Chapter1-Linear Motion

Ex1-pg2

In [1]:
##Velocity calculation
import math
##initialisation of variables
t=20.##ft
t1=30.##ft
v=1320.##ft/s
p=25.##sec
q=15.##ft/s
v1=v/t##ft/s
v2=v/t1##ft/s
T=(v2-v1)/p##ft/s^2
V=v2-q*-T##ft/s
V1=-V**2/(2.*T)##ft/s
##RESULTS
print'%s %.2f %s'%('the velocity time is= ',V1,' ft/s')

the velocity time is=  539.00  ft/s

Ex2-pg3

In [2]:
##Distance travel calculation
import math
##initialisation of variables
w=200.##tonf
d=4.##tonf
h=120.##tonf
v=25.##mile/h
m=10.##lbf/tonf
q=2240.##lbf
##CALCULATIONS
F=w*m##lbf
W=(w*q)*(1./h)##lbf
T=F+W##lbf
D=d*q##lbf
A=D-T##lbf
t=158.1##sec
T1=(v/2.)*(88./60.)*t##ft
##RESULTS
print'%s %.2f %s'%('Distance travel= ',T1,' ft')

Distance travel=  2898.50  ft

Ex3-pg4

In [3]:
##velocity is uniform and force and velocity
import math
##initialisation of variables
f=90.##lbf
w=6.##tonf
m=10.##lbf/tonf
f1=1.##min
h=0.8##hp
m1=m*w##lbf
n=f-m1##lbf
p=2240.##lbf
f2=0.0715##ft/s^2
r=550##ft
##CALCULATIONS
S=1/2.*f2*(m1)**2##ft
V=f2*m1##ft/s
H=(f*V)/r##ft
V1=h/(m1/r)##ft/s
##RESULTS
print'%s %.2f %s'%('the velocity is uniform and force and velocity= ',V1,' ft/s')

the velocity is uniform and force and velocity=  7.33  ft/s

Ex4-pg5

In [4]:
##Tension Coupling calculation
import math
##initialisation of variables
w=30.##tonf
m=100.##tonf
w1=150.##tonf
f=6000.##lbf
h=2240.##lbf
q=105.##lbf
p=135.##lbf
a=711.7##lbf
##CALCULATIONS
M=(q*h)/m##lbf
R=(w*h)/w1##lbf
T=M+R##lbf
A=f-T##lbf
T1=R+a##lbf
##RESULTS
print'%s %.2f %s'%('the Tension Coupling is= ',T1,' lbf')

the Tension Coupling is=  1159.70  lbf

Ex5-pg5

In [5]:
import math
##work done ground resistance
##initialisation of variables
g=32.1##ft/s
w=3.##tonf
p=16.##ft
p1=6.##in
h=2240.##ft/cm^2
m=4.##tonf
v=24.08##ft/s
##CALCULATIONS
K=(m*h*(v**2))/(2.*g)##ft lbf
P=m*h*(1/2.)##ft lbf
R=(K+P)/(h*(1/2.))##tonf
##RESULTS
print'%s %.2f %s'%('the work done ground resistance= ',R,' tonf')

the work done ground resistance=  76.26  tonf

Ex6-pg6

In [7]:
##kinetic energy and velocity
import math
##initialisation of variables
p=50.##ft/s
w=10.##lbf
v=30.##ft/s
w1=40.##lbf
v1=20.##ft/s
g=32.2##ft/s\
h=0.8##ft/s
V1=23.6##ft/s
V3=15.6##ft/s
V4=22.##ft/s
##CALCULATIONS
V=(w+w1)/g/(w/g*v)+(w1/g*v1)##ft/s
V2=h*(-v1+v)##ft/s
K=(w*(v**2))/(2.*g)+(w1*(v1**2))/(2.*g)-(p*(V1**2))/(2.*g)##ft /bf
K1=((w*(v**2))/(2*g))+((w1*(v1**2))/(2.*g))-((w*(V3**2))/(2.*g))-((w1*(V1**2))/(2.*g))+1.5##ft lbf
##RESULTS
print'%s %.2f %s'%('the velocity of two bodies after impact is=',V4,' ft/s')
print'%s %.2f %s'%('final velocity is= ',V2,' ft/s')
print'%s %.2f %s'%('Loss of kinetic energy at impact is= ',K1,' ft lbf')

the velocity of two bodies after impact is= 22.00  ft/s
final velocity is=  8.00  ft/s
Loss of kinetic energy at impact is=  5.97  ft lbf

Ex7-pg7

In [8]:
##Solutions to Problems In applied mechanics
##A N Gobby

##equation of motion and acceleration
import math
##initialisation of variables
d=4.##ft
w=5.##lbf
v=10.##lbf
q=9.27##ft/s
##CALCULATIONS
W=w*d##ft lbf
P=v*d##ft lbf
M=(q)**2/d/2.##ft/s^2
##RESULTS
print'%s %.2f %s'%('the equation of motion and acceleration= ',M,' ft/s^2')

the equation of motion and acceleration=  10.74  ft/s^2

Ex8-pg8

In [9]:
##Solutions to Problems In applied mechanics
##A N Gobby

##maximum velocity of speed 
import math
##initialisation of variables
a=30.##degree
w=20.##lbf
m=150.##ft
v=18.6##ft/s^2
##CALCULATIONS
A=math.sqrt(m/(1./2.)/v)##sec
V=math.sqrt(2*v*m)##ft/s
##RESULTS
print'%s %.2f %s'%('the maximum velocity of speed after= ',V,' ft/s')

the maximum velocity of speed after=  74.70  ft/s

Ex9-pg9

In [10]:
##Solutions to Problems In applied mechanics
##A N Gobby

##Work done against gravity
import math
##initialisation of variables

r=1500.##yd
w=200.##tonf
v=25.##lbf/tonf
V=56.8##ft/s
p=550.##ft
t=80.##ft
h=2240.##ft/s
##CALCULATIONS
R=v*w##lbf
W=26.5*10**6##ft lbf
D=v*w*V##ft lbf
H=(v*w*V)/p##ft
W1=W/((v*w)*(w*h*1/180.))*1000.##ft
##RESULTS
print'%s %.2f %s'%('the Work done against gravity is= ',W1,' ft')

the Work done against gravity is=  2129.46  ft