha) //here alpha=theta
diff --git a/3204/CH2/EX2.9/Ex2_9.sce b/3204/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..6187b650e --- /dev/null +++ b/3204/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,13 @@ +//Initilization of variables
+F=1000 //N
+lAB=0.5 //m
+lDC=0.25 //m //length of the perpendicular drawn from point C to AB
+//Calculations
+lAC=sqrt((0.3)^2+(0.25)^2) //m
+lBC=sqrt((0.20)^2+(0.25)^2) //m
+Sac=(lAC*F)/(lAB) //N //by law of concurrent forces
+Sbc=(lBC*F)/(lAB) //N //by law of concurrent forces
+//Results
+clc
+printf('The axial force in the bar AC(by aw of concurrent forces) is %f N \n',Sac)
+printf('The axial force in the bar BC(by aw of concurrent forces) is %f N \n',Sbc)
diff --git a/3204/CH20/EX20.1/Ex20_1.sce b/3204/CH20/EX20.1/Ex20_1.sce new file mode 100644 index 000000000..da3deade4 --- /dev/null +++ b/3204/CH20/EX20.1/Ex20_1.sce @@ -0,0 +1,19 @@ +// Initilization of variables
+v_o=500 // m/s // velocity of the projectile
+alpha=30 // angle at which the projectile is fired
+t=30 // seconds
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+v_x=v_o*cosd(alpha) // m/s // Initial velocity in the horizontal direction
+v_y=v_o*sind(alpha) // m/s // Initial velocity in the vertical direction
+// MOTION IN HORIZONTA DIRECTION:
+V_x=v_x // m/s // V_x=Horizontal velocity after 30 seconds
+// MOTION IN VERTICAL DIRECTION: // using the eq'n v=u+a*t
+V_y=v_y-(g*t) // m/s // -ve sign denotes downward motion
+// Let the Resultant velocity be v_R. It is given as,
+v_R=sqrt((V_x)^2+(-V_y)^2) // m/s
+theta=atand((-V_y)/V_x) // degree // direction of the projectile
+// Results
+clc
+printf('The velocity of the projectile is %f m/s \n',v_R) // The answer of velocity is wrong in the text book.
+printf('The direction of the projectile is %f degree \n',theta) // -ve value of theta indicates that the direction is in downward direction
diff --git a/3204/CH20/EX20.10/Ex20_10.sce b/3204/CH20/EX20.10/Ex20_10.sce new file mode 100644 index 000000000..eb4aceb26 --- /dev/null +++ b/3204/CH20/EX20.10/Ex20_10.sce @@ -0,0 +1,19 @@ +// Initilization of variables
+v_o=200 // m/s // initial velocity
+theta=60 // degree // angle of the incline
+y=5 // rise of incline
+x=12 // length of incline
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// The angle of the inclined plane with respect to horizontal
+beta=atand(y/x) // degree
+// The angle of projection with respect to horizontal
+alpha=90-theta // degree
+// Range is given by eq'n (ref. fig.20.14)
+AB=(2*v_o^2*(sind(alpha-beta))*cosd(alpha))/(g*(cosd(beta))^2) // m
+// Range AC when the short is fired down the plane
+AC=(2*v_o^2*(sind(alpha+beta))*cosd(alpha))/(g*(cosd(beta))^2) // m
+BC=AB+AC // m
+// Results
+clc
+printf('The range covered (i.e BC) is %f m \n',BC)
diff --git a/3204/CH20/EX20.2/Ex20_2.sce b/3204/CH20/EX20.2/Ex20_2.sce new file mode 100644 index 000000000..a50b34963 --- /dev/null +++ b/3204/CH20/EX20.2/Ex20_2.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+v_A=10 // m/s // velocity of body A
+alpha_A=60 // degree // direction of body A
+alpha_B=45 // degree // direction of body B
+// Calculations
+// (a) The velocity (v_B) for the same range is given by eq'n;
+v_B=sqrt((v_A^2*sind(2*alpha_A))/(sind(2*alpha_B))) // m/s
+// (b) Now velocity v_B for the same maximum height is given as,
+v_b=sqrt((v_A^2)*((sind(alpha_A))^2/(sind(alpha_B))^2)) // m/s
+// (c) Now the velocity (v) for the equal time of flight is;
+v=(v_A*sind(alpha_A))/(sind(alpha_B)) // m/s
+// Results
+clc
+printf('(a) The velocity of body B for horizontal range is %f m/s \n',v_B)
+printf('(b) The velocity of body B for the maximum height is %f m/s \n',v_b)
+printf('(c) The velocity of body B for equal time of flight is %f m/s \n',v)
diff --git a/3204/CH20/EX20.3/Ex20_3.sce b/3204/CH20/EX20.3/Ex20_3.sce new file mode 100644 index 000000000..4902cab34 --- /dev/null +++ b/3204/CH20/EX20.3/Ex20_3.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+y=3.6 // m // height of the wall
+x_1=4.8 // m // position of the boy w.r.t the wall
+x_2=3.6 // m // distance from the wall where the ball hits the ground
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// The range of the projectile is r, given as,
+r=x_1+x_2 // m
+// Let the angle of the projection be alpha, which is derived and given as,
+alpha=atand((y)/(x_1-(x_1^2/r))) // degree
+// Now substuting the value of alpha in eq'n 3 we get the least velocity (v_o) as;
+v_o=sqrt((g*r)/(sind(2*alpha))) // m/s
+// Results
+clc
+printf('The least velocity with which the ball can be thrown is %f m/s \n',v_o)
+printf('The angle of projection for the same is %f degree \n',alpha)
diff --git a/3204/CH20/EX20.5/Ex20_5.sce b/3204/CH20/EX20.5/Ex20_5.sce new file mode 100644 index 000000000..bccf5db74 --- /dev/null +++ b/3204/CH20/EX20.5/Ex20_5.sce @@ -0,0 +1,25 @@ +// Initilization of variables
+v_o=400 // m/s // initial velocity of each gun
+r=5000 // m // range of each of the guns
+g=9.81 // m/s^2 // acc due to gravity
+pi=180 // degree
+// Calculations
+// now from eq'n 1
+theta_1=(asind((r*g)/(v_o^2)))/(2) // degree // angle at which the 1st gun is fired
+// from eq'n 3
+theta_2=(pi-(2*theta_1))/2 // degree
+// For 1st & 2nd gun, s is
+s=r // m
+// For 1st gun
+v_x=v_o*cosd(theta_1) // m/s
+// Now the time of flight for 1st gun is t_1, which is given by relation,
+t_1=s/(v_x) // seconds
+// For 2nd gun
+V_x=v_o*cosd(theta_2)
+// Now the time of flight for 2nd gun is t_2
+t_2=s/(V_x) // seconds
+// Let the time difference between the two hits be delta.T. Then,
+delta.T=t_2-t_1 // seconds
+// Results
+clc
+printf('The time difference between the two hits is %f seconds \n',delta.T)
diff --git a/3204/CH20/EX20.6/Ex20_6.sce b/3204/CH20/EX20.6/Ex20_6.sce new file mode 100644 index 000000000..d5893d35c --- /dev/null +++ b/3204/CH20/EX20.6/Ex20_6.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+h=2000 // m/ height of the plane
+v=540*(1000/3600) // m/s // velocity of the plane
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Time t required to travel down a height 2000 m is given by eq'n,
+u=0 // m/s // initial velocity
+t=sqrt((2*h)/(g)) // seconds
+// Now let s be the horizonta distance travelled by the bomb in time t seconds, then
+s=v*t // m
+// angle is given as theta,
+theta=atand(h/s) // degree
+// Results
+clc
+printf('The pilot should release the bomb from a distance of %f m \n',s)
+printf('The angle at which the target would appear is %f degree \n',theta)
diff --git a/3204/CH20/EX20.7/Ex20_7.sce b/3204/CH20/EX20.7/Ex20_7.sce new file mode 100644 index 000000000..7fac0e018 --- /dev/null +++ b/3204/CH20/EX20.7/Ex20_7.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+theta=30 // degree // angle at which the bullet is fired
+s=-50 // position of target below hill
+v=100 // m/s // velocity at which the bullet if fired
+g=9.81 // m/s^2
+// Calculations
+v_x=v*cosd(theta) // m/s // Initial velocity in horizontal direction
+v_y=v*sind(theta) // m/s // Initial velocity in vertical direction
+// (a) Max height attained by the bullet
+h=v_y^2/(2*g) // m
+// (b)Let the vertical Velocity with which the bullet will hit the target be V_y. Then,
+V_y=sqrt((2*-9.81*s)+(v_y)^2) // m/s // the value of V_y is +ve & -ve
+// Let V be the velocity with wich it hits the target
+V=sqrt((v_x)^2+(V_y)^2) // m/s
+// (c) The time required to hit the target
+a=g // m/s^2
+t=(v_y-(-V_y))/a // seconds
+// Results
+clc
+printf('(a) The maximum height to which the bullet will rise above the soldier is %f m \n',h)
+printf('(b) The velocity with which the bullet will hit the target is %f m/s \n',V)
+printf('(c) The time required to hit the target is %f seconds \n',t)
diff --git a/3204/CH20/EX20.8/Ex20_8.sce b/3204/CH20/EX20.8/Ex20_8.sce new file mode 100644 index 000000000..f33f781ae --- /dev/null +++ b/3204/CH20/EX20.8/Ex20_8.sce @@ -0,0 +1,29 @@ +// Initilization of variables
+W=30 // N // Weight of the hammer
+theta=30 // degree // ref fig.20.12
+mu=0.18 // coefficient of friction
+s=10 // m // distance travelled by the hammer // fig 20.12
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// The acceleration of the hammer is given as,
+a=g*((sind(theta))-(mu*cosd(theta))) // m/s^2
+// The velocity of the hammer at point B is,
+v=sqrt(2*a*s) // m/s
+// Let the initial velocity of the hammer in horizontal direction be v_x & v_y in vertical direction, Then,
+v_x=v*cosd(theta) // m/s
+v_y=v*sind(theta) // m/s
+// MOTION IN VERTICAL DIRECTION
+// Now, let time required to travel vertical distance (i.e BB'=S=5 m) is given by finding the roots of the second degree eq'n as,
+// From the eq'n 4.9*t^2+4.1*t-5=0,
+a=4.9
+b=4.1
+c=-5
+// The roots of the eq'n are,
+t=((-b)+(sqrt(b^2-(4*a*c))))/(2*a)
+// MOTION IN HORIZONTAL DIRECTION
+// Let the horizotal distance travelled by the hammer in time t be s_x.Then,
+s_x=v_x*cosd(theta)*t // m
+x=1+s_x // m
+// Results
+clc
+printf('The distance x where the hammer hits the round is %f m \n',x)
diff --git a/3204/CH20/EX20.9/Ex20_9.sce b/3204/CH20/EX20.9/Ex20_9.sce new file mode 100644 index 000000000..f39ea22c1 --- /dev/null +++ b/3204/CH20/EX20.9/Ex20_9.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+s=1000 // m // distance OB (ref fig.20.13)
+h=19.6 // m // height of shell from ground
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// MOTION OF ENTIRE SHELL FROM O to A.
+v_y=sqrt(2*(g)*h) // m/s // initial velocity of shell in vertical direction
+t=v_y/g // seconds // time taken by the entire shell to reach point A
+v_x=s/t // m/s // velocity of shell in vertical direction
+// VELOCITIES OF THE TWO PARTS OF THE SHELL AFTER BURSTING AT A:
+// Let v_x2 be the horizontal velocity of 1st & the 2nd part after bursting which is given as,
+v_x2=v_x*2 // m/s
+// Now distance BC travelled by part 2 is
+BC=v_x2*t // m
+// Distance from firing point OC
+OC=s+BC // m
+// Results
+clc
+printf('(a) The velocity of shell just before bursting is %f m/s \n',v_x)
+printf('(b) The velocity of first part immediately after the shell burst is %f m/s \n',v_x2)
+printf('(c) The velocity of second part immediately after the shell burst is %f m/s \n',v_x2)
+printf('(b) The distance between the firing point & the point where the second part of the shell hit the ground is %f m \n',OC)
diff --git a/3204/CH21/EX21.1/Ex21_1.sce b/3204/CH21/EX21.1/Ex21_1.sce new file mode 100644 index 000000000..787a98152 --- /dev/null +++ b/3204/CH21/EX21.1/Ex21_1.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+N=1800 // r.p.m // Speed of the shaft
+t=5 // seconds // time taken to attain the rated speed // case (a)
+T=90 // seconds // time taken by the unit to come to rest // case (b)
+// Calculations
+omega=(2*%pi*N)/(60)
+// (a)
+// we take alpha_1,theta_1 & n_1 for case (a)
+alpha_1=omega/t // rad/s^2 //
+theta_1=(omega^2)/(2*alpha_1) // radian
+// Let n_1 be the number of revolutions turned,
+n_1=theta_1*(1/(2*%pi))
+// (b)
+// similarly we take alpha_1,theta_1 & n_1 for case (b)
+alpha_2=(omega/T) // rad/s^2 // However here alpha_2 is -ve
+theta_2=(omega^2)/(2*alpha_2) // radians
+// Let n_2 be the number of revolutions turned,
+n_2=theta_2*(1/(2*%pi))
+// Results
+clc
+printf('(a) The no of revolutions the unit turns to attain the rated speed is %f \n',n_1)
+printf('(b) The no of revolutions the unit turns to come to rest is %f \n',n_2)
diff --git a/3204/CH21/EX21.11/Ex21_11.sce b/3204/CH21/EX21.11/Ex21_11.sce new file mode 100644 index 000000000..4a132581c --- /dev/null +++ b/3204/CH21/EX21.11/Ex21_11.sce @@ -0,0 +1,10 @@ +// Initilization of variables
+v_a=2 // m/s // velocity at end A
+r=0.05 // m // radius of the disc
+alpha=30 // degree // angle made by the bar with the horizontal
+// Calculations
+// Soving eqn's 1 & 2 and substuting eqn 1 in it we get eq'n for omega as,
+omega=(v_a*(sind(alpha))^2)/(r*cosd(alpha)) // radian/second
+// Results
+clc
+printf('The anguar veocity of the bar is %f radian/second \n',omega)
diff --git a/3204/CH21/EX21.12/Ex21_12.sce b/3204/CH21/EX21.12/Ex21_12.sce new file mode 100644 index 000000000..b456765ee --- /dev/null +++ b/3204/CH21/EX21.12/Ex21_12.sce @@ -0,0 +1,20 @@ +// Initilization of variables
+l=0.6 // m
+r=0.12 // m
+theta=30 // degree // angle made by OA with the horizontal
+phi=5.7 // degree // from EX 21.5
+N=300
+// Calculations
+// Let the angular velocity of the connecting rod be (omega_ab) which is given from eqn's 1 & 4 as,
+omega_oa=(2*%pi*N)/(60) // radian/ second
+// Now,in triangle IBO.
+IB=(l*cosd(phi)*tand(theta))+(r*sind(theta)) // m
+IA=(l*cosd(phi))/(cosd(theta)) // m
+// from eq'n 5
+v_b=(r*omega_oa*IB)/(IA) // m/s
+// From eq'n 6
+omega_ab=(r*omega_oa)/(IA) // radian/second
+// Results
+clc
+printf('The velocity at B is %f m/s \n',v_b)
+printf('The angular velocity of the connecting rod is %f radian/second \n',omega_ab)
diff --git a/3204/CH21/EX21.13/Ex21_13.sce b/3204/CH21/EX21.13/Ex21_13.sce new file mode 100644 index 000000000..749c3b68e --- /dev/null +++ b/3204/CH21/EX21.13/Ex21_13.sce @@ -0,0 +1,21 @@ +// Initilization of variables
+omega_ab=5 // rad/s // angular veocity of the bar
+AB=0.20 // m
+BC=0.15 // m
+CD=0.3 // m
+theta=30 // degree // where theta= angle made by AB with the horizontal
+alpha=60 // degree // where alpha=angle made by CD with the horizontal
+// Calculations
+// Consider triangle BIC
+IB=sind(alpha)*BC*1 // m
+IC=sind(theta)*BC*1 // m
+v_b=omega_ab*AB // m/s
+// let the angular velocity of the bar BC be omega_bc
+omega_bc=v_b/IB // radian/second
+v_c=omega_bc*IC // m/s
+// let the angular velocity of bar DC be omega_dc
+omega_dc=v_c/CD // radian/second
+// Results
+clc
+printf('The angular velocity of bar BC is %f rad/s \n',omega_bc)
+printf('The angular velocity of bar CD is %f rad/s \n',omega_dc)
diff --git a/3204/CH21/EX21.2/Ex21_2.sce b/3204/CH21/EX21.2/Ex21_2.sce new file mode 100644 index 000000000..a312deff7 --- /dev/null +++ b/3204/CH21/EX21.2/Ex21_2.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+r=1 // m // radius of the cylinder
+v_c=20 // m/s // velocity of the cylinder at its centre
+// Calculations
+// The velocity of point E is given by using the triangle law as,
+v_e=sqrt(2)*v_c // m/s
+// Similarly the velocity at point F is given as,
+v_f=2*v_c // m/s
+// Results
+clc
+printf('The velocity of point E is %f m/s \n',v_e)
+printf('The velocity of point F is %f m/s \n',v_f)
diff --git a/3204/CH21/EX21.3/Ex21_3.sce b/3204/CH21/EX21.3/Ex21_3.sce new file mode 100644 index 000000000..0798e4724 --- /dev/null +++ b/3204/CH21/EX21.3/Ex21_3.sce @@ -0,0 +1,14 @@ +// Initilization of Variables
+v_1=3 // m/s // uniform speed of the belt at top
+v_2=2 // m/s // uniform speed of the belt at the bottom
+r=0.4 // m // radius of the roller
+// Calculations
+// equating eq'ns 2 & 4 and solving for v_c & theta' (angular velocity). We use matrix to solve the eqn's
+A=[1 r;1 -r]
+B=[v_1;v_2]
+C=inv(A)*B
+// Results
+clc
+printf('The linear velocity (v_c) at point C is %f m/s \n',C(1))
+printf('The angular velocity at point C is %f radian/seconds \n', C(2))
+// NOTE: The answer of angular velocity is incorrect in the book
diff --git a/3204/CH21/EX21.4/Ex21_4.sce b/3204/CH21/EX21.4/Ex21_4.sce new file mode 100644 index 000000000..12c6ad880 --- /dev/null +++ b/3204/CH21/EX21.4/Ex21_4.sce @@ -0,0 +1,20 @@ +// Initilization of Variables
+l=1 // m // length of bar AB
+v_a=5 // m/s // velocity of A
+theta=30 // degree // angle made by the bar with the horizontal
+// Calculations
+// From the vector diagram linear velocity of end B is given as,
+v_b=v_a/tand(theta) // m/s
+// Now let the relative velocity be v_ba which is given as,
+v_ba=v_a/sind(theta) // m/s
+// Now let the angular velocity of the bar be theta_a which is given as,
+theta_a=(v_ba)/l // radian/second
+// Velocity of point A
+v_a=(l/2)*theta_a // m/s
+// Magnitude of velocity at point C is,
+v_c=v_a // m/s // from the vector diagram
+// Results
+clc
+printf('(a) The angular velocity of the bar is %f radian/second \n',theta_a)
+printf('(b) The velocity of end B is %f m/s \n',v_b)
+printf('(c) The velocity of mid point C is %f m/s \n',v_c)
diff --git a/3204/CH21/EX21.5/Ex21_5.sce b/3204/CH21/EX21.5/Ex21_5.sce new file mode 100644 index 000000000..b4b18bb55 --- /dev/null +++ b/3204/CH21/EX21.5/Ex21_5.sce @@ -0,0 +1,22 @@ +// Initilization of Variables
+r=0.12 // m // length of the crank
+l=0.6 // m // length of the connecting rod
+N=300 // r.p.m // angular velocity of the crank
+theta=30 // degree // angle made by the crank with the horizontal
+// Calculations
+// Now let the angle between the connecting rod and the horizontal rod be phi
+phi=asind((r*sind(theta))/(l)) // degree
+// Now let the angular velocity of crank OA be omega_oa, which is given by eq'n
+omega_oa=(2*%pi*N)/(60) // radian/second
+// Linear velocity at A is given as,
+v_a=r*omega_oa // m/s
+// Now using the sine rule linear velocity at B can be given as,
+v_b=(v_a*sind(35.7))/(sind(84.3)) // m/s
+// Similarly the relative velocity (assume v_ba) is given as,
+v_ba=(v_a*sind(60))/(sind(84.3))
+// Angular velocity (omega_ab) is given as,
+omega_ab=v_ba/l // radian/second
+// Results
+clc
+printf('(a) The angular velocity of the connecting rod is %f radian/second \n',omega_ab)
+printf('(b) The velocity of the piston when the crank makes an angle of 30 degree is %f m/s \n',v_b)
diff --git a/3204/CH21/EX21.6/Ex21_6.sce b/3204/CH21/EX21.6/Ex21_6.sce new file mode 100644 index 000000000..9541acd61 --- /dev/null +++ b/3204/CH21/EX21.6/Ex21_6.sce @@ -0,0 +1,14 @@ +// Initiization of variables
+r=1 // m // radius of the cylinder
+v_c=20 // m/s // velocity at the centre
+// Calculations
+// Angular velocity is given as,
+omega=v_c/r // radian/second
+// Velocity at point D is
+v_d=omega*sqrt(2)*r // m/s // from eq'n 1
+// Now, the velocity at point E is,
+v_e=omega*2*r // m/s
+// Results
+clc
+printf('The velocity at point D is %f m/s \n',v_d)
+printf('The velocity at point E is %f m/s \n',v_e)
diff --git a/3204/CH21/EX21.7/Ex21_7.sce b/3204/CH21/EX21.7/Ex21_7.sce new file mode 100644 index 000000000..f5807c877 --- /dev/null +++ b/3204/CH21/EX21.7/Ex21_7.sce @@ -0,0 +1,20 @@ +// Initilization of Variables
+r=5 // cm // radius of the roller
+AB=0.1 // m
+v_a=3 // m/s // velocity at A
+v_b=2 // m/s // velocity at B
+// Calculations
+// Solving eqn's 1 & 2 using matrix for IA & IB we get,
+A=[-2 3;1 1]
+B=[0;AB]
+C=inv(A)*B
+d1=C(2)*10^2 // cm // assume d1 for case 1
+// Similary solving eqn's 3 & 4 again for IA & IB we get,
+P=[-v_b v_a;1 -1]
+Q=[0;AB]
+R=inv(P)*Q
+d2=R(2)*10^2 // cm // assume d2 for case 2
+// Results
+clc
+printf('The distance d when the bars move in the opposite directions are %f cm \n',d1)
+printf('The distance d when the bars move in the same directions are %f cm \n',d2)
diff --git a/3204/CH21/EX21.8/Ex21_8.sce b/3204/CH21/EX21.8/Ex21_8.sce new file mode 100644 index 000000000..99cc2f9d4 --- /dev/null +++ b/3204/CH21/EX21.8/Ex21_8.sce @@ -0,0 +1,21 @@ +// Initilization of Variables
+v_c=1 // m/s // velocity t the centre
+r1=0.1 // m
+r2=0.20 // m
+EB=0.1 // m
+EA=0.3 // m
+ED=sqrt(r1^2+r2^2) // m
+// Calculations
+// angular velocity is given as,
+omega=v_c/r1 // radian/seconds
+// Velocit at point B
+v_b=omega*EB // m/s
+// Velocity at point A
+v_a=omega*EA // m/s
+// Velocity at point D
+v_d=omega*ED // m/s
+// Results
+clc
+printf('The velocity at point A is %f m/s \n',v_a)
+printf('The velocity at point B is %f m/s \n',v_b)
+printf('The velocity at point D is %f m/s \n',v_d)
diff --git a/3204/CH21/EX21.9/Ex21_9.sce b/3204/CH21/EX21.9/Ex21_9.sce new file mode 100644 index 000000000..e1bb298c8 --- /dev/null +++ b/3204/CH21/EX21.9/Ex21_9.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+l=1 // m // length of bar AB
+v_a=5 // m/s // velocity at A
+theta=30 // degree // angle made by the bar with the horizontal
+// Calculations
+IA=l*sind(theta) // m
+IB=l*cosd(theta) // m
+IC=0.5 // m // from triangle IAC
+// Angular veocity is given as,
+omega=v_a/(IA) // radian/second
+v_b=omega*IB // m/s
+v_c=omega*IC // m/s
+// Results
+clc
+printf('The velocity at point B is %f m/s \n',v_b)
+printf('The velocity at point C is %f m/s \n',v_c)
diff --git a/3204/CH22/EX22.1/Ex22_1.sce b/3204/CH22/EX22.1/Ex22_1.sce new file mode 100644 index 000000000..2874f73c7 --- /dev/null +++ b/3204/CH22/EX22.1/Ex22_1.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+N=1500 // r.p.m
+r=0.5 // m // radius of the disc
+m=300 // N // weight of the disc
+t=120 // seconds // time in which the disc comes to rest
+omega=0
+g=9.81 // m/s^2
+// Calculations
+omega_0=(2*%pi*N)/60 // rad/s
+// angular deceleration is given as,
+alpha=-(omega_0/t) // radian/second^2
+theta=(omega_0^2)/(2*(-alpha)) // radian
+// Let n be the no of revolutions taken by the disc before it comes to rest, then
+n=theta/(2*%pi)
+// Now,
+I_G=((1/2)*m*r^2)/g
+// The frictional torque is given as,
+M=I_G*alpha // N-m
+// Results
+clc
+printf('(a) The no of revolutions executed by the disc before coming to rest is %f \n',n)
+printf('(b) The frictional torque is %f N-m \n',M)
diff --git a/3204/CH22/EX22.10/Ex22_10.sce b/3204/CH22/EX22.10/Ex22_10.sce new file mode 100644 index 000000000..72f4eddf5 --- /dev/null +++ b/3204/CH22/EX22.10/Ex22_10.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+L=1 // m // length of rod AB
+m=10 // kg // mass of the rod
+g=9.81
+theta=30 // degree
+// Calculations
+// solving eq'n 4 for omega we get,
+omega=sqrt(2*16.82*sind(theta)) // rad/s
+// Now solving eq'ns 1 &3 for alpha we get,
+alpha=(12/7)*g*cosd(theta) // rad/s
+// Components of reaction are given as,
+R_t=((m*g*cosd(theta))-((m*alpha*L)/4)) // N
+R_n=((m*omega^2*L)/(4))+(m*g*sind(theta)) // N
+R=sqrt(R_t^2+R_n^2) // N
+// Results
+clc
+printf('(a) The angular velocity of the rod is %f rad/sec \n',omega)
+printf('(b) The reaction at the hinge is %f N \n',R)
diff --git a/3204/CH22/EX22.2/Ex22_2.sce b/3204/CH22/EX22.2/Ex22_2.sce new file mode 100644 index 000000000..f0b99440a --- /dev/null +++ b/3204/CH22/EX22.2/Ex22_2.sce @@ -0,0 +1,15 @@ +// Initilization of variables
+s=1 // m
+mu=0.192 // coefficient of static friction
+g=9.81 // m/s^2
+// Calculations
+// The maximum angle of the inclined plane is given as,
+theta=atand(3*mu) // degree
+a=(2/3)*g*sind(theta) // m/s^2 // by solving eq'n 4
+v=sqrt(2*a*s) // m/s
+// Let the acceleration at the centre be A which is given as,
+A=g*sind(theta) // m/s^2 // from eq'n 1
+// Results
+clc
+printf('(a) The acceleration at the centre is %f m/s^2 \n',A)
+printf('(b) The maximum angle of the inclined plane is %f degree \n',theta)
diff --git a/3204/CH22/EX22.5/Ex22_5.sce b/3204/CH22/EX22.5/Ex22_5.sce new file mode 100644 index 000000000..cba34089a --- /dev/null +++ b/3204/CH22/EX22.5/Ex22_5.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+W_a=25 // N
+W_b=25 // N
+W=200 // N // weight of the pulley
+i_g=0.2 // m // radius of gyration
+g=9.81 // m/s^2
+// Calculations
+// Solving eqn's 1 & 2 for acceleration of weight A (assume a)
+a=(0.15*W_a*g)/(((W*i_g^2)/(0.45))+(0.45*W_a)+((0.6*W_b)/(3))) // m/s^2
+// Results
+clc
+printf('The acceleration of weight A is %f m/s^2 \n',a)
diff --git a/3204/CH22/EX22.8/Ex22_8.sce b/3204/CH22/EX22.8/Ex22_8.sce new file mode 100644 index 000000000..829ce65b7 --- /dev/null +++ b/3204/CH22/EX22.8/Ex22_8.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+r_1=0.075 // m
+r_2=0.15 // m
+P=50 // N
+W=100 // N
+i_g=0.05 // m
+theta=30 // degree
+g=9.81 // m/s^2
+// Calculations
+// The eq'n for acceleration of the pool is given by solving eqn's 1,2 &3 as,
+a=(50*g*(r_2*cosd(theta)-r_1))/(100*((i_g^2/r_2)+r_2)) // m/s^2
+// Results
+clc
+printf('The acceleration of the pool is %f m/s^2 \n',a)
diff --git a/3204/CH23/EX23.2/Ex23_2.sce b/3204/CH23/EX23.2/Ex23_2.sce new file mode 100644 index 000000000..32542c01a --- /dev/null +++ b/3204/CH23/EX23.2/Ex23_2.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+m=600 // kg // mass of the roller
+r=0.25 // m // radius of the roller
+P=850 // N // Force
+v=3 // m/s // velocity to be acquired
+theta=30 // degree // angle made by v with the force P
+// Calculations
+// The distance required to be rolled is given by equating the Work done between positions 1 & 2 as,
+x=((3/4)*m*v^2)/(P*cosd(theta)) // m
+// Results
+clc
+printf('The distance required to be rolled is %f m \n',x)
diff --git a/3204/CH24/EX24.1/Ex24_1.sce b/3204/CH24/EX24.1/Ex24_1.sce new file mode 100644 index 000000000..8d0d7ed9e --- /dev/null +++ b/3204/CH24/EX24.1/Ex24_1.sce @@ -0,0 +1,17 @@ +// Initilization of variables
+f=1/6 // oscillations/second
+x=8 // cm // distance from the mean position
+// Calculations
+omega=2*%pi*f
+// Amplitude is given by eq'n
+r=sqrt((25*x^2)/16) // cm
+// Maximum acceleration is given as,
+a_max=(%pi/3)^2*10 // cm/s^2
+// Velocity when it is at a dist of 5 cm (assume s=5 cm) is given by
+s=5 // cm
+v=omega*sqrt(r^2-s^2) // cm/s
+// Results
+clc
+printf('(a) The amplitude of oscillation is %f cm \n',r)
+printf('(b) The maximum acceleration is %f cm/s^2 \n',a_max)
+printf('(c) The velocity of the particle at 5 cm from mean position is %f cm/s \n',v)
diff --git a/3204/CH24/EX24.10/Ex24_10.sce b/3204/CH24/EX24.10/Ex24_10.sce new file mode 100644 index 000000000..8d659d749 --- /dev/null +++ b/3204/CH24/EX24.10/Ex24_10.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+l=1 // m // length of the simple pendulum
+g=9.81 // m/s^2
+// Calculations
+// Let t_s be the time period when the elevator is stationary
+t_s=2*%pi*sqrt(l/g) /// seconds
+// Let t_u be the time period when the elevator moves upwards. Then from eqn 1
+t_u=2*%pi*sqrt((l)/(g+(g/10))) // seconds
+// Let t_d be the time period when the elevator moves downwards.
+t_d=2*%pi*sqrt(l/(g-(g/10))) // seconds
+// Results
+clc
+printf('The time period of oscillation of the pendulum for upward acc of the elevator is %f seconds \n',t_u)
+printf('The time period of oscillation of the pendulum for downward acc of the elevator is %f seconds \n',t_d)
diff --git a/3204/CH24/EX24.11/Ex24_11.sce b/3204/CH24/EX24.11/Ex24_11.sce new file mode 100644 index 000000000..10b50fe6a --- /dev/null +++ b/3204/CH24/EX24.11/Ex24_11.sce @@ -0,0 +1,14 @@ +// Initilization of variables
+t=1 // second // time period of the simple pendulum
+g=9.81 // m/s^2
+// Calculations
+// Length of pendulum is given as,
+l=(t/(2*%pi)^2)*g // m
+// Let t_u be the time period when the elevator moves upwards. Then the time period is given as,
+t_u=2*%pi*sqrt((l)/(g+(g/10))) // seconds
+// Let t_d be the time period when the elevator moves downwards.
+t_d=2*%pi*sqrt(l/(g-(g/10))) // seconds
+// Results
+clc
+printf('The time period of oscillation of the pendulum for upward acc of the elevator is %f seconds \n',t_u)
+printf('The time period of oscillation of the pendulum for downward acc of the elevator is %f seconds \n',t_d)
diff --git a/3204/CH24/EX24.12/Ex24_12.sce b/3204/CH24/EX24.12/Ex24_12.sce new file mode 100644 index 000000000..255e2f320 --- /dev/null +++ b/3204/CH24/EX24.12/Ex24_12.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+m=15 // kg // mass of the disc
+D=0.3 // m // diameter of the disc
+R=0.15 // m // radius
+l=1 // m // length of the shaft
+d=0.01 // m // diameter of the shaft
+G=30*10^9 // N-m^2 // modulus of rigidity
+// Calculations
+// M.I of the disc about the axis of rotation is given as,
+I=(m*R^2)/2 // kg-m^2
+// Stiffness of the shaft
+k_t=(%pi*d^4*G)/(32*l) // N-m/radian
+t=2*%pi*sqrt(I/k_t) // seconds
+// Results
+clc
+printf('The time period of oscillations of the disc is %f seconds \n',t)
diff --git a/3204/CH24/EX24.2/Ex24_2.sce b/3204/CH24/EX24.2/Ex24_2.sce new file mode 100644 index 000000000..8c33c4385 --- /dev/null +++ b/3204/CH24/EX24.2/Ex24_2.sce @@ -0,0 +1,22 @@ +// Initilization of variables
+x_1=0.1 // m // assume the distance of the particle from mean position as (x_1 & x_2)
+x_2=0.2// m
+// assume velocities as v_1 & v_2
+v_1=1.2 // m/s
+v_2=0.8 // m/s
+// Calculations
+// The amplitude of oscillations is given by dividing eq'n 1 by 2 as,
+r=sqrt(0.32/5) // m
+omega=v_1/(sqrt(r^2-x_1^2)) // radians/second
+t=(2*%pi)/omega // seconds
+v_max=r*omega // m/s
+// let the max acceleration be a which is given as,
+a=r*omega^2 // m/s^2
+// Results
+clc
+printf('(a) The amplitude of oscillations is %f m \n',r)
+printf('(b) The time period of oscillations is %f seconds \n',t)
+printf('(c) The maximum velocity is %f m/s \n',v_max)
+printf('(d) The maximum acceleration is %f m/s^2 \n',a) // the value of max acc is incorrect in the textbook
+// NOTE: the value of t is incorrect in the text book
+// The values may differ slightly due to decimal point accuracy
diff --git a/3204/CH24/EX24.5/Ex24_5.sce b/3204/CH24/EX24.5/Ex24_5.sce new file mode 100644 index 000000000..308fd2173 --- /dev/null +++ b/3204/CH24/EX24.5/Ex24_5.sce @@ -0,0 +1,17 @@ +// Initilization of variabes
+W=50 // N // weight
+x_0=0.075 // m // amplitude
+f=1 // oscillation/sec // frequency
+g=9.81
+// Calculations
+omega=2*%pi*f
+K=(((2*%pi)^2*W)/g)*(10^-2) // N/cm
+// let the total extension of the string be delta which is given as,
+delta=(W/K)+(x_0*10^2) // cm
+T=K*delta // N // Max Tension
+v=omega*x_0 //m/s // max velocity
+// Results
+clc
+printf('(a) The stiffness of the spring is %f N/cm \n',K)
+printf('(b) The maximum Tension in the spring is %f N \n',T)
+printf('(c) The maximum velocity is %f m/s \n',v)
diff --git a/3204/CH25/EX25.5/Ex25_5.sce b/3204/CH25/EX25.5/Ex25_5.sce new file mode 100644 index 000000000..da0ca1a0c --- /dev/null +++ b/3204/CH25/EX25.5/Ex25_5.sce @@ -0,0 +1,33 @@ +// Initilization of variables
+L_AB=3 // m // length of the beam
+L_AC=1 // m
+L_BC=2 // m
+M_C=12 // kNm // clockwise moment at C
+// Calculations
+// REACTIONS
+R_B=M_C/L_AB // kN // moment at A
+R_A=-M_C/L_AB // kN // moment at B
+// S.F
+F_A=R_A // kN
+F_B=R_A // kN
+// B.M
+M_A=0 // kNm
+M_C1=R_A*L_AC // kNm // M_C1 is the BM just before C
+M_C2=(R_A*L_AC)+M_C // kNm // M_C2 is the BM just after C
+M_B=0 // kNm
+// Plotting SFD & BMD
+x=[0;0.99;1;3]
+y=[-4;-4;-4;-4]
+a=[0;0.99;1;3]
+b=[0;-4;8;0]
+subplot(221)
+xlabel("Span (m)")
+ylabel("Shear Force (kN)")
+plot(x,y)
+subplot(222)
+plot(a,b)
+xlabel("Span (m)")
+ylabel("Bending Moment (kNm)")
+// Results
+clc
+printf('The graphs are the solutions')
diff --git a/3204/CH25/EX25.7/Ex25_7.sce b/3204/CH25/EX25.7/Ex25_7.sce new file mode 100644 index 000000000..8d4c7776c --- /dev/null +++ b/3204/CH25/EX25.7/Ex25_7.sce @@ -0,0 +1,47 @@ +// Initilization of variables
+L_AD=8 // m // length of the beam
+L_AB=2 // m
+L_BC=4 // m
+L_CD=2 // m
+UDL=1 // kN/m
+P=2 // kN // point load at A
+// Caalculations
+// REACTIONS
+// solving eqn's 1&2 using matrix to get R_B & R_C as,
+A=[1 1;1 3]
+B=[8;30]
+C=inv(A)*B
+// SHEAR FORCE
+// the term F with suffixes 1 & 2 indicates SF just to left and right
+F_A=-P // kN
+F_B1=-P // kN
+F_B2=-P+C(1) // kN
+F_C1=-P+C(1)-(UDL*L_BC) // kN
+F_C2=-P+C(1)-(UDL*L_BC)+C(2) // kN
+F_D=0
+// BENDING MOMENT
+// the term F with suffixes 1 & 2 indicates BM just to left and right
+M_A=0 // kNm
+M_B=(-P*L_CD) // kNm
+M_C=(-P*(L_AB+L_BC))+(C(1)*L_BC)-(UDL*L_BC*(L_BC/2)) // kNm
+M_D=0 // kNm
+// LOCATION OF MAXIMUM BM
+// Max BM occurs at E at a distance of 2.5 m from B i.e x=L_AE=4.5 m from free end A. Thus max BM is given by taking moment at B
+L_AE=4.5 // m // given
+M_E=(-2*L_AE)+(4.5*(L_AE-2))-((1/2)*(L_AE-2)^2) // kNm
+// PLOTTING SFD & BMD
+x=[0;1.99;2;4.5;5.99;6;8]
+y=[-2;-2;2.5;0;-1.5;2;0]
+a=[0;2;4.5;6;8]
+b=[0;-4;-0.875;-2;0]
+subplot(221)
+xlabel("Span (m)")
+ylabel("Shear Force (kN)")
+plot(x,y)
+subplot(222)
+plot(a,b)
+xlabel("Span (m)")
+ylabel("Bending Moment (kNm)")
+// Results
+clc
+printf('The graphs are the solutions')
diff --git a/3204/CH26/EX26.1/Ex26_1.sce b/3204/CH26/EX26.1/Ex26_1.sce new file mode 100644 index 000000000..19093874f --- /dev/null +++ b/3204/CH26/EX26.1/Ex26_1.sce @@ -0,0 +1,16 @@ +// 1 APPENDIX. Ex no 1. Page no 638
+//Initilization of variables
+P=[-5,2,14] //Point co-ordinates
+//Calculations
+r=sqrt(P(1)^2+P(2)^2+P(3)^2) //Magnitude of the poistion vector
+//Direction cosines
+l=P(1)/r
+m=P(2)/r
+n=P(3)/r
+//Unit Vector calculations
+r_unit=P/r
+//Results
+clc
+printf("The Position vector is %fi+%fj+%fk \n",P(1),P(2),P(3))
+printf('The value of r is %f*l i + %f*m j + %f*n k \n',r,r,r)
+printf("The unit vector in the direction of r is %fi+%fj+%fk",r_unit(1),r_unit(2),r_unit(3))
diff --git a/3204/CH26/EX26.10/Ex26_10.sce b/3204/CH26/EX26.10/Ex26_10.sce new file mode 100644 index 000000000..338d36e3b --- /dev/null +++ b/3204/CH26/EX26.10/Ex26_10.sce @@ -0,0 +1,16 @@ +// 1 APPENDIX. Ex no 10. Page no 647
+// Initilization of variables
+// Points as matrices
+O=[-2,3,5]
+P=[1,-2,4]
+Q=[5,2,3]
+F=[4,4,-1] // Force vector
+// Calculations
+// Positon vector ( we will solve only by using r_1 as r_2 also gives the same answer)
+r_1=[P(1)-O(1),P(2)-O(2),P(3)-O(3)]
+// Moment
+// Calculating the cross product
+M=[(r_1(2)*F(3)-r_1(3)*F(2)),(r_1(3)*F(1)-r_1(1)*F(3)),(r_1(1)*F(2)-r_1(2)*F(1))]
+// Results
+clc
+printf('The Moment about point O is %fi %fj+%fk \n',M(1),M(2),M(3))
diff --git a/3204/CH26/EX26.11/Ex26_11.sce b/3204/CH26/EX26.11/Ex26_11.sce new file mode 100644 index 000000000..662a634a8 --- /dev/null +++ b/3204/CH26/EX26.11/Ex26_11.sce @@ -0,0 +1,14 @@ +// 1 APPENDIX. Ex no 11. Page no 647
+// Initilization of variables
+// Points as matrices
+P=[1,-1,2] // Point where force is applied
+O=[2,-1,3] // point where moment is to be found
+F=[3,2,-4] // Force vector
+// Calculations
+// Position vector of point P wrt O
+r=[P(1)-O(1),P(2)-O(2),P(3)-O(3)]
+// Moment
+M=[(r(2)*F(3)-r(3)*F(2)),(r(3)*F(1)-r(1)*F(3)),(r(1)*F(2)-r(2)*F(1))]
+// Resuts
+clc
+printf('The moment of the force is %fi %fj %fk \n',M(1),M(2),M(3))
diff --git a/3204/CH26/EX26.12/Ex26_12.sce b/3204/CH26/EX26.12/Ex26_12.sce new file mode 100644 index 000000000..5b9ba12d7 --- /dev/null +++ b/3204/CH26/EX26.12/Ex26_12.sce @@ -0,0 +1,20 @@ +// 1 APPENDIX. Ex no 12. Page no 648
+// Initilization of variables
+// Different notations have been used at some places
+f=22 // N
+// Points as matrices
+A=[4,-1,7]
+O=[1,-3,2]
+V=[9,6,-2] // Given vector
+// Calculations
+// Unit vector in the direction of the vector
+v=[V(1),V(2),V(3)]/sqrt(V(1)^2+V(2)^2+V(3)^2)
+// Force
+F=f*[v(1),v(2),v(3)]
+// Position vector of point A wrt O
+r=[A(1)-O(1),A(2)-O(2),A(3)-O(3)]
+// Moment
+M=[(r(2)*F(3)-r(3)*F(2)),(r(3)*F(1)-r(1)*F(3)),(r(1)*F(2)-r(2)*F(1))]
+// Results
+clc
+printf('The moment is %fi + %fj + %f k \n',M(1),M(2),M(3))
diff --git a/3204/CH26/EX26.13/Ex26_13.sce b/3204/CH26/EX26.13/Ex26_13.sce new file mode 100644 index 000000000..5a20f1d56 --- /dev/null +++ b/3204/CH26/EX26.13/Ex26_13.sce @@ -0,0 +1,26 @@ +// 1 APPENDIX. Ex no 13. Page no 648
+// Initilization of vriables
+// Notations have been changed
+// Force Vector
+F=[50,-80,30]
+// from fig.13
+theta1=30 // angles by which the axis is rotated ( all in degrees)
+theta2=60
+theta3=90
+theta4=120
+theta5=0
+// Calculations
+// Unit vector in u-direction
+u_unit=[1*cosd(theta1),1*cosd(theta2),1*cosd(theta3)]
+// Unit vector in v-direction
+v_unit=[1*cosd(theta4),1*cosd(theta1),1*cosd(theta3)]
+// Unit vector in w-direction
+w_unit=[1*cosd(theta3),1*cosd(theta3),1*cosd(theta5)]
+// Components of force
+// finding the dot product as
+u=F(1)*u_unit(1)+F(2)*u_unit(2)+F(3)*u_unit(3) // N
+v=F(1)*v_unit(1)+F(2)*v_unit(2)+F(3)*v_unit(3) // N
+w=F(1)*w_unit(1)+F(2)*w_unit(2)+F(3)*w_unit(3) // N
+// Results
+clc
+printf('The components of the force is,along u=%f N,along v=%f N,along w=%f N \n',u,v,w)
diff --git a/3204/CH26/EX26.14/Ex26_14.sce b/3204/CH26/EX26.14/Ex26_14.sce new file mode 100644 index 000000000..444dae94c --- /dev/null +++ b/3204/CH26/EX26.14/Ex26_14.sce @@ -0,0 +1,32 @@ +// 1 APPENDIX. Ex no 14. Page no 649
+// Initilization of variables
+// Some notations have been assumed
+f=100 // N // magnitude of force
+// Co-ordinates of corners of the box as matrices
+A=[0,0,0]
+B=[0.5,0,0]
+C=[0.5,0,1]
+D=[0,0,1]
+E=[0,0.5,0]
+F=[0.5,0.5,0]
+G=[0.5,0.5,1]
+H=[0,0.5,1]
+// Calculations
+// Force vector
+Fmag=f/sqrt((F(1)-C(1))^2+(F(2)-C(2))^2+(F(3)-C(3))^2)
+F=Fmag*[F(1)-C(1),F(2)-C(2),F(3)-C(3)]
+// Position vector
+r_EC=[C(1)-E(1),C(2)-E(2),C(3)-E(3)]
+// Moment about point E
+// Calculating the cross product
+M_E=[(r_EC(2)*F(3)-r_EC(3)*F(2)),(r_EC(3)*F(1)-r_EC(1)*F(3)),(r_EC(1)*F(2)-r_EC(2)*F(1))] // N.m // The value taken for F is incorrect in textbook.
+// Unit vector
+n_AE=[E(1)-A(1),E(2)-A(2),E(3)-A(3)]/sqrt((E(1)-A(1))^2+(E(2)-A(2))^2+(E(3)-A(3))^2)
+// Moment of force about axis AE
+// finding the dot product
+M_AE=M_E(1)*n_AE(1)+M_E(2)*n_AE(2)+M_E(3)*n_AE(3) // N.m
+// Results
+clc
+printf('The moment of the force about point E is %fi - %fj + %fk N.m \n',M_E)
+printf('The moment of force about axis AE is -%f N.m \n',M_AE)
+// The value of M_AE & M_E is incorrect in the textbook.Incorrect value of force vector is taken in calculation of M_E
diff --git a/3204/CH26/EX26.15/Ex26_15.sce b/3204/CH26/EX26.15/Ex26_15.sce new file mode 100644 index 000000000..4a3910c4f --- /dev/null +++ b/3204/CH26/EX26.15/Ex26_15.sce @@ -0,0 +1,35 @@ +// 1 APPENDIX. Ex no 15. Page no 652
+// Initilization of variables
+// Consider fig. 16. The co-ordinates of various points are
+// Co-ordinates as matrices
+// Some of the notations have been changed
+f1=5 // kN
+f2=7.5 // kN
+f3=10 // kN
+A=[0,0,0]
+E=[0,1,0]
+D=[0,0,2]
+F=[3,1,0]
+G=[3,1,2]
+H=[0,1,2] // co-ordinates of H not given in book. ref fig.16 for the same
+// Calculations
+// Force vectors
+F1=(f1/sqrt((F(1)-E(1))^2+(F(2)-E(2))^2+(F(3)-E(3))^2))*[F(1)-E(1),F(2)-E(2),F(3)-E(3)]
+F2=(f2/sqrt((D(1)-H(1))^2+(D(2)-H(2))^2+(D(3)-H(3))^2))*[D(1)-H(1),D(2)-H(2),D(3)-H(3)]
+F3=(f3/sqrt((G(1)-E(1))^2+(G(2)-E(2))^2+(G(3)-E(3))^2))*[G(1)-E(1),G(2)-E(2),G(3)-E(3)]
+// Resultant force
+R=F1+F2+F3 // kN
+// Position vectors
+r_AE=[E(1)-A(1),E(2)-A(2),E(3)-A(3)]
+r_AD=[D(1)-A(1),D(2)-A(2),D(3)-A(3)]
+// Moment of forces about A
+// Calculating the cross product
+M1=[(r_AE(2)*F1(3)-r_AE(3)*F1(2)),(r_AE(3)*F1(1)-r_AE(1)*F1(3)),(r_AE(1)*F1(2)-r_AE(2)*F1(1))]
+M2=[(r_AD(2)*F2(3)-r_AD(3)*F2(2)),(r_AD(3)*F2(1)-r_AD(1)*F2(3)),(r_AD(1)*F2(2)-r_AD(2)*F2(1))]
+M3=[(r_AE(2)*F3(3)-r_AE(3)*F3(2)),(r_AE(3)*F3(1)-r_AE(1)*F3(3)),(r_AE(1)*F3(2)-r_AE(2)*F3(1))]
+// Rseultant moment
+M_R=M1+M2+M3 // KN.m
+// Results
+clc
+printf('The resultant force is %fi %fj + %fk kN \n',R)
+printf('The resultant moment of all the forces about point A is %fi + %fj %fk kN.m \n',M_R)
diff --git a/3204/CH26/EX26.16/Ex26_16.sce b/3204/CH26/EX26.16/Ex26_16.sce new file mode 100644 index 000000000..78074d267 --- /dev/null +++ b/3204/CH26/EX26.16/Ex26_16.sce @@ -0,0 +1,11 @@ +// 1 APPENDIX. Ex no 16. Page no 654
+// Initiization of variables
+F=20 // kN // Force acting at O
+M_x=76 // kNm
+M_y=82 // kNm
+// Calculations
+x=M_x/F // m
+z=M_y/F // m
+// Results
+clc
+printf('The point of application should be shifted to: x=%f m and z=%f m \n',x,z)
diff --git a/3204/CH26/EX26.17/Ex26_17.sce b/3204/CH26/EX26.17/Ex26_17.sce new file mode 100644 index 000000000..fe9be14b0 --- /dev/null +++ b/3204/CH26/EX26.17/Ex26_17.sce @@ -0,0 +1,20 @@ +// 1 APPENDIX. Ex no 17. Page no 655
+// Initilization of variables
+W=5000 // N
+// Co-ordinates of various points
+A=[0,4.5,0]
+B=[2.8,0,0]
+C=[0,0,-2.4]
+D=[-2.6,0,1.8]
+// Calculations
+// Ref textbook for the values of tenion in the cable AB, AC & AD. The values consist of variables which cannot be defined here
+// We re-arrange and define the equations of equilibrium as matrices and solve them as,
+P=[0.528,0.0,-0.472;0.0,0.47,-0.327;0.85,0.88,0.818]
+Q=[0;0;5000]
+X=inv(P)*Q
+// Results
+clc
+printf('Tension in cable AD is %f N \n',X(3))
+printf('Tension in cable AB is %f N \n',X(1))
+printf('Tension in cable AC is %f N \n',X(2))
+// The answer may vary slightly due to decimal point error. Ans for T_AB is incorrect in textbook.
diff --git a/3204/CH26/EX26.18/Ex26_18.sce b/3204/CH26/EX26.18/Ex26_18.sce new file mode 100644 index 000000000..cc441356c --- /dev/null +++ b/3204/CH26/EX26.18/Ex26_18.sce @@ -0,0 +1,21 @@ +// 1 APPENDIX. Ex no 18. Page no 656
+// Initilization of variables
+P=5 // kN
+Q=3 // kN
+C=5 // kNm // couple
+// ref fig.20 // Notations have been assumed
+z1=1.5 // m
+z2=0.625 // m
+z3=0.5 // m
+x1=3.5 // m
+x2=0.625 // m
+// Calculations
+// sum M_x=0
+R_A=((P*z2)+(Q*z3)+C)/z1 // kN
+// M_z=0
+R_C=((Q*x1)+(P*x2))/x1 // kN
+// sum F_y=0
+R_B=P+Q-R_A-R_C // kN
+// Results
+clc
+printf('The reactions are: R_A=%f kN ,R_C=%f kN and R_B=%f kN \n',R_A,R_C,R_B)
diff --git a/3204/CH26/EX26.19/Ex26_19.sce b/3204/CH26/EX26.19/Ex26_19.sce new file mode 100644 index 000000000..cceb47fc2 --- /dev/null +++ b/3204/CH26/EX26.19/Ex26_19.sce @@ -0,0 +1,29 @@ +// 1 APPENDIX. Ex no 19. Page no 657 +// Initilization of variables +F=2 // kN +W=1 // kN +// Co-ordinates as matrices +A=[0,0,0] +C=[0,0,1.2] +B=[0,0,2.5] +D=[-1,1,0] +E=[1,1,0] +F=[0,0,1] +G=[0,0,2] +// Force vector +f=[0,-2,0] +// Weight vector +w=[0,-1,0] +// Calculations +// we have 5 unknowns: A_x,A_y,A_z,T_FE & T_GD +// we define and solve eqn's 1,2,3,4&5 using matrix as, +P=[1,0,0,0.58,-0.41;0,1,0,0.58,0.41;0,0,1,-0.58,-0.82;0,0,0,0.58,0.82;0,0,0,0.58,-0.82] +Q=[0;3;0;6.25;0] + +X=inv(P)*Q + +// Results +clc +printf('The components of reaction at A are: A_x=%f kN , A_y=%f kN and A_z=%f kN \n',X(1),X(2),X(3)) +printf('The tensions in the cable are: T_FE=%f kN and T_GD=%f kN \n',X(4),X(5)) +// The solution in the textbook is incorrect and yeilds singularity in matrix calculation. diff --git a/3204/CH26/EX26.2/Ex26_2.sce b/3204/CH26/EX26.2/Ex26_2.sce new file mode 100644 index 000000000..733735b94 --- /dev/null +++ b/3204/CH26/EX26.2/Ex26_2.sce @@ -0,0 +1,27 @@ +// 1 APPENDIX. Ex no 2. Page no 639
+//Initilizatin of variable
+F=10 //N
+P_1=[2,4,3]
+P_2=[1,-5,2]
+
+//Calculations
+d_x=P_2(1)-P_1(1)
+d_y=P_2(2)-P_1(2)
+d_z=P_2(3)-P_1(3)
+d=sqrt(d_x^2+d_y^2+d_z^2)
+Fx=(F/d)*d_x //N
+Fy=(F/d)*d_y //N
+Fz=(F/d)*d_z //N
+//Direction cosines
+l=Fx/F
+m=Fy/F
+n=Fz/F
+//Angles
+theta_x=acosd(l) //degrees
+theta_y=acosd(m) //degrees
+theta_z=acosd(n) //degrees
+
+//Result
+clc
+printf("The force in vector notation is %fi%fj%fk\n",Fx,Fy,Fz)
+printf("Thetax=%f degrees,Thetay=%f degrees,Thetaz=%f degrees",theta_x,theta_y,theta_z)
diff --git a/3204/CH26/EX26.3/Ex26_3.sce b/3204/CH26/EX26.3/Ex26_3.sce new file mode 100644 index 000000000..95418aa40 --- /dev/null +++ b/3204/CH26/EX26.3/Ex26_3.sce @@ -0,0 +1,20 @@ +// 1 APPENDIX. Ex no 3. Page no 640 +//initiliation of variables +T=2500 //N +//Co-ordinates +Q=[40,0,-30] +P=[0,80,0] + +//Calculations +mag_QP=sqrt((P(1)-Q(1))^2+(P(2)-Q(2))^2+(P(3)-Q(3))^2) //Magnitude +QP=[(P(1)-Q(1)),(P(2)-Q(2)),(P(3)-Q(3))] +F=(T/mag_QP)*QP //N +thetax=acosd(F(1)/T) //degrees +thetay=acosd(F(2)/T) //degrees +thetaz=acosd(F(3)/T) //degrees + +//Result +clc +printf("The force vector is %fi+%fj+%fk N\n",F(1),F(2),F(3)) +//Answer in the textbook is printed as 1600 which is incorrect +printf("The angles are thetax=%f,thetay=%f and thetaz=%f degrees",thetax,thetay,thetaz) diff --git a/3204/CH26/EX26.4/Ex26_4.sce b/3204/CH26/EX26.4/Ex26_4.sce new file mode 100644 index 000000000..c746c284e --- /dev/null +++ b/3204/CH26/EX26.4/Ex26_4.sce @@ -0,0 +1,14 @@ +// 1 APPENDIX. Ex no 4. Page no 642 +//initilization of variables +A=[2,-1,1] +B=[1,1,2] +C=[3,-2,4] +//Calculations +R=[A(1)+B(1)+C(1),A(2)+B(2)+C(2),A(3)+B(3)+C(3)] //Resultant +mag=sqrt(R(1)^2+R(2)^2+R(3)^2) +//Unit vector +U=R/mag +//Result +clc +printf("The unit vector is %fi%fj+%fk",U(1),U(2),U(3)) +//Answer for k part is incorrect in the textbook diff --git a/3204/CH26/EX26.5/Ex26_5.sce b/3204/CH26/EX26.5/Ex26_5.sce new file mode 100644 index 000000000..41c627ca2 --- /dev/null +++ b/3204/CH26/EX26.5/Ex26_5.sce @@ -0,0 +1,14 @@ +// 1 APPENDIX. Ex no 5. Page no 642 +//initilization of variables +A=[2,-6,-3] +B=[4,3,-1] +//Calculations +AdotB=A(1)*B(1)+A(2)*B(2)+A(3)*B(3) +magA=sqrt(A(1)^2+A(2)^2+A(3)^2) +magB=sqrt(B(1)^2+B(2)^2+B(3)^2) +theta=acosd(AdotB/(magA*magB)) //degrees + +//Result +clc +printf("The product of both the vectors is %f\n",AdotB) +printf("The angle between them is %f degrees",theta) diff --git a/3204/CH26/EX26.6/Ex26_6.sce b/3204/CH26/EX26.6/Ex26_6.sce new file mode 100644 index 000000000..c2b1e718e --- /dev/null +++ b/3204/CH26/EX26.6/Ex26_6.sce @@ -0,0 +1,13 @@ +// 1 APPENDIX. Ex no 6. Page no 643 +//initilization of variables +A=[4,-3,1] +P=[2,3,-1] +Q=[-2,-4,3] +//Calculations +B=[Q(1)-P(1),Q(2)-P(2),Q(3)-P(3)] +AdotB=A(1)*B(1)+A(2)*B(2)+A(3)*B(3) +magB=sqrt(B(1)^2+B(2)^2+B(3)^2) +Acostheta=AdotB/magB +//Result +clc +printf("The value of A.costheta is %f",Acostheta) diff --git a/3204/CH26/EX26.7/Ex26_7.sce b/3204/CH26/EX26.7/Ex26_7.sce new file mode 100644 index 000000000..82d92ed38 --- /dev/null +++ b/3204/CH26/EX26.7/Ex26_7.sce @@ -0,0 +1,12 @@ +// 1 APPENDIX. Ex no 7. Page no 643 +//Initilization of variables +F=[5,10,-15] +a=[1,0,3] +b=[3,-1,-6] +//Calculations +d=b-a +work=F.*d +Work=work(1)+work(2)+work(3) +//Result +clc +printf("The work done is %f units",Work) diff --git a/3204/CH26/EX26.8/Ex26_8.sce b/3204/CH26/EX26.8/Ex26_8.sce new file mode 100644 index 000000000..948f16730 --- /dev/null +++ b/3204/CH26/EX26.8/Ex26_8.sce @@ -0,0 +1,16 @@ +// 1 APPENDIX. Ex no 8. Page no 644 +//initilization of variables +A=[2,-6,-3] +B=[4,3,-1] +//Calculations +AcrossB=[(A(2)*B(3)-B(2)*A(3)),A(3)*B(1)-A(1)*B(3),A(1)*B(2)-A(2)*B(1)] +mag=sqrt(AcrossB(1)^2+AcrossB(2)^2+AcrossB(3)^2) +n=AcrossB/mag +magA=sqrt(A(1)^2+A(2)^2+A(3)^2) +magB=sqrt(B(1)^2+B(2)^2+B(3)^2) +theta=asind(mag/(magA*magB)) +//Result +clc +printf("The cross prcoduct of the two vectors is %fi %fj+%fk\n",AcrossB(1),AcrossB(2),AcrossB(3)) // the answer for j part is wrong in textbook +printf("The angle between the two is %f degrees",theta) +// Only 1 value for theta has been solved. Ref textbook for the other value diff --git a/3204/CH26/EX26.9/Ex26_9.sce b/3204/CH26/EX26.9/Ex26_9.sce new file mode 100644 index 000000000..5d26dd827 --- /dev/null +++ b/3204/CH26/EX26.9/Ex26_9.sce @@ -0,0 +1,28 @@ +// 1 APPENDIX. Ex no 9. Page no 645
+// Initilization of Variable
+// NOTE:Some Notation has been change to avoid conflict
+// Points As martices
+A=[0,1,2]
+B=[1,3,-2]
+P=[3,6,4]
+a_s=2 // Angular speed in rad/s
+
+// Calculations
+C=[B(1)-A(1),B(2)-A(2),B(3)-A(3)]
+magC=(C(1)^2+C(2)^2+C(3)^2)^0.5 // Magnitude of the Vector C
+// Unit vector
+C_unit=[C(1)/magC,C(2)/magC,C(3)/magC] // Unit vector
+// Position Vector
+r=[P(1)-A(1),P(2)-A(2),P(3)-A(3)]
+// Velocity Vector
+// Calculating the cross product as,
+V=[(C(2)*r(3)-C(3)*r(2)),(C(3)*r(1)-C(1)*r(3)),(C(1)*r(2)-C(2)*r(1))]
+// Vector notation
+V_n=(a_s/magC)*[V(1),V(2),V(3)]
+// Velocity Magnitude
+magV=sqrt(V(1)^2+V(2)^2+V(3)^2)
+v=(a_s/magC)*magV
+// Result
+clc
+printf("The vector notation of velocity is %fi %fj %fk \n",V_n(1),V_n(2),V_n(3))
+printf("The magnitude of the Velocity Vector is %f \n",v)
diff --git a/3204/CH3/EX3.1/Ex3_1.sce b/3204/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..bbb9837f1 --- /dev/null +++ b/3204/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,12 @@ +//Initilization of variables
+W=1000 //N
+Lab=1 //m
+Lac=0.6 //m
+theta=60 //degree //angle made by the beam with the horizontal
+//Calculations
+Q=(W*Lac*cosd(theta))/(Lab*cosd(theta)) //N // from eq'n 2
+P=W-Q //N // from eq'n 1
+//Results
+clc
+printf('The load taken by man P is %f N \n',P)
+printf('The load taken by man Q is %f N \n',Q)
diff --git a/3204/CH3/EX3.10/Ex3_10.sce b/3204/CH3/EX3.10/Ex3_10.sce new file mode 100644 index 000000000..a40f78aab --- /dev/null +++ b/3204/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,13 @@ +//Initilization of variables
+w=2000 //N/m
+Lab=3 //m
+//Calculations
+W=w*Lab/2 //N// Area under the curve
+Lac=(2/3)*Lab //m//centroid of the triangular load system
+Rb=(W*Lac)/Lab //N //sum of moment at A
+Ra=W-Rb //N
+//Results
+clc
+printf('The resultant of the distibuted load lies at %f m \n',Lac)
+printf('The reaction at support A is %f N \n',Ra)
+printf('The reaction at support B is %f N \n',Rb)
diff --git a/3204/CH3/EX3.11/Ex3_11.sce b/3204/CH3/EX3.11/Ex3_11.sce new file mode 100644 index 000000000..b08f1cbaf --- /dev/null +++ b/3204/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,13 @@ +//Initiization of variables
+w=1500 //N/m
+x=4 //m
+L=4 //m
+//Calculations
+k=x^2/w //m^3/N
+//Solving the intergral we get
+W=L^3/(3*k) //N
+x_bar=L^4/(4*k*W) //m
+//Result
+clc
+printf("The resultant is %f N and the line of action of the force is %f m",W,x_bar)
+
diff --git a/3204/CH3/EX3.12/Ex3_12.sce b/3204/CH3/EX3.12/Ex3_12.sce new file mode 100644 index 000000000..f401d9f62 --- /dev/null +++ b/3204/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+w1=1.5 //kN/m // intensity of varying load at the starting point of the beam
+w2=4.5 //kN/m // intensity of varying load at the end of the beam
+l=6 //m // ength of the beam
+// Calculations
+// The varying load distribution is divided into a rectangle and a right angled triangle
+W1=w1*l //kN // where W1 is the area of the load diagram(rectangle ABED)
+x1=l/2 //m // centroid of the rectangular load system
+W2=(w2-w1)*l/2 //kN // where W1 is the area of the load diagram(triangle DCE)
+x2=2*l/3 //m // centroid of the triangular load system
+W=W1+W2 //kN // W is the resultant
+x=((W1*x1)+(W2*x2))/W //m // where x is the distance where the resultant lies
+//Results
+clc
+printf('The resultant of the distributed load system is %f kN \n',W)
+printf('The line of action of the resulting load is %f m \n',x)
diff --git a/3204/CH3/EX3.13/Ex3_13.sce b/3204/CH3/EX3.13/Ex3_13.sce new file mode 100644 index 000000000..48bd7f9ab --- /dev/null +++ b/3204/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,25 @@ +// Initiization of variables
+W1=10 //kN //point load acting at D
+W2=20 //kN // point load acting at C at an angle of 30 degree
+W3=5 //kN/m // intensity of udl acting on span EB of 4m
+W4=10 //kN/m // intensity of varying load acting on span BC of 3m
+M=25 //kN-m // moment acting at E
+theta=30 //degree // angle made by 20 kN load with the beam
+Lad=2 //m
+Leb=4 //m
+Laf=6 //m //distance between the resultant of W3 & point A
+Lac=11 //m
+Lag=9 //m //distance between the resultant of W4 and point A
+Lbc=3 //m
+Lab=8 //m
+// Calculations
+Xa=20*cosd(theta) //kN // sum Fx=0
+Rb=((W1*Lad)+(-M)+(W3*Leb*Laf)+(W2*sind(theta)*Lac)+((W4*Lbc*Lag)/2))/Lab //kN // taking moment at A
+Ya=W1+(W2*sind(theta))+(W3*Leb)+(W4*Lbc/2)-Rb //kN // sum Fy=0
+Ra=sqrt(Xa^2+Ya^2) //kN // resultant at A
+//Results
+clc
+printf('The horizontal reaction at A i.e Xa is %f kN \n',Xa)
+printf('The vertical reaction at A i.e Ya is %f kN \n',Ya)
+printf('The reaction at A i.e Ra is %f kN \n',Ra)
+printf('The reaction at B i.e Rb is %f kN \n',Rb)
diff --git a/3204/CH3/EX3.14/Ex3_14.sce b/3204/CH3/EX3.14/Ex3_14.sce new file mode 100644 index 000000000..888bd92b1 --- /dev/null +++ b/3204/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,12 @@ +// Initilization of variables
+h=4 //m //height of the dam wall
+rho_w=1000 // kg/m^3 // density of water
+rho_c=2400 // kg/m^3 // density of concrete
+g=9.81 // m/s^2
+// Calculations
+P=(rho_w*g*h^2)/2 // The resultant force due to water pressure per unit length of the dam
+x=(2/3)*h //m // distance at which the resutant of the triangular load acts
+b=sqrt((2*P*h)/(3*h*rho_c*g)) // m // eq'n required to find the minimum width of the dam
+// Results
+clc
+printf('The minimum width which is to be provided to the dam to prevent overturning about point B is %f m \n',b)
diff --git a/3204/CH3/EX3.2/Ex3_2.sce b/3204/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..c202bafa8 --- /dev/null +++ b/3204/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,12 @@ +//Initilization of variables
+F=1000 //N
+Lab=1 //m
+Lbc=0.25 //m
+Lac=1.25 //m
+//Calculations
+Rb=(F*Lac)/Lab //N // from eq'n 2
+Ra=Rb-F //N // fom eq'n 1
+//Results
+clc
+printf('The reaction (downwards)at support A is %f N \n',Ra)
+printf('The reaction (upwards)at support B is %f N \n',Rb)
diff --git a/3204/CH3/EX3.3/Ex3_3.sce b/3204/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..2425806d8 --- /dev/null +++ b/3204/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,15 @@ +//Inilitization of variables
+Lab=12 //m
+Mc=40 //kN-m
+Md=10 //kN-m
+Me=20 //kN-m
+Fe=20 //kN //force acting at point E
+//Calculations
+Xa=-(Fe) //kN //take sum Fx=0
+Rb=(Md+Me-Mc)/Lab //N //take moment at A
+Ya=-Rb //N //take sum Fy=0
+//Results
+clc
+printf('The vertical reaction (upwards) at A is %f kN \n',Ya)
+printf('The horizontal reaction (towards A) is %f kN \n',Xa)
+printf('The reaction (downwards) at B is %f kN \n',Rb)
diff --git a/3204/CH3/EX3.5/Ex3_5.sce b/3204/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..caa984962 --- /dev/null +++ b/3204/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,22 @@ +//Initilization of variables
+W=1000 //N
+Lad=7.5 //m
+Lae=1.5 //m
+La1=3.75 //m //distance of 1st 1000N load from pt A
+La2=5 //m //distance of 2nd 1000N load from pt A
+La3=6 //m // distance of 3rd 1000N load from pt A
+// Calculations (part1)
+//using matrix to solve the given eqn's 1 & 2
+A=[1 -2.5;3.5 -5]
+B=[1000;7250]
+C=inv(A)*B
+//Calculations (part 2)
+//Consider combined F.B.D of beams AB,BC &CD. Take moment at A
+Re=((W*La1)+(W*La2)+(W*La3)+(C(2)*Lad)-(C(1)*La3))/Lae //N
+Ra=C(2)-Re-C(1)+(3*W) //N //Taking sum of forces in Y direction
+//Results
+clc
+printf('The reaction at F i.e Rf is %f N \n',C(1))
+printf('The reaction at D i.e Rd is %f N \n',C(2))
+printf('The reaction at pt E i.e Re is %f N \n',Re)
+printf('The reaction at pt A i.e Ra is %f N \n',Ra) //acting vertically downwards
diff --git a/3204/CH3/EX3.6/Ex3_6.sce b/3204/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..857b1acc7 --- /dev/null +++ b/3204/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,10 @@ +// Initilization of variables
+W=100 // N //force acting at D
+AB=50 // N // weight of bar ab
+CD=50 // N // weight of bar cd
+// Calculations
+// From the derived expression the value of the angle is given as,
+theta=atand(5/17.5) //degrees
+// Results
+clc
+printf('The angle theta is %f degrees \n',theta)
diff --git a/3204/CH3/EX3.7/Ex3_7.sce b/3204/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..df53e6556 --- /dev/null +++ b/3204/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,15 @@ +//Initilization of variables
+Ws=2 //kN //weight of scooter
+Wd=0.5 //kN //weight of driver
+Lab=1 //m
+Led=0.8 //m
+Leg=0.1 //m
+//Calculations
+Rc=((2*Leg)+(Wd*Led))/Lab //kN //take moment at E
+Ra=(2+Wd-Rc)/2 // kN // as Ra=Rb,(Ra+Rb=2*Ra)
+Rb=Ra // kN
+//Results
+clc
+printf('The reaction at wheel A is %f kN \n',Ra)
+printf('The reaction at wheel B is %f kN \n',Rb)
+printf('The reaction at wheel C is %f kN \n',Rc)
diff --git a/3204/CH3/EX3.8/Ex3_8.sce b/3204/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..8f4e82ed1 --- /dev/null +++ b/3204/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,25 @@ +//Initilization of variables
+W1=15 //N //up
+W2=60 //N //down
+W3=10 //N //up
+W4=25 //N //down
+Lab=1.2 //m
+Lac=0.4 //m
+Lcd=0.3 //m
+Ldb=0.5 //m
+Lad=0.7 //m
+Leb=0.417 //m //Leb=Lab-x
+//Calculations
+//(a) A single force
+Ry=W1-W2+W3-W4 //N //take sum Fy=0
+x=((-W2*Lac)+(W3*Lad)-(W4*Lab))/(Ry) //m
+// (b) Single force moment at A
+Ma=(Ry*x) //N-m
+// Single force moment at B
+Mb=W2*Leb //N-m
+//Results
+clc
+printf('The reaction for single force (a) is %f N \n',Ry)
+printf('The distance of Ry from A is %f m \n',x)
+printf('The moment at A is %f N-m \n',Ma)
+printf('The moment at B is %f N-m \n',Mb)
diff --git a/3204/CH3/EX3.9/Ex3_9.sce b/3204/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..faabd512e --- /dev/null +++ b/3204/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,20 @@ +//Initilization of variables
+Ra=5000 //N
+Ma=10000 //Nm
+alpha=60 //degree //angle made by T1 with the pole
+beta=45 //degree //angle made by T2 with the pole
+theta=30 //degree //angle made by T3 with the pole
+Lab=6 //m
+Lac=1.5 //m
+Lcb=4.5 //m
+//Calculations
+T3=Ma/(4.5*sind(theta)) //N //take moment at B
+// Now we use matrix to solve eqn's 1 & 2 simultaneously,
+A=[-0.707 0.8666;0.707 0.5]
+B=[2222.2;8848.8]
+C=inv(A)*B
+//Results
+clc
+printf('Tension in wire 1 i.e T1 is %f N \n',C(2))
+printf('Tension in wire 2 i.e T2 is %f N \n',C(1))
+printf('Tension in wire 3 i.e T3 is %f N \n',T3)
diff --git a/3204/CH4/EX4.8/Ex4_8.sce b/3204/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..0d53855aa --- /dev/null +++ b/3204/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+b1=20 //cm // width of top flange
+t1=5 //cm // thickness of top flange
+b2=5 //cm // width of web
+t2=15 //cm // thickness or height of the web
+b3=30 //cm // width of bottom flange
+t3=5 //cm // thickness of bottom flange
+// Calculations
+A1=b1*t1 //cm^2 // area of bottom flange
+A2=b2*t2 //cm^2 // area of the web
+A3=b3*t3 //cm^2 // area of top flange
+y1=t3+t2+(t1/2) //cm // y co-ordinate of the centroid of top flange
+y2=t3+(t2/2) //cm // y co-ordinate of the centroid of the web
+y3=t3/2 //cm // y co-ordinate of the centroid of the bottom flange
+y_c=((A1*y1)+(A2*y2)+(A3*y3))/(A1+A2+A3) //cm // where y_c is the centroid of the un-symmetrycal I-section
+// Results
+clc
+printf('The centroid of the un equal I-section is %f cm \n',y_c)
diff --git a/3204/CH4/EX4.9/Ex4_9.sce b/3204/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..99b137e4f --- /dev/null +++ b/3204/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,27 @@ +// Initilization of variables
+// The given section is Z-section which is un-symmetrycal about both the axis
+b1=20 //cm // width of bottom flange
+t1=5 //cm // thickness of the bottom flange
+b2=2.5 //cm // thickness of the web of the flange
+t2=15 //cm // depth of the web
+b3=10 //cm // width of the top flange
+t3=2.5 //cm // thickness of the top flange
+// Calculations
+// Respective areas
+A1=b1*t1 // cm^2 // area of the bottom flange
+A2=b2*t2 // cm^2 // area of the web
+A3=b3*t3 // cm^2 // area of the top-flange
+// first we calculate the x co-ordinate of the centroid
+x1=b3-b2+(b1/2) //cm // for the bottom flange
+x2=b3-(b2/2) //cm // for the web
+x3=b3/2 //cm // for the top flange
+x_c=((A1*x1)+(A2*x2)+(A3*x3))/(A1+A2+A3) //cm
+// secondly we calculate the y co-ordinate of the centroid
+y1=t1/2 //cm // for the bottom flange
+y2=t1+(t2/2) //cm // for the web
+y3=t1+t2+(t3/2) //cm // for the top flange
+y_c=((A1*y1)+(A2*y2)+(A3*y3))/(A1+A2+A3) // cm
+// Results
+clc
+printf('The centroid of the cross-sectional area of a Z-section about x-axis is %f cm \n',x_c)
+printf('The centroid of the cross-sectional area of a Z-section about y-axis is %f cm \n',y_c)
diff --git a/3204/CH5/EX5.2/Ex5_2.sce b/3204/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..921cdc9d8 --- /dev/null +++ b/3204/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,14 @@ +//Inilization of variables
+W=2000 //N
+Lab=2 //m //length of the member from the vertical to the 1st load of 2000 N
+Lac=5 //m //length of the member from the vertical to the 2nd load of 2000 N
+Lpq=3.5 //m
+//Calculations
+Rq=((W*Lab)+(W*Lac))/Lpq //N //take moment abt. pt P
+Xp=Rq //N //sum Fx=0
+Yp=2*W //N //sum Fy=0
+Rp=sqrt(Xp^2+Yp^2) //N
+//Resuts
+clc
+printf('The reaction at P is %f N \n',Rp)
+printf('The reaction at Q is %f N \n',Rq)
diff --git a/3204/CH5/EX5.3/Ex5_3.sce b/3204/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..780c792c3 --- /dev/null +++ b/3204/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,20 @@ +//Initilization of vaiables
+W=25 //N // self weight of the ladder
+M=75 //N // weight of the man standing o the ladder
+theta=63.43 //degree // angle which the ladder makes with the horizontal
+alpha=30 //degree // angle made by the string with the horizontal
+Loa=2 //m // spacing between the wall and the ladder
+Lob=4 //m //length from the horizontal to the top of the ladder touching the wall(vertical)
+//Calculations
+//Using matrix to solve the simultaneous eqn's 3 & 4
+A=[2 -4;1 -0.577]
+B=[100;100]
+C=inv(A)*B
+//Results
+clc
+printf('The reaction at A i.e Ra is %f N \n',C(1))
+printf('The reaction at B i.e Rb is %f N \n',C(2))
+//Calculations
+T=C(2)/cosd(alpha) //N // from (eqn 1)
+//Results
+printf('The required tension in the string is %f N \n',T)
diff --git a/3204/CH5/EX5.4/Ex5_4.sce b/3204/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..6624e8ed5 --- /dev/null +++ b/3204/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,17 @@ +//Initilization of variables
+W=100 //N
+theta=60 //degree //angle made by the ladder with the horizontal
+alpha=30 //degree //angle made by the ladder with the vertical wall
+Lob=4 //m // length from the horizontal to the top of the ladder touching the wall(vertical)
+Lcd=2 //m // length from the horizontal to the centre of the ladder where the man stands
+//Calculations
+Lab=Lob*secd(alpha) //m //length of the ladder
+Lad=Lcd*tand(alpha) //m
+Rb=(W*Lad)/Lab //N //take moment at A
+Xa=Rb*sind(theta) //N // From eq'n 1
+Ya=W+Rb*cosd(theta) //N From eq'n 2
+//Results
+clc
+printf('The reaction at B i.e Rb is %f N \n',Rb)
+printf('The horizontal reaction at A i.e Xa is %f N \n',Xa)
+printf('The vertical reaction at A i.e Ya is %f N \n',Ya)
diff --git a/3204/CH5/EX5.5/Ex5_5.sce b/3204/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..b1297171f --- /dev/null +++ b/3204/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,16 @@ +//Initilization of variables
+W=100 //N //self weight of the man
+alpha=30 //degree // angle made by the ladder with the wall
+Lob=4 //m // length from the horizontal to the top of the ladder touching the wall(vertical)
+Lcd=2 //m
+//Calculations
+// using the equiblirium equations
+Ya=W //N // From eq'n 2
+Lad=Lcd*tand(alpha) //m //Lad is the distance fom pt A to the point where the line from the cg intersects the horizontal
+Rb=(W*Lad)/Lob //N // Taking sum of moment abt A
+Xa=Rb //N // From eq'n 1
+//Results
+clc
+printf('The horizontal reaction at A i.e Xa is %f N \n',Xa)
+printf('The vertical reaction at A i.e Ya is %f N \n',Ya)
+printf('The reaction at B i.e Rb is %f N \n',Rb)
diff --git a/3204/CH5/EX5.6/Ex5_6.sce b/3204/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..dbe1e5ba2 --- /dev/null +++ b/3204/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,18 @@ +//Initilization of variables
+d=0.09 //m //diametre of the right circular cylinder
+h=0.12 //m //height of the cyinder
+W=10 //N // self weight of the bar
+l=0.24 //m //length of the bar
+//Calculations
+theta=atand(h/d) // angle which the bar makes with the horizontal
+Lad=sqrt(d^2+h^2) //m // Lad is the length of the bar from point A to point B
+Rd=(W*h*cosd(theta))/Lad //N // Taking moment at A
+Xa=Rd*sind(theta) //N // sum Fx=0.... From eq'n 1
+Ya=W-(Rd*cosd(theta)) //N // sum Fy=0..... From eq'n 2
+Ra=sqrt(Xa^2+Ya^2) //resultant of Xa & Ya
+//Results
+clc
+printf('The horizontal reaction at A i.e Xa is %f N \n',Xa)
+printf('The vertical reaction at A i.e Ya is %f N \n',Ya)
+printf('Therefore the reaction at A i.e Ra is %f N \n',Ra)
+printf('The reaction at D i.e Rd is %f N \n',Rd)
diff --git a/3204/CH6/EX6.1/Ex6_1.sce b/3204/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..5a5e56d2c --- /dev/null +++ b/3204/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,21 @@ +// Initilization of variables
+m=5 //kg // mass of the bock
+g=9.81 // m/s^2 // acceleration due to gravity
+theta=15 // degree // angle made by the forces (P1 & P2) with the horizontal of the block
+mu=0.4 //coefficient of static friction
+//Calculations
+// Case 1. Where P1 is the force required to just pull the bock
+// Solving eqn's 1 & 2 using matrix
+A=[cosd(theta) -mu;sind(theta) 1]
+B=[0;(m*g)]
+C=inv(A)*B
+// Calculations
+// Case 2. Where P2 is the force required to push the block
+// Solving eqn's 1 & 2 using matrix
+P=[-cosd(theta) mu;-sind(theta) 1]
+Q=[0;(m*g)]
+R=inv(P)*Q
+// Results
+clc
+printf('The required pull force P1 is %f N \n',C(1))
+printf('The required push force P2 is %f N \n',R(1))
diff --git a/3204/CH6/EX6.13/Ex6_13.sce b/3204/CH6/EX6.13/Ex6_13.sce new file mode 100644 index 000000000..b52fbcd06 --- /dev/null +++ b/3204/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,15 @@ +// Initilization of variables
+P=100 //N // force acting at 0.2 m from A
+Q=200 //N // force acting at any distance x from B
+l=1 //m // length of the bar
+theta=45 //degree //angle made by the normal reaction at A&B with horizontal
+// Calculations
+// solving eqn's 1 & 2 using matrix for Ra & Rb,
+A=[1 -1;sind(theta) sind(theta)]
+B=[0;(P+Q)]
+C=inv(A)*B
+// Now take moment about B
+x=((C(1)*l*sind(theta))-(P*(l-0.2)))/200 //m // here 0.2 is the distance where 100 N load lies from A
+// Results
+clc
+printf('The minimum value of x at which the load Q=200 N may be applied before slipping impends is %f m \n',x)
diff --git a/3204/CH6/EX6.4/Ex6_4.sce b/3204/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..0d7dc3297 --- /dev/null +++ b/3204/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,11 @@ +// Initilization of variables
+W1=50 // N // weight of the first block
+W2=50 // N // weight of the second block
+mu_1=0.3 // coefficient of friction between the inclined plane and W1
+mu_2=0.2 // coefficient of friction between the inclined plane and W2
+// Calculations
+// On adding eq'ns 1&3 and substuting the values of N1 & N2 from eqn's 2&4 in this and on solving for alpha we get,
+alpha=atand(((mu_1*W1)+(mu_2*W2))/(W1+W2)) // degrees
+// Results
+clc
+printf('The inclination of the plane is %f degree \n',alpha)
diff --git a/3204/CH6/EX6.7/Ex6_7.sce b/3204/CH6/EX6.7/Ex6_7.sce new file mode 100644 index 000000000..4ab756816 --- /dev/null +++ b/3204/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,10 @@ +// Initilization of variables
+M=2000 // kg // mass of the car
+mu=0.3 // coefficient of static friction between the tyre and the road
+g=9.81 // m/s^2 // acc. due to gravity
+// Calculations
+// Divide eqn 1 by eqn 2, We get
+theta=atand(mu) //degree
+// Results
+clc
+printf('The angle of inclination is %f degree \n',theta)
diff --git a/3204/CH6/EX6.9/Ex6_9.sce b/3204/CH6/EX6.9/Ex6_9.sce new file mode 100644 index 000000000..040ffb556 --- /dev/null +++ b/3204/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,26 @@ +// Initilization of variabes
+Wa=1000 //N // weight of block A
+Wb=500 //N // weight of block B
+theta=15 // degree // angle of the wedge
+mu=0.2 // coefficient of friction between the surfaces in contact
+phi=7.5 // degrees // used in case 2
+pie=3.14
+// Caculations
+// CASE (a)
+// consider the equilibrium of upper block A
+// rearranging eq'ns 1 &2 and solving them using matrix for N1 & N2
+A=[1 -0.4522;-0.2 0.914]
+B=[0;1000]
+C=inv(A)*B
+// Now consider the equilibrium of lower block B
+// From eq'n 4
+N3=Wb+(C(2)*cosd(theta))-(mu*C(2)*sind(theta)) //N
+// Now from eq'n 3
+P=(mu*N3)+(mu*C(2)*cosd(theta))+(C(2)*sind(theta)) // N
+// CASE (b)
+// The eq'n for required coefficient for the wedge to be self locking is,
+mu_req=(theta*pie)/(360) // multiplying with (pie/180) to convert it into radians
+// Results
+clc
+printf('The minimum horizontal force (P) which should be applied to raise the block is %f N \n',P)
+printf('The required coefficient for the wedge to be self locking is %f \n',mu_req)
diff --git a/3204/CH7/EX7.1/Ex7_1.sce b/3204/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..eb8b6159a --- /dev/null +++ b/3204/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,18 @@ +// Initilization of variables
+d1=24 // cm // diameter of larger pulley
+d2=12 // cm // diameter of smaller pulley
+d=30 //cm // seperation betweem 1st & the 2nd pulley
+pie=3.14
+// Calcuations
+r1=d1/2 //cm // radius of 1st pulley
+r2=d2/2 //cm // radius of 2nd pulley
+theta=asind((r1-r2)/d) //degrees
+// Angle of lap
+beta_1=180+(2*theta) //degree // for larger pulley
+beta_2=180-(2*theta) //degree //for smaller pulley
+L=pie*(r1+r2)+(2*d)+((r1-r2)^2/d) //cm // Length of the belt
+// Results
+clc
+printf('The angle of lap for the larger pulley is %f degree \n',beta_1)
+printf('The angle of lap for the smaller pulley is %f degree \n',beta_2)
+printf('The length of pulley required is %f cm \n',L)
diff --git a/3204/CH7/EX7.10/Ex7_10.sce b/3204/CH7/EX7.10/Ex7_10.sce new file mode 100644 index 000000000..192db4cbb --- /dev/null +++ b/3204/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,21 @@ +// Initilization of variables
+p=0.0125 // m // pitch of screw
+d=0.1 //m // diameter of the screw
+r=0.05 //m // radius of the screw
+l=0.5 //m // length of the lever
+W=50 //kN // load on the lever
+mu=0.20 // coefficient of friction
+pie=3.14 //constant
+// Calculations
+theta=atand(p/(2*pie*r)) //degree // theta is the Helix angle
+phi=atand(mu) // degree // phi is the angle of friction
+// Taking the leverage due to handle into account,force F1 required is,
+F1=(W*(tand(theta+phi)))*(r/l) //kN
+// To lower the load
+F2=(W*(tand(theta-phi)))*(r/l) //kN // -ve sign of F2 indicates force required is in opposite sense
+E=(tand(theta)/tand(theta+phi))*100 // % // here E=eata=efficiency in %
+// Results
+clc
+printf('The force required (i.e F1) to raise the weight is %f kN \n',F1)
+printf('The force required (i.e F2) to lower the weight is %f kN \n',F2)
+printf('The efficiency of the jack is %f percent \n',E)
diff --git a/3204/CH7/EX7.11/Ex7_11.sce b/3204/CH7/EX7.11/Ex7_11.sce new file mode 100644 index 000000000..c2b601aec --- /dev/null +++ b/3204/CH7/EX7.11/Ex7_11.sce @@ -0,0 +1,12 @@ +// Initilization of variabes
+P=20000 //N //Weight of the shaft
+D=0.30 //m //diameter of the shaft
+R=0.15 //m //radius of the shaft
+mu=0.12 // coefficient of friction
+// Calculations
+// Friction torque T is given by formulae,
+T=(2/3)*P*R*mu //N-m
+M=T //N-m
+// Results
+clc
+printf('The frictional torque is %f N-m \n',M)
diff --git a/3204/CH7/EX7.2/Ex7_2.sce b/3204/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..781b1f118 --- /dev/null +++ b/3204/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,16 @@ +// Initilization of variables
+d1=0.6 //m // diameter of larger pulley
+d2=0.3 //m // diameter of smaller pulley
+d=3.5 //m // separation between the pulleys
+pie=3.14
+// Calculations
+r1=d1/2 //m // radius of larger pulley
+r2=d2/2 //m // radius of smaller pulley
+theta=asind((r1+r2)/d) //degree
+// Angle of lap for both the pulleys is same, i.e
+beta=180+(2*theta) // degree
+L=((pie*(r1+r2))+(2*d)+((r1-r2)^2/d)) //cm // Length of the belt
+// Results
+clc
+printf('The angle of lap for the pulley is %f degree \n',beta)
+printf('The length of pulley required is %f m \n',L)
diff --git a/3204/CH7/EX7.4/Ex7_4.sce b/3204/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..b1efe5796 --- /dev/null +++ b/3204/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+W1=1000 //N
+mu=0.25 //coefficient of friction
+pie=3.14
+beta=pie
+T1=W1 // Tension in the 1st belt carrying W1
+e=2.718 //constant
+// Calculations
+T2=T1/(e^(mu*beta)) //N // Tension in the 2nd belt
+W2=T2 //N
+// Results
+clc
+printf('The minimum weight W2 to keep W1 in equilibrium is %f N \n',W2)
diff --git a/3204/CH7/EX7.5/Ex7_5.sce b/3204/CH7/EX7.5/Ex7_5.sce new file mode 100644 index 000000000..8d1c140ff --- /dev/null +++ b/3204/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,20 @@ +// Initilization of variables
+mu=0.5 // coefficient of friction between the belt and the wheel
+W=100 //N
+theta=45 //degree
+e=2.718
+Lac=0.75 //m // ength of the lever
+Lab=0.25 //m
+Lbc=0.50 //m
+r=0.25 //m
+pie=3.14 // constant
+// Calculations
+beta=((180+theta)*pie)/180 // radian // angle of lap
+// from eq'n 2
+T1=(W*Lbc)/Lab //N
+T2=T1/(e^(mu*beta)) //N // from eq'n 1
+// consider the F.B.D of the pulley and take moment about its center, we get Braking Moment (M)
+M=r*(T1-T2) //N-m
+// Results
+clc
+printf('The braking moment (M) exerted by the vertical weight W is %f N-m \n',M)
diff --git a/3204/CH7/EX7.6/Ex7_6.sce b/3204/CH7/EX7.6/Ex7_6.sce new file mode 100644 index 000000000..3f1cca717 --- /dev/null +++ b/3204/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,14 @@ +// Initiization of variables
+W= 1000 //N // or 1kN
+mu=0.3 // coefficient of friction between the rope and the cylinder
+e=2.718 // constant
+pie=3.14 // constant
+alpha=90 // degree // since 2*alpha=180 egree
+// Calculations
+beta=2*pie*3 // radian // for 3 turn of the rope
+// Here T1 is the larger tension in that section of the rope which is about to slip
+T1=W //N
+F=W/e^(mu*(1/(sind(alpha)))*(beta)) //N // Here T2=F
+// Results
+clc
+printf('The force required to suport the weight of 1000 N i.e 1kN is %f N \n',F)
diff --git a/3204/CH7/EX7.7/Ex7_7.sce b/3204/CH7/EX7.7/Ex7_7.sce new file mode 100644 index 000000000..fe01df722 --- /dev/null +++ b/3204/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,21 @@ +// Initilization of variables
+Pw=50 //kW
+T_max=1500 //N
+v=10 // m/s // velocity of rope
+w=4 // N/m // weight of rope
+mu=0.2 // coefficient of friction
+g=9.81 // m/s^2 // acceleration due to gravity
+e=2.718 // constant
+pie=3.14 // constant
+alpha=30 // degree // since 2*alpha=60
+// Calcuations
+beta=pie // radian // angle of lap
+T_e=(w*v^2)/g // N // where T_e is the centrifugal tension
+T1=(T_max)-(T_e) //N
+T2=T1/(e^(mu*(1/sind(alpha))*(beta))) //N // From eq'n T1/T2=e^(mu*cosec(alpha)*beta)
+P=(T1-T2)*v*(10^-3) //kW // power transmitted by a single rope
+N=Pw/P // Number of ropes required
+// Results
+clc
+printf('The number of ropes required to transmit 50 kW is %f Nos \n',N)
+// approx no of ropes is 5
diff --git a/3204/CH7/EX7.8/Ex7_8.sce b/3204/CH7/EX7.8/Ex7_8.sce new file mode 100644 index 000000000..694a12be4 --- /dev/null +++ b/3204/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,24 @@ +// Initilization of variables
+d1=0.45 //m // diameter of larger pulley
+d2=0.20 //m // diameter of smaller pulley
+d=1.95 //m // separation between the pulley's
+T_max=1000 //N // or 1kN which is the maximum permissible tension
+mu=0.20 // coefficient of friction
+N=100 // r.p.m // speed of larger pulley
+e=2.718 // constant
+pie=3.14 // constant
+T_e=0 //N // as the data for calculating T_e is not given we assume T_e=0
+// Calculations
+r1=d1/2 //m // radius of larger pulley
+r2=d2/2 //m // radius of smaller pulley
+theta=asind((r1+r2)/d) // degree
+// for cross drive the angle of lap for both the pulleys is same
+beta=((180+(2*(theta)))*pie)/180 //radian
+T1=T_max-T_e //N
+T2=T1/(e^(mu*(beta))) //N // from formulae, T1/T2=e^(mu*beta)
+v=((2*pie)*N*r1)/60 // m/s // where v=velocity of belt which is given as, v=wr=2*pie*N*r/60
+P=(T1-T2)*v*(10^-3) //kW // Power
+// Results
+clc
+printf('The power transmitted by the cross belt drive is %f kW \n',P)
+// the approx answer is 1.3 kW The answer given in the book (i.e 1.81kW) is wrong.
diff --git a/3204/CH7/EX7.9/Ex7_9.sce b/3204/CH7/EX7.9/Ex7_9.sce new file mode 100644 index 000000000..d6441877b --- /dev/null +++ b/3204/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,32 @@ +// Initilization of variabes
+b=0.1 //m //width of the belt
+t=0.008 //m //thickness of the belt
+v=26.67 // m/s // belt speed
+pie=3.14 // constant
+beta=165 // radian // angle of lap for the smaller belt
+mu=0.3 // coefficient of friction
+sigma_max=2 // MN/m^2 // maximum permissible stress in the belt
+m=0.9 // kg/m // mass of the belt
+g=9.81 // m/s^2
+e=2.718 // constant
+// Calculations
+A=b*t // m^2 // cross-sectional area of the belt
+T_e=m*v^2 // N // where T_e is the Centrifugal tension
+T_max=(sigma_max)*(A)*(10^6) // N // maximum tension in the belt
+T1=(T_max)-(T_e) // N
+T2=T1/(e^((mu*pie*beta)/180)) //N // from formulae T1/T2=e^(mu*beta)
+P=(T1-T2)*v*(10^-3) //kW // Power transmitted
+T_o=(T1+T2)/2 // N // Initial tension
+// Now calculations to transmit maximum power
+Te=T_max/3 // N // max tension
+u=sqrt(T_max/(3*m)) // m/s // belt speed for max power
+T_1=T_max-Te // N // T1 for case 2
+T_2=T_1/(e^((mu*pie*beta)/180)) // N
+P_max=(T_1-T_2)*u*(10^-3) // kW // Max power transmitted
+// Results
+clc
+printf('The initial power transmitted is %f kW \n',P)
+printf('The initial tension in the belt is %f N \n',T_o)
+printf('The maximum power that can be transmitted is %f kW \n',P_max)
+printf('The maximum power is transmitted at a belt speed of %f m/s \n',u)
+
diff --git a/3204/CH8/EX8.1/Ex8_1.sce b/3204/CH8/EX8.1/Ex8_1.sce new file mode 100644 index 000000000..1b376902b --- /dev/null +++ b/3204/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,26 @@ +// Initilization of variables
+V.R=6 // Velocity ratio
+P=20 //N // Effort
+W=100 //N // Load lifted
+// Calculations
+//(a)
+P_actual=P //N
+W_actual=W //N
+M.A=W/P // where, M.A= Mechanical advantage
+E=(M.A/V.R)*100 //% // Where E= efficiency
+//(b)
+// Now ideal effort required is,
+P_ideal=W/V.R //N
+// Effort loss in friction is, (Le)
+Le=P_actual-P_ideal //N // Effort loss in friction
+//(c)
+// Ideal load lifted is,(W_ideal)
+W_ideal=P*V.R //N
+// Frictional load/resistance,
+F=W_ideal-W_actual // N
+// Results
+clc
+printf('(a) The efficiency of the machine is %f percent \n',E)
+printf('(b) The effort loss in friction of the machine is %f N \n',Le)
+printf('(c) The Frictional load of the machine is %f N \n',F)
+
diff --git a/3204/CH8/EX8.2/Ex8_2.sce b/3204/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..0829fe80a --- /dev/null +++ b/3204/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,17 @@ +// Initilization of variables
+V_r=20 // Velocity ratio
+// Refer the table given in the textbook for values of W,P,M.A & efficiency (eta)
+// Calculations
+// Part (a)- Realtionship between W & P
+// Here part a cannot be solved as it has variables which cannot be defined in Scilab. Ref.textbook for the solution
+// Part (b)- Graph between W & efficiency (eta)
+x=[0;30;40;50;60;70;80;90;100] // values for W // N
+y=[0;21.4;23.5;25;26.1;25.9;27.6;28.1;28.6] // values for efficiency (eta)
+subplot(221)
+xlabel("W (N)")
+ylabel("efficiency (eta)")
+plot(x,y)
+// Results
+clc
+printf('The graph is the solution')
+// The value of m is found by drawing straight line on the graph and by taking its slope. Ref textbook for the solution
diff --git a/3204/CH8/EX8.3/Ex8_3.sce b/3204/CH8/EX8.3/Ex8_3.sce new file mode 100644 index 000000000..c725ecce0 --- /dev/null +++ b/3204/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,21 @@ +// Calculations
+W_actual=1360 //N //Load lifted
+P_actual=100 //N // Effort
+n=4 // no of pulleys
+// Calculations
+// for 1st system of pulleys having 4 movable pulleys, Velocity ratio is
+V.R=2^(n) // Velocity Ratio
+// If the machine were to be ideal(frictionless)
+M.A=V.R // Here, M.A= mechanical advantage
+// For a load of 1360 N, ideal effort required is
+P_ideal=W_actual/V.R //N
+// Effort loss in friction is,
+P_friction=P_actual-P_ideal //N
+// For a effort of 100 N, ideal load lifted is,
+W_ideal=V.R*100 //N
+// Load lost in friction is,
+W_friction=W_ideal-W_actual // N
+// Results
+clc
+printf('(a) The effort wasted in friction is %f N \n',P_friction)
+printf('(b) The load wasted in friction is %f N \n',W_friction)
diff --git a/3204/CH8/EX8.4/Ex8_4.sce b/3204/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..d2c954bbc --- /dev/null +++ b/3204/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,13 @@ +// Initilization of variables
+W=1000 //N // Load to be lifted
+n=5 // no. of pulleys
+E=75 //% // Efficiency
+// Calculations
+// Velocity Ratio is given as,
+V.R=n
+// Mechanical Advantage (M.A) is,
+M.A=(E/100)*V.R // from formulae, Efficiency=E=M.A/V.R
+P=W/M.A //N // Effort required
+// Results
+clc
+printf('The effort required to lift the load of 1000 N is %f N \n',P)
diff --git a/3204/CH8/EX8.5/Ex8_5.sce b/3204/CH8/EX8.5/Ex8_5.sce new file mode 100644 index 000000000..ef37ead7e --- /dev/null +++ b/3204/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,23 @@ +// Initilization of variables
+W=2000 //N // Load to be raised
+l=0.70 //m // length of the handle
+d=0.05 //m // diameter of the screw
+p=0.01 //m // pitch of the screw
+mu=0.15 // coefficient of friction at the screw thread
+pie=3.14 //constant
+E=1 // efficiency
+// Calculations
+phi=atand(mu) //degree
+theta=atand(p/(pie*d)) //degree // where theta is the Helix angle
+// Force required at the circumference of the screw is,
+P=W*tand(theta+phi) // N //
+// Force required at the end of the handle is,
+F=(P*(d/2))/l //N
+// Force required (Ideal case)
+V.R=2*pie*l/p
+M.A=E*V.R // from formulae E=M.A/V,R
+P_ideal=W/M.A //N // From formulae, M.A=W/P
+// Results
+clc
+printf('The force required at the end of the handle is %f N \n',F)
+printf('The force required if the screw jack is considered to be an ideal machine is %f N \n',P_ideal)
diff --git a/3204/CH9/EX9.1/Ex9_1.sce b/3204/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..efd5e44d0 --- /dev/null +++ b/3204/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,33 @@ +// Initilization of variables
+W1=2000 //N // load at joint D of the truss
+W2=4000 //N // load at joint E of the truss
+Lac=6 //m // length of the tie
+Lab=3 //m
+Lbc=3 //m
+theta=60 //degree // interior angles of the truss
+// Calculations
+// Here A is simply supported & B is roller support. Now the SUPPORT REACTIONS are given as,
+Rc=((W1*(Lab/2))+(W2*(Lab+(Lbc/2))))/Lac //N // Taking moment at A
+Ra=W1+W2-Rc //N // Take sum Fy=0
+// ANALYSIS OF TRUSS BY METHOD OF JOINT
+// ASSUMPTION- we consider the,(1) Forces moving towards each other as +ve i.e TENSILE (T) & (2) Forces moving away from each other as -ve i.e COMPRESSIVE (C)
+// (1) JOINT A
+Fad=Ra/(sind(theta)) //N //(C) // Using eq'n 2
+Fab=Fad*cosd(theta) //N // (T) // Using eq'n 1
+// (2) JOINT C
+Fce=Rc/(sind(theta)) //N // (C) // Using eq'n 4
+Fcb=Fce*cosd(theta) //N // (T) // Using eq'n 3
+// (3) JOINT D
+Fdb=((Fad*sind(theta))-(W1))/sind(theta) //N // (T) // Using eq'n 6
+Fde=(Fdb*cosd(theta))+(Fad*cosd(theta)) //N // (C) // Using eq'n 5
+// (4) JOINT E
+Feb=((Fce*cosd(theta))-(Fde))/cosd(theta) //N // (C) // Using eq'n 7
+// Results
+clc
+printf('The Axial Force in member AD (Fad) is %f N \n',Fad)
+printf('The Axial Force in member AB (Fab) is %f N \n',Fab)
+printf('The Axial Force in member CE (Fce) is %f N \n',Fce)
+printf('The Axial Force in member CB (Fcb) is %f N \n',Fcb)
+printf('The Axial Force in member DB (Fdb) is %f N \n',Fdb)
+printf('The Axial Force in member DE (Fde)is %f N \n',Fde)
+printf('The Axial Force in member EB (Feb) is %f N \n',Feb)
diff --git a/3204/CH9/EX9.10/Ex9_10.sce b/3204/CH9/EX9.10/Ex9_10.sce new file mode 100644 index 000000000..7f0b7ad70 --- /dev/null +++ b/3204/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,26 @@ +// Initilization of variables
+W=24 // kN // Load acting at pt C
+Laf=12 // m // length of the tie beam
+l=4 // m// length of each member in the tie
+h=3 // m // height of the slings
+Lae=8 // m
+// Calculations
+s=sqrt((l^2)+(h^2)) // m // sloping length
+// From triangle BCD,
+theta=acosd(h/s)
+// SUPPORT REACTIONS
+Rf=(W*l)/Laf // kN // take moment at A
+Ra=W-Rf // kN // sum Fy=0
+// now pass a sectio mn through the truss and consider te equilibrium of the left hand portion
+Fce=(Ra*l)/h // kN (T) // Take moment at B
+Fbd=((W*l)-(Ra*Lae))/h // kN (C) // take moment at E
+Fbe=(Ra-W)/cosd(theta) // kN
+Fbd=(-Ra*l)/h // kN // take moment at C
+Fce=((Ra*Lae)-(W*l))/h // kN (T) // take moment at D
+Fcd=(W-Ra)/cosd(theta) // kN (T) // sum Fy=0
+// Resuts
+clc
+printf('(1) The axial force in the bar CE (Fce) is %f kN \n',Fce)
+printf('(2) The axial force in the bar BD (Fbd) is %f kN \n',Fbd)
+printf('(3) The axial force in the bar BE (Fbe) is %f kN \n',Fbe)
+printf('(4) The axial force in the bar CD (Fcd) is %f kN \n',Fcd)
diff --git a/3204/CH9/EX9.12/Ex9_12.sce b/3204/CH9/EX9.12/Ex9_12.sce new file mode 100644 index 000000000..d5ca013f9 --- /dev/null +++ b/3204/CH9/EX9.12/Ex9_12.sce @@ -0,0 +1,29 @@ +// Initilization of variables
+W1=4 // kN // load acting at a distance of 5 m from C
+W2=3 // kN // load acting at a distance of 7.5 m from C
+L=30 //m // distance AB
+L1=15 // dist AC
+L2=15 //m //dist BC
+l1=10 //m // distance between A and 4 kN load
+l2=22.5 //m // distance between A and 3 kN load
+// Calculations
+// (1) Reactions
+Yb=((W1*l1)+(W2*l2))/L // kN // Take moment at A
+Ya=W1+W2-Yb // kN // sum Fy=0
+// Xa=Xb........(eq'n 1) // sum Fx=0
+// (2) Dismember
+// Member AC. Consider equilibrium of member AC
+// Xa=Xc ... Consider thus as eq'n 2 // sum Fx=0
+Yc=W1-Ya // kN // sum Fy=0
+// Take moment about A
+Xc=((W1*l1)-(Yc*L1))/L1 // kN
+// now from eq'n 1 & 2
+Xa=Xc // kN
+Xb=Xa // kN
+// The components of reactions at A & B are,
+Ra=sqrt(Xa^2+Ya^2) // kN
+Rb=sqrt(Xb^2+Yb^2) // kN
+// Results
+clc
+printf('The reaction at A ( Ra) is %f kN \n',Ra)
+printf('The reaction at B ( Rb) is %f kN \n',Rb)
diff --git a/3204/CH9/EX9.13/Ex9_13.sce b/3204/CH9/EX9.13/Ex9_13.sce new file mode 100644 index 000000000..072d40fd9 --- /dev/null +++ b/3204/CH9/EX9.13/Ex9_13.sce @@ -0,0 +1,24 @@ +// Initilization of variables
+W1=2 // kN // load acting at a distance of 1m from point A
+W2=1 // kN // load acting at a distance of 1m from point B
+theta=30 // degree
+L=4 // m // length of the tie beam
+l=1 //m // length of each member in the tie
+// Calculations
+// (a) Reactions
+Yb=((W1*l)+(W2*3*l))/L // kN // Taking moment about A
+Ya=W1+W2-Yb // kN // sum Fy=0
+// (b) Dismember
+// MEMBER AB
+// Xa=Xb........ (eq'n 1) // sum Fx=0
+// MEMBER AC
+// Xa=Xc.........(eq'n 2) // sum Fx=0
+Yc=W1-Ya // kN // sum Fy=0
+// Taking moment about A
+Xc=((W1*l)-(Yc*2*l))/(2*tand(theta)) // kN
+// From eq'n 1 & 2
+Xa=Xc // kN
+Xb=Xa // kN
+// Results
+clc
+printf('The force in tie bar AB is %f kN \n',Xb)
diff --git a/3204/CH9/EX9.14/Ex9_14.sce b/3204/CH9/EX9.14/Ex9_14.sce new file mode 100644 index 000000000..a178508d9 --- /dev/null +++ b/3204/CH9/EX9.14/Ex9_14.sce @@ -0,0 +1,29 @@ +// Initilization of variables
+W=1000 // N
+r=0.25 // radius of pulley at E
+Lab=2 //m
+Lad=1 // m
+Lbd=1 // m
+Ldc=0.75 // m
+l1=0.5 //m // c/c distance between bar AB and point E
+l2=1.25 // m // dist between rigid support and the weight
+// Calculations
+// (a) Reactions
+Xa=W // N // sum Fx=0
+Yb=((W*l1)+(W*l2))/Lab // N // Take moment about A
+Ya=W-Yb // N // sum Fy=0
+// Dismember
+// MEMBER ADB
+// consider triangle BCD to find theta, where s= length of bar BC,
+s=sqrt(Lbd^2+Ldc^2) // m
+theta=acosd(Lbd/s) // degree
+// equilibrium eq'n of member ADB
+Yd=(Ya*Lab)/Lad // take moment about B
+Fbc=(Yb+Ya-Yd)/sind(theta) // N // sum Fy=0
+Xd=(Fbc*cosd(theta))+(Xa) // N // sum Fx=0
+// PIN D
+Rd=sqrt(Xd^2+Yd^2) // N // shear force on the pin
+// Results
+clc
+printf('The compressive force in bar BC (Fbc) is %f N \n',Fbc)
+printf('The shear force on the pin is %f N \n',Rd)
diff --git a/3204/CH9/EX9.15/Ex9_15.sce b/3204/CH9/EX9.15/Ex9_15.sce new file mode 100644 index 000000000..a1681d098 --- /dev/null +++ b/3204/CH9/EX9.15/Ex9_15.sce @@ -0,0 +1,28 @@ +// Initiliztion of variables
+P=5000 // N
+theta=45 // degree // angle made by Rd & Re with the horizontal
+Lab=3 // m
+Lac=3 // m
+Lbd=2 // m
+Lce=2 // m
+l=1.5 // m // dist of load P from B
+// Calculations (BEAM AB )
+// Consider the equilibrium of beams
+// We are using matrix to solve the simultaneous eqn's
+A=[(Lbd*sind(theta)) Lab;(Lce*sind(theta)) -Lac]
+B=[(P*l) 0]
+C=B*inv(A)
+// Calculations (BEAM AC)
+Re=C(1) // N (C) // from eq'n 1
+Ya=(Re*Lce*sind(theta))/Lac // N // from eq'n 7
+Xa=C(1)*cosd(theta) // N // from eq'n 2
+Ra=sqrt(Xa^2+Ya^2) // N (C)
+Yb=P-Ya-(C(1)*sind(theta)) // N (C) // eq'n 3
+Yc=Ya-(Re*sind(theta)) // N (T)
+// Results
+clc
+printf('(1) The value of axial force (Rd) in bar 2 is %f N \n',C(1))
+printf('(2) The value of axial force (Re) in bar 3 is %f N \n',Re)
+printf('(3) The value of axial force (Yb) in bar 1 is %f N \n',Yb)
+printf('(4) The value of axial force (Yc) in bar 4 is %f N \n',Yc)
+// here consider Yc as +ve coz the assumed direction of the force was compressive which is to be reversed
diff --git a/3204/CH9/EX9.2/Ex9_2.sce b/3204/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..dac26bcfa --- /dev/null +++ b/3204/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,19 @@ +// Initilization of variables
+W1=2000 //N (or 2 kN)// load at joint D of the truss
+W2=4000 //N (or 4 kN)// load at joint E of the truss
+Lac=6 //m // length of the tie
+Lab=3 //m
+Lbc=3 //m
+theta=60 //degree // interior angles of the truss
+// Calculations
+// Here A is simply supported & B is roller support. Now the SUPPORT REACTIONS are given as,
+Rc=((W1*(Lab/2))+(W2*(Lab+(Lbc/2))))/Lac //N // Taking moment at A
+Ra=W1+W2-Rc //N // Take sum Fy=0
+// Calculations
+// Calculating the axial forces in the respective members by METHOD OF SECTION
+// A section is drawn passing through member DE such that it cuts the respective member. Now consider the equilibrium of the left hand portion of the truss. The three unknown forces are Fde, Fdb, & Fab
+// Take moment about B
+Fde=((3*Ra)-(W1*Lab*sind(30)))/(3*cosd(30)) //N // (T)
+// Results
+clc
+printf('The axial force in the member DE (Fde)is %f N \n',Fde)
diff --git a/3204/CH9/EX9.3/Ex9_3.sce b/3204/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..48b2bfe5c --- /dev/null +++ b/3204/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,25 @@ +// Initilization of variables
+W=1 //kN // load on the truss at joint D
+theta=45 //degree // angle made by the members AC & BD with the horizontal
+Lab=1 //m
+Lcd=1 //m // here Lcd= the distance from B to the line of extension drawn from 1kN force on the horizontal
+// Calculations
+// (1) JOINT E
+// Here the joint E is in equilibrium under two forces Fec & Fed which are non-collinear. Hence they must be 0. i.e Fec=Fed=0
+Fec=0
+Fed=0
+// (2) JOINT D
+Fdb=W/sind(theta) // kN // (C)// sum Fy=0
+Fdc=Fdb*cosd(theta) // kN // (T) // sum Fx=0
+// (3) JOINT C
+Fca=Fdc/sind(theta) // kN // (T) // sum Fx=0
+Fcb=-(Fca*sind(theta)) // kN // (C) // sum Fy=0
+// Results
+clc
+printf('The axial force in the member DC (Fdc) is %f kN \n',Fdc)
+printf('The axial force in the member DB (Fdb) is %f kN \n',Fdb)
+printf('The axial force in the member CA (Fca) is %f kN \n',Fca)
+printf('The axial force in the member CB (Fcb) is %f kN \n',Fcb)
+printf('The axial force in the member EC (Fec) is %f kN \n',Fec)
+printf('The axial force in the member ED (Fed) is %f kN \n',Fed)
+// Here -ve sign indicates COMPRESSIVE force & +ve indicates TENSILE force
diff --git a/3204/CH9/EX9.5/Ex9_5.sce b/3204/CH9/EX9.5/Ex9_5.sce new file mode 100644 index 000000000..52a571b5a --- /dev/null +++ b/3204/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,27 @@ +// Initilization of variables
+W1=1000 //N // Load acting at the end pannels and the ridge
+W2=2000 //N // Load acting at the intermidiate pannels
+Laf=1 //m
+Lgf=1 //m
+Lag=2 //m
+Lbg=1 //m
+Lab=3 //m
+theta=30 //degree // angle made by the principal rafter with the tie beam
+beta=60 //degree // angle made by the slings (i.e members CF & CG) with the tie beam
+// Calculations
+// consider the equilibrium of the entire truss as a F.B.D
+Xa=2*(W1*sind(theta))+(W2*sind(theta)) //N // sum Fx=0
+Rb=((W2*Laf*cosd(theta))+(W1*Lag*cosd(theta)))/Lab // N // Moment at A=0
+Ya=2*(W1*cosd(theta))+(W2*cosd(theta))-(Rb) //N // sum Fy=0
+// Now pass a section through the truss such that it cuts the members CE,CG & FG. Now consider the equilibrium of the right hand side of the truss
+// Take moment about C
+Ffg=(Rb*(Lbg+0.5))/(0.5*tand(beta)) // N // (T) // Here 0.5 is the half distance of Lgf
+// Take moment about G
+Fce=(-Rb*Lbg)/(Lbg*sind(theta)) // N // (C)
+// Take moment about B
+Fcg=0/(Lbg*sind(beta)) // N
+// Results
+clc
+printf('The axial force in the member FG (Ffg) is %f N \n',Ffg)
+printf('The axial force in the member CE (Fce) is %f N \n',Fce)
+printf('The axial force in the member CG (Fcg) is %f N \n',Fcg)
diff --git a/3204/CH9/EX9.6/Ex9_6.sce b/3204/CH9/EX9.6/Ex9_6.sce new file mode 100644 index 000000000..ecce6dc27 --- /dev/null +++ b/3204/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,24 @@ +// Initilization of variables
+W1=100 //N // load acting at pt. C vertically
+W2=50 //N // load acting at point B horizontaly
+L=2 //m // length of each bar in the hexagonal truss
+theta=60 //degree // internal angle of the truss
+// Calculations
+// We calculate the values of different members of the truss
+HG=L*sind(theta)
+AF=L
+// Support A is hinged whereas support F is a roller support. Firstly we find the support reactios as follows,
+Rf=(W2*HG)/AF //N // moment at F
+Xa=W2 //N // sum Fx=0
+Ya=W1-Rf //N // sum Fy=0
+// Now pass a section through the truss cutting the members CD,GD,GE & GF and consider equilibrium of right hand portion of the truss
+Fcd=(Rf*(L/2))/(L*sind(theta)) // N (C) // Taking moment about G
+// Now pass a scetion pq cutting the members CB,GB & GA
+Fga=((Rf*(L+(L/2)))-(W1*(L/2)))/(L*sind(theta)) // N (T) // Taking moment about B
+// take moment about G
+Fcb=((W1*(L/2))+(Rf*(L/2)))/(L*sind(theta)) // N (C)
+Fgb=(Fcb*cosd(theta))-(Fga*cosd(theta)) // N (T) // sum Fx=0
+// Results
+clc
+printf('The axial force in the member CD (Fcd) is %f N \n',Fcd)
+printf('The axial force in the member GB (Fgb) is %f N \n',Fgb)
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