initial commit / add all books

26 files changed, 334 insertions, 0 deletions

diff --git a/1325/CH2/EX2.1/2_1.PNG b/1325/CH2/EX2.1/2_1.PNG
new file mode 100644
index 000000000..1382aabc4
--- /dev/null
+++ b/1325/CH2/EX2.1/2_1.PNG
diff --git a/1325/CH2/EX2.1/ex2_1.sce b/1325/CH2/EX2.1/ex2_1.sce
new file mode 100644
index 000000000..130dbf689
--- /dev/null
+++ b/1325/CH2/EX2.1/ex2_1.sce
@@ -0,0 +1,49 @@
+//to find velocity and change in kinetic energy when impact between two spheres moving in a same line is a) INELASTIC , b) ELASTIC , c) e = 0.6

+clc

+//a) INELASTIC

+//for sphere 1 ,mass=m1 and initial velocity=u1 

+//for sphere 2 ,mass=m2 and initial velocity=u2

+m1=100//lb

+u1=10//ft/s

+m2=50//lb

+u2=5//ft/s

+v=(m1*u1+m2*u2)/(m1+m2)

+//change in kinetic energy

+//initial kinetic energy = ke1

+ke1=(m1*(u1^2)+m2*(u2^2))/(2*32.2)

+//Kinetic Energy after inelastic colision = ke2

+ke2=((m1+m2)*8.333^2)/(2*32.2)

+//Change in Kinetic Energy =l

+l=ke1-ke2

+//b) Elastic

+// for a very short time bodies will have a common velocity given by v=8.333 ft/s

+// for a very short time bodies will have a common velocity given by v=8.333 ft/s

+//immidiately after impact ends the velocities for both the bodies are given by v1 and v2

+v1=2*v-u1

+v2=2*v-u2

+//c) Coeeficient of Restitution=0.6

+e=0.6

+ve1=(1+e)*v-e*u1

+ve2=(1+e)*v-e*u2

+ke3=(m1*(ve1^2)+m2*(ve2^2))/(2*32.2)

+loss=ke1-ke3

+printf("kinetic energy before collisio0n is %f ft lb\n",ke1)

+printf("\n")

+printf("a) INELASTIC\n")

+printf("\n")

+printf("velocity after collision is %f ft/s\n",v)

+printf("the Kinetic Energy after collision is %f ft lb\n",ke2) 

+printf("the change in Kinetic Energy after collision is %f ft lb\n",l)

+printf("\n")

+printf("b) ELASTIC\n")

+printf("\n")

+printf("velocity of 1 after collision is %f ft/s\n",v1)

+printf("velocity of 2 after collision is %f ft/s\n",v2)

+printf("there is no loss of kinetic energy in case of elastic collision\n")

+printf("\n")

+printf("c) e=0.6\n")

+printf("\n")

+printf("velocity of 1 after collision is %f ft/s\n",ve1)

+printf("velocity of 2 after collision is %f ft/s\n",ve2)

+printf("the Kinetic Energy after collision is %f ft lb\n",ke3) 

+printf("the change in Kinetic Energy after collision is %f ft lb\n",loss)

diff --git a/1325/CH2/EX2.10/2_10.PNG b/1325/CH2/EX2.10/2_10.PNG
new file mode 100644
index 000000000..f1e993b2e
--- /dev/null
+++ b/1325/CH2/EX2.10/2_10.PNG
diff --git a/1325/CH2/EX2.10/ex2_10.sce b/1325/CH2/EX2.10/ex2_10.sce
new file mode 100644
index 000000000..bfbf4826a
--- /dev/null
+++ b/1325/CH2/EX2.10/ex2_10.sce
@@ -0,0 +1,22 @@
+//to find the acceleration if mass is allowed to fall freely and when efficiency of the gearing were 90%

+//gravitaional force (g)=32.2 ft/s^2

+clc

+//given

+Ia=200//lb ft2

+Ib=15//lb ft2

+G=5//wb==5*wa

+m=150//lb

+r=8//in

+printf("\n")

+//the equivalent mass of the geared system referred to the circumference of the drum is given by

+//Me=(1/r)^2*(Ia+(G^2*Ib))

+Me=(12/r)^2*(Ia+(G^2*Ib))

+M=m+Me

+a=(m/M)*32.2//acceleration

+//if efficiency of gearing is 90% then Me=(1/r^2)*(Ia+(G^2*Ib)/n)

+n=.9

+Me1=(12/r)^2*(Ia+(G^2*Ib)/n)

+M1=Me1+m

+a1=(m/M1)*32.2

+printf("acceleration = %.2f ft/s2\n",a)

+printf("acceleration when gear efficiency is 0.9= %.2f ft/s2\n",a1)

diff --git a/1325/CH2/EX2.11/2_11.PNG b/1325/CH2/EX2.11/2_11.PNG
new file mode 100644
index 000000000..c238d61a0
--- /dev/null
+++ b/1325/CH2/EX2.11/2_11.PNG
diff --git a/1325/CH2/EX2.11/ex2_11.sce b/1325/CH2/EX2.11/ex2_11.sce
new file mode 100644
index 000000000..359a2b030
--- /dev/null
+++ b/1325/CH2/EX2.11/ex2_11.sce
@@ -0,0 +1,37 @@
+//to find the maximum acceleration of car on each gear

+//gravitaional force (g)=32.2 ft/s^2

+clc

+printf("\n")

+//let

+//S=displacement of car from rest with uniform acceleration a, the engine torque T assumed to remain ocnstant

+//v=final speed ofcar

+//G=gear ratio

+//r=effective radius

+//n=efficiency of transmission

+//M=mass of the car

+//Ia and Ib=moments of inertia of road whels and engine 

+//formulas => F=29.5nG ; Me= 1648+$.54nG^2 ; a=32.2 F/Me

+//given

+G1=22.5

+G2=12.5

+G3=7.3

+G4=5.4

+n=.82//for 1st ,2nd and 3rd gear

+n4=.9//for 4th gear

+F1=29.5*n*G1

+F2=29.5*n*G2

+F3=29.5*n*G3

+F4=29.5*n4*G4

+//on reduction and putting values we get

+Me1=1648+4.54*n*G1^2

+Me2=1648+4.54*n*G2^2

+Me3=1648+4.54*n*G3^2

+Me4=1648+4.54*n4*G4^2

+a1=32.2*F1/Me1

+a2=32.2*F2/Me2

+a3=32.2*F3/Me3

+a4=32.2*F4/Me4

+printf("Maximum acceleration of car on top gear is %.2f ft/s^2 \n",a4)

+printf("Maximum acceleration of car on third gear is %.2f ft/s^2 \n",a3)

+printf("Maximum acceleration of car on second gear is %.2f ft/s^2 \n",a2)

+printf("Maximum acceleration of car on first gear is %.2f ft/s^2 \n",a1)

diff --git a/1325/CH2/EX2.12/2_12.PNG b/1325/CH2/EX2.12/2_12.PNG
new file mode 100644
index 000000000..eade451e8
--- /dev/null
+++ b/1325/CH2/EX2.12/2_12.PNG
diff --git a/1325/CH2/EX2.12/ex2_12.sce b/1325/CH2/EX2.12/ex2_12.sce
new file mode 100644
index 000000000..8282be87b
--- /dev/null
+++ b/1325/CH2/EX2.12/ex2_12.sce
@@ -0,0 +1,12 @@
+//to find the couple supplied to shaft

+//gravitaional force (g)=32.2 ft/s^2

+clc

+printf("\n")

+//given

+I=40//lb ft2

+n=500//rpm

+w=%pi*n/30//angular velocity

+wp=2*%pi/5//angular velocity of precession

+I1=I/32.2

+T=I1*w*wp//gyroscopic couple

+printf("the couple supplied to the shaft= %.2f lb ft\n",T)

diff --git a/1325/CH2/EX2.13/2_13.PNG b/1325/CH2/EX2.13/2_13.PNG
new file mode 100644
index 000000000..442f75d2f
--- /dev/null
+++ b/1325/CH2/EX2.13/2_13.PNG
diff --git a/1325/CH2/EX2.13/ex2_13.sce b/1325/CH2/EX2.13/ex2_13.sce
new file mode 100644
index 000000000..3e764705d
--- /dev/null
+++ b/1325/CH2/EX2.13/ex2_13.sce
@@ -0,0 +1,22 @@
+//to find the gyroscopic reaction of the airscrew on the aeroplane when it has a) three blades and b)two blades

+//gravitaional force (g)=32.2 ft/s^2

+clc

+//given

+printf("\n")

+I=250//lb ft2

+n=1600//rpm

+v=150//mph

+r=500//ft

+w=%pi*160/3//angular velocity of rotation

+wp=(150*88)/(60*500)//angular velocity of precession

+//a) with three bladed screw

+//T=I*w*wp

+T=(250/32.2)*%pi*(160/3)*wp

+//b)with two bladed air screw

+//T1=2*I*w*wp*sin(o)

+printf("The magnitude of gyroscopic couple is given by %.0f lb ft\n",T)

+//Tix=T(1-cos(2o)) lb ft

+//T1y=Tsin(2o)) lb ft

+printf("The component gyroscopic couple in the vertical plane =%.0f(1-cos(2x)) lb ft\n",T)

+printf("The component gyroscopic couple in the horizontal plane =%.0f(sin(2x)) lb ft\n",T)

+// for direction refer the book example 

diff --git a/1325/CH2/EX2.2/2_2.PNG b/1325/CH2/EX2.2/2_2.PNG
new file mode 100644
index 000000000..52b96bdab
--- /dev/null
+++ b/1325/CH2/EX2.2/2_2.PNG
diff --git a/1325/CH2/EX2.2/ex2_2.sce b/1325/CH2/EX2.2/ex2_2.sce
new file mode 100644
index 000000000..d4c24527f
--- /dev/null
+++ b/1325/CH2/EX2.2/ex2_2.sce
@@ -0,0 +1,42 @@
+//to find speed of truck immidiately after collision and the maximum deflection of spring during impact. Moreover if k=0.5 then determine hoe the final speedsw will be affected and amount of dissipated energy 

+clc

+//given

+m1=15//tons

+u1=12//m/h

+m2=5//tons

+u2=8//m/h

+k=2//ton/in

+e1=0.5//coefficient of restitution

+printf("\n")

+//conservation of linear momentum

+v=(m1*u1+m2*u2)/(m1+m2)

+printf("velocity at the instant of  collision is %.2f mph",v)

+e=(m1*m2*(88/60)^2*(u1-u2)^2)/(2*32.2*(u1+u2))

+printf("\n")

+printf("The difference between the kinetic energy before and during the impact is %.2f ft tons\n",e)

+//energy stored in spring equals energy dissipated

+//s=(1/2)*k*x^2

+//s=e

+//since there are 4 buffer springs ,4x^2=24 inches (2 ft=24 inches)

+x=((e*12)/4)^.5

+printf("Maximum deflection of the spring is %.2f in\n",x)

+// maximum force acting between pair of buffer = stiffness of spring*deflection

+f=k*x

+printf("Maximum force acting between each buffer is %.2f tons\n",f)

+//assuming perfectly elastic collision

+//for loaded truck 

+v1=2*11-12

+//for unloaded truck 

+v2=2*11-8

+printf("Speed of loaded truck after impact %.2f mph\n",v1)

+printf("speed of unloaded truck after impact %.2f mph\n",v2)

+//if coefficient of restitution =o.5

+//for loaded truck 

+ve1=(1+.5)*11-.5*12

+//for unloaded truck 

+ve2=(1+.5)*11-.5*8

+printf("Speed of loaded truck after impact when e=0.5 %.2f mph\n",ve1)

+printf("Speed of unloaded truck after impact when e=0.5 %.2f mph\n",ve2)

+//net loss of kinetic energy=(1-e^2)*energy stored in spring

+l=(1-(e1^2))*2//ft tons

+printf("Net loss of kinetic energy is %.2f ft tons\n",l)

diff --git a/1325/CH2/EX2.3/2_3.PNG b/1325/CH2/EX2.3/2_3.PNG
new file mode 100644
index 000000000..4f2e09edc
--- /dev/null
+++ b/1325/CH2/EX2.3/2_3.PNG
diff --git a/1325/CH2/EX2.3/ex2_3.sce b/1325/CH2/EX2.3/ex2_3.sce
new file mode 100644
index 000000000..091910bcb
--- /dev/null
+++ b/1325/CH2/EX2.3/ex2_3.sce
@@ -0,0 +1,29 @@
+//to find maximum twist,apeed of flywheels when twist is maximum and when springs regains its shape

+clc

+//given

+m1=500//lb ft^2

+m2=1500//lb ft^2

+k=150//lb ft^2

+w1=150//rpm

+//angular momentum will be conserved as net external force is zero

+//let final angular velocity be N then (m1+m2)N=w1*m1

+N=(w1*m1)/(m1+m2)

+printf("Angular velocity at the instant when speeds of the flywheels are equalised is given by %.2f r.p.m\n",N)

+//kinetic energy at this instance 

+ke1=(1/2)*((m1+m2)/32.2)*((%pi*N)/30)^2

+printf("The kinetic energy of the system at this instance is %.2f ft lb\n",ke1)

+printf("which is almost equal to 480 ft lb \n")

+//initial kinetic energy

+ke0=(1/2)*((m1)/32.2)*((%pi*w1)/30)^2

+printf("The initial kinetic energy of the system is %.2f ft lb\n",ke0)

+printf("which is almost equal to 1915 ft lb \n")

+//strain energy = s

+s=ke0-ke1

+printf("strain energy stored in the spring is %.2f ft lb which is approximately 1435 ft lb\n",s)

+//if x is the maximum anglular displacement of wheel and the mean torque applied by spring is i/2*k*x then work done or strain energy is given by 1/2 *k*x^2

+x=((1435*2)/150)^.5

+printf("Maximum angular displacement is %.2f in radians which is equal to 250 degrees\n",x)

+//na1 and na are initial and final speeds of the flywheel 1 and same nb1 and nb for flywheel 2 

+na=2*N-w1//w1=na1

+nb=2*N-0//nb1=0

+printf ("Speed of flywheel a and b when spring regains its unstrained position are %.2f rpm and %.2f rpm respectively\n",na,nb)

diff --git a/1325/CH2/EX2.4/2_4.PNG b/1325/CH2/EX2.4/2_4.PNG
new file mode 100644
index 000000000..10921a480
--- /dev/null
+++ b/1325/CH2/EX2.4/2_4.PNG
diff --git a/1325/CH2/EX2.4/ex2_4.sce b/1325/CH2/EX2.4/ex2_4.sce
new file mode 100644
index 000000000..f0f9e39dd
--- /dev/null
+++ b/1325/CH2/EX2.4/ex2_4.sce
@@ -0,0 +1,18 @@
+//to find length of equivalent simple pendulum

+//gravitaional force (g)=32.2 ft/s^2

+clc

+//given

+m1=150 //lb

+l=3//ft

+//number of oscillation per second is given by n

+printf("\n")

+n=(50/92.5)

+printf ("number of oscillation per second = %.2f\n",n)

+//length of simple pendulum is given by L=g/(2*%pi*n)^2

+L=32.2/(2*%pi*n)^2

+printf ("length of simple pendulum = %.2f ft\n",L)

+// distance of cg from point of suspension is given by a

+a=25/12

+k=(a*(L-a))^.5//radius of gyration

+moi=m1*k^2

+printf("The moment of inertia of rod is %.2f lb ft^2",moi)

diff --git a/1325/CH2/EX2.5/2_5.PNG b/1325/CH2/EX2.5/2_5.PNG
new file mode 100644
index 000000000..93d992588
--- /dev/null
+++ b/1325/CH2/EX2.5/2_5.PNG
diff --git a/1325/CH2/EX2.5/ex2_5.sce b/1325/CH2/EX2.5/ex2_5.sce
new file mode 100644
index 000000000..a075a31cd
--- /dev/null
+++ b/1325/CH2/EX2.5/ex2_5.sce
@@ -0,0 +1,24 @@
+//to find moment of inertia and distance of cg from small end centre

+clc

+//let l1 and l2 be length of equivalent simple pendulum when axis coincides with small end and big end respectively

+//n1 and n2 =corresponding frequencies of oscillation per second

+n1=50/84.4

+n2=50/80.3

+//let a1 and a2 = distances of cg from small end and big end centers respectively

+//gravitaional force (g)=32.2 ft/s^2

+//L=g/(2*%pi*n)

+L1=(32.2*12)*(84.4/(100*%pi))^2

+L2=(32.2*12)*(80.3/(100*%pi))^2

+//a1(L1-a1)=k^2=a2(L2-a2) and a1+a2=30 inches

+//substituting and solving for a we get 

+a1=141/6.8

+a2=30-a1

+k=(a1*(L1-a1))^.5

+moi=90*(149/144)//moi=m*k^2

+printf("length of equivalent simple pendulum when axis coincides with small end and big end respectively-\n")

+printf("L1=%.1f in\n",L1)

+printf("L2=%.1f in\n",L2)

+printf("distances of cg from small end and big end centers respectively are-\n")

+printf("a1=%.1f in\n",a1)

+printf("a2=%.1f in\n",a2)

+printf("Moment of inertia of rod =%.2f lb ft^2",moi)

diff --git a/1325/CH2/EX2.6/2_6.PNG b/1325/CH2/EX2.6/2_6.PNG
new file mode 100644
index 000000000..06f59b237
--- /dev/null
+++ b/1325/CH2/EX2.6/2_6.PNG
diff --git a/1325/CH2/EX2.6/ex2_6.sce b/1325/CH2/EX2.6/ex2_6.sce
new file mode 100644
index 000000000..800f4f6be
--- /dev/null
+++ b/1325/CH2/EX2.6/ex2_6.sce
@@ -0,0 +1,16 @@
+//to find radius of gyration about the mass centre

+//gravitaional force (g)=32.2 ft/s^2

+clc

+//given

+printf("\n")

+m1=150

+l=8.5

+g=32.2

+a=83.2

+n=25

+//k=(a/2*%pi*n)*(g/l)^0.5

+k=(14*a*((g)^0.5))/(2*%pi*n*(l^0.5))

+k1=14.5/12

+printf("radius of gyration is %.2f inches which is equal to %.2f ft \n",k,k1)

+moi=m1*(k1^2)

+printf("moment of inertia=%.2f lb ft^2",moi)

diff --git a/1325/CH2/EX2.7/2_7.PNG b/1325/CH2/EX2.7/2_7.PNG
new file mode 100644
index 000000000..2eb72ebde
--- /dev/null
+++ b/1325/CH2/EX2.7/2_7.PNG
diff --git a/1325/CH2/EX2.7/ex2_7.sce b/1325/CH2/EX2.7/ex2_7.sce
new file mode 100644
index 000000000..eeb018df4
--- /dev/null
+++ b/1325/CH2/EX2.7/ex2_7.sce
@@ -0,0 +1,26 @@
+//to find the equivalent dynamical system 

+//gravitaional force (g)=32.2 ft/s^2

+clc

+printf("\n")

+//given

+m=2.5//lb

+a=6//in

+k=3.8//in

+l=9//in

+c=3//in

+w=22500

+//k^2=ab

+//case a) to find equivalent dynamic system

+b=(k^2)/a

+ma=(2.5*6)/8.42//m*a/a+b

+mb=m-ma

+printf("Mass ma =%.2f lb will be situated at 6 inches from cg and mb =%.2f lb will be situated at %.2f inches from cg in the equivalent dynamical system",ma,mb,b)

+printf("\n")

+//if two masses are situated at the bearing centres 

+ma1=(2.5*6)/9

+mb1=m-ma1

+k1=(a*c)^.5

+//t=m*((k1^2)-(k^2))*w

+t=((2.5*(18-3.8^2))*22500)/(32.2*12*12)

+printf("correction couple which must be applied in order that the two mass system is dynamically equivalent to the rod is given by %.2f lb ft\n",t)

+

diff --git a/1325/CH2/EX2.8/2_8.PNG b/1325/CH2/EX2.8/2_8.PNG
new file mode 100644
index 000000000..a36f35853
--- /dev/null
+++ b/1325/CH2/EX2.8/2_8.PNG
diff --git a/1325/CH2/EX2.8/ex2_8.sce b/1325/CH2/EX2.8/ex2_8.sce
new file mode 100644
index 000000000..9a7fb5f5e
--- /dev/null
+++ b/1325/CH2/EX2.8/ex2_8.sce
@@ -0,0 +1,17 @@
+//to find  forces throught pin A and B in order to accelerate the link 

+//gravitaional force (g)=32.2 ft/s^2

+clc

+printf("\n")

+m=20//lb

+g=32.2

+a=200//ft/s^2

+w=120//rad/s^2

+k=7//in

+f=(m/g)*a//effective force appllied to the link

+//this force acts parallel to the acceleration fg

+t=(m/g)*(k/12)^2*w//couple required in order to provide the angular acceleration

+//the line of action of F is therefore at a distance from G given by

+x=t/f

+printf("Effective force applied to the link is %.3f lb and the line of action of F is therefore at a distance from G given by %.3f ft \n",f,x)

+printf("F is the resultant of Fa and Fb, using x as shown in figure.25 , the force F may then be resolved along the appropriate lines of action to give the magnitudes of Fa and Fb\n")

+printf("From the scaled diagram shown in figure we get,Fa=65 lb and Fb=91 lb\n")

diff --git a/1325/CH2/EX2.9/2_9.PNG b/1325/CH2/EX2.9/2_9.PNG
new file mode 100644
index 000000000..29081ea64
--- /dev/null
+++ b/1325/CH2/EX2.9/2_9.PNG
diff --git a/1325/CH2/EX2.9/ex2_9.sce b/1325/CH2/EX2.9/ex2_9.sce
new file mode 100644
index 000000000..02a79a83e
--- /dev/null
+++ b/1325/CH2/EX2.9/ex2_9.sce
@@ -0,0 +1,20 @@
+//to find force that must be exerted in oeder to give an acceleration of 3ft/s^2 and smallest value of u(friction coefficient)

+//gravitaional force (g)=32.2 ft/s^2

+clc

+printf("\n")

+//given

+m=10//ton

+m2=1000//lb

+a=3//ft/s^2

+//the addition to actual mass in order to allow for the rotational inertia of the wheels and axles

+m1=2*(1000/2240)*(15/21)^2//m1=m2*k^2/r^2 and 1 ton=2240 lbs

+M=m+m1

+F=3*(10.46/32.2)//F=M.a

+f=F*2240//lb

+Fa=(2*1000/2240)*(3/32.2)*(15/21)^2//total tangential force required in order to provide the angular acceleration of the wheels and axles

+//Limiting friction force =uW 

+//u*10>0.042

+u=0.042/10

+printf("The total tangential force required in order to provide the angular acceleration of the wheels and axles is %.4f ton\n",Fa)

+printf("If there is to be pure rolling ,u>%.4f",u)

+