initial commit / add all books

21 files changed, 325 insertions, 0 deletions

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)