# Chapter 16: Dynamics of a Rigid Body in Plane Motion¶

## Example 16.16-2, Page No 335¶

In [3]:
import math
import numpy as np

#Intilization of variables
W=600 #lb
d=30 #in
# as theta=25 degrees,
sintheta=0.422
costheta=0.906
g=32.2 #ft/s**2

#Calculations
m=W/g #lb-s**2/ft
#Moment of inertia
I=0.5*m*((d/2)/12)**2 #lb-s**2-ft
#Applying Newtons law and coservation of angular momentum and rolling
#Solving by matrix method
A=np.array([[1,m,0,0],[0,0,0,1],[((d/2)/12),0,-I,0],[0,1,-((d/2)/12),0]])
B=np.array([[W*sintheta],[W*costheta],[0],[0]])
C=np.linalg.solve(A,B)

#Result
print'The Frictional Force is',round(C[0],1),"lb",'and the acceleration is',round(C[1],2),"ft/s**2"

# The answers wary due to decimal point descrepancy.

The Frictional Force is 84.4 lb and the acceleration is 9.06 ft/s**2


## Example 16.16-3, Page No 336¶

In [5]:
import math
import numpy as np

#Initilization of variables
m=18 #kg
d=0.6 #m
vo=3 #m/s
# as theta=20 degrees,
sintheta=0.342
costheta=0.939
g=9.8 #m/s**2

#Calculations
#Moment of Inertia
I=0.5*m*(d/2)**2
#Applying Newtons second Law a
A=np.array([[1,m,0,0],[0,0,1,0],[d/2,0,0,-I],[0,1,0,(-d/2)]])
B=np.array([[g*m*sintheta],[g*m*costheta],[0],[0]])
C=np.linalg.solve(A,B)
F=C[0] #N
ax=C[1] #m/s**2
Na=C[2] #N
#Time Calculations
v=0 #m/s**2
t=(vo)/ax #s

#Result
print'It takes',round(t,2),"s to reach the highest point of travel"

# The ans is off by 0.01 sec.

It takes 1.34 s to reach the highest point of travel


## Example 16.16-5, Page No 337¶

In [6]:
import math
import numpy as np

#Initilization of variables
m=20 #kg
F1=40 #N
ro=0.6 #m
ri=0.45 #m
g=9.8 #m/s**2

#Calculations
#Moment of inertia
I=(2*5**-1)*m*ro**2 #kg-m**2
#Applying Newtons Law and conservation of angular Momentum
#Solving by matrix method
A=np.array([[1,m],[ro,-I/ro]])
B=np.array([[F1],[F1*ri]])
C=np.linalg.solve(A,B)
F=C[0] #N
a=C[1] #m/s**2

#Result
print'The acceleration is',round(a,2),"m/s**2"
print'The force is',round(F,2),"N"
#The solution in the textbook is incorrect

The acceleration is 0.36 m/s**2
The force is 32.86 N


## Example 16.16-6, Page No 338¶

In [9]:
import math

#Initilization of variables
W=16.1 #lb
u=0.10 #co-efficient of friction
g=32.2 #ft/s**2
# as theta=30 degrees,
sintheta=2**-1
F=1.39 #lb

#Calculations
#Applying Newtons Second Law
#Using F=1.39 lb
a=((W*sintheta)-F)/(W*g**-1) #ft/s**2

#Result
print'The value of a is',round(a,1),"ft/s**2",'and alpha is ',round(alpha,1),"rad/s**2."
print'Hence the sphere will both,roll and slip'

The value of a is 13.3 ft/s**2 and alpha is  13.9 rad/s**2.
Hence the sphere will both,roll and slip


## Example 16.16-8, Page No 339¶

In [10]:
import math
import numpy as np

#Initilization of variables
# as theta=30 degrees
sintheta=2**-1
W=80 #lb
Ww=100 #lb
I=4 #slug-ft**2
r=0.5 #ft
v= 20 #ft/s
vo=0 #ft/s
g=32.2 #ft/s**2

#Calculations
#Using Equations of motion
#Solving the system of linear equatinons by matrix method
A=np.array([[-1,0,-W/g],[1,-1,-Ww/g],[0,r,-2*I]])
B=np.array([[-W],[Ww*sintheta],[0]])
C=np.linalg.solve(A,B)
#Storing values in variables
T=C[0] #lb
F=C[1] #lb
a=C[2] #ft/s**2
#Time calculations
t=(v-vo)/a #s

#Result
print'The time required is',round(t,1),"s"

The time required is 14.4 s


## Example 16.16-9, Page No 340¶

In [11]:
import math
import numpy as np

#Initilization of variables
M=70 #kg
ko=0.4 #m
ri=0.45 #m
ro=0.6 #m
# as theta=30 degrees,
sintheta=2**-1
costheta=(3**0.5)*2**-1
m=35 #kg
g= 9.8 #m/s**2

#Calculations
I=M*ko**2 #kg-m**2
#Using Equations of motion
#Solving the equations by matrix method
A=np.array([[-1,-m*0.15,0],[1,-M*ro,-1],[-ri,-I,ro]])
B=np.array([[-m*g],[M*g*sintheta],[0]])
C=np.linalg.solve(A,B)
F=C[2] #N
Na=M*g*costheta #N
#Required coefficient of friction
u=F/Na #coefficient of friction

#Result
print'The value of alpha is',round(C[1],2),"rad/s**2",'and tension is',round(C[0]),"N"
print'F=',round(F),"N,",'Na=',round(Na),"N",'and u=',round(u,2),"(coefficient of friction)"

The value of alpha is -4.15 rad/s**2 and tension is 365.0 N
F= 196.0 N, Na= 594.0 N and u= 0.33 (coefficient of friction)


## Exampe 16.16-10, Page No 341¶

In [13]:
import math
import numpy as np

#Initilization of variables
m=200 #kg
g=9.8 #m/s**2
r=1.2 #m
F1=1000 #N
F2=1400 #N

#Calculations
N=m*g #N
I=(2*5**-1)*(m)*r**2 #kg-m**2
#Using equations of motion
#Solving for F and alpha using matrix method
#Applying equations of motion
A=np.array([[1,-m],[-r,-I/r]])
B=np.array([[F1-F2],[F1*r]])
C=np.linalg.solve(A,B)
#Storing values
F=C[0] #N
a=r*alpha #m/s**2

#Result
print'The value of a is',round(a,2),"m/s**2",'and F is',round(F),"N"
#The negative signs indicate that the direction is opposite to what was origninally assumed
# The answers wary due to decimal point descrepancy.

The value of a is -2.57 m/s**2 and F is -829.0 N


## Example 16.16-11, Page No 341¶

In [16]:
import math
import numpy as np

#Initilization of variables
Wa=161 #lb
Wb=193.2 #lb
Wc=300 #lb
ka=3 #ft
kb=2.5 #ft
# as theta1=30 degrees & theta2=45 degrees,
sintheta1=2**-1
sintheta2=(2**0.5)**-1
costheta2=(2**0.5)**-1
g=32.2 #ft/s**2

#Calculations
#Moment of inertia Calculations
Ia=(Wa/g)*ka**2 #lb-s**2-ft
Ib=(Wb/g)*kb**2 #lb-s**2-ft
#Using equations of motion for A and B and C
#Solving by matrix method
A=np.array([[1,1,-Wa/g,0,0],[1,-4,-Ia*(4**-1),0,0],[-2,0,-Ib*(5*8**-1),4,0],[0,0,-(Wc/g)*(5*2**-1),-1,-0.25],[0,0,0,0,1]])
B=np.array([[Wa*sintheta1],[0],[0],[-Wc*costheta2],[Wc*sintheta2]])
C=np.linalg.solve(A,B)
#Storing values in the variables
T1=C[0] #lb
T2=C[3] #lb
a=C[2] #ft/s**2

#Result
print'The values are:'
print'a=',round(a,2),"ft/s**2 ,",'T1=',round(T1),"lb",'and T2=',round(T2,1),"lb"

The values are:
a= 3.93 ft/s**2 , T1= 89.0 lb and T2= 67.5 lb


## Example 16.16-12, Page No 342¶

In [18]:
import math
import numpy as np

#Initilization of variables
W=644 #lb
F=30 #lb
# as theta=30 degrees,
sintheta=(2)**-1
costheta=(3**0.5)*2**-1
r=1.5 #ft
g=32.2 #ft/s**2

#Calculations
#Using equations of motion
#Solving by matrix method
A=np.array([[1,-W/g],[-r,-(2**-1)*(W/g)*(2*2)*(r**-1)]])
B=np.array([[(W*sintheta)-(F*costheta)],[-F*2]])
C=np.linalg.solve(A,B)
a=C[1] #ft/s**2

#Result
print'The value of a is',round(a,2),"ft/s**2"
# The negative sign indicates that the cylinder will roll down the plane.

The value of a is -6.78 ft/s**2


## Example 16.16-14, Page No 344¶

In [20]:
import math
import numpy as np

#Initilization of variables
W=20 #lb
g=32.2 #ft/s**2

#Calculations
#Using equations of motion
#Solving the three equations simultaneously by matrix method
X=np.array([[0,1,-(W*g**-1)*5.2],[-1,0,-(W*g**-1)*3],[3,-3,-(12**-1)*(W*g**-1)*12**2]])
Y=np.array([[-0.75*(W*g**-1)],[(W*g**-1)*1.3-W],[0]])
C=np.linalg.solve(X,Y)
A=C[0] #lb
B=C[1] #lb

#Result
print'The value of alpha is',round(alpha,1),"rad/s**2 countercockwise",'and of A and B are',round(A,1),"lb up and",round(B,2),"lb to the right"

The value of alpha is 2.6 rad/s**2 countercockwise and of A and B are 14.4 lb up and 7.91 lb to the right


## Example 16.16-16, Page No 345¶

In [22]:
import math
import numpy as np

#Initilization of variables
mc=7.25 #kg
d=0.9 #m
la=0.2 #m
ma=9 #kg
F=45 #N
ay=0 #m/s**2
g=9.8 #m/s**2

#Calculations
I=2*(0.5*mc*(d*2**-1)**2)+0.5*ma*(la*2**-1)**2 #kg-m**2
#Using the equations of motion
Na=(2*mc+ma)*g #N
#Solving the two equations using matrix method
A=np.array([[-1,-(2*mc+ma)],[(d*2**-1),-I/(d*2**-1)]])
B=np.array([[-F],[F*(la*2**-1)]])
C=np.linalg.solve(A,B)
F=C[0] #N
ax=C[1] #m/s**2

#Result
print'The computation yields ax=',round(ax,2),"m/s**2 to the right."

The computation yields ax= 1.13 m/s**2 to the right.


## Example 16.16-18, Page No 347¶

In [23]:
import math

#Initilization of variables
g=9.8 #m/s**2

#Calculations
#Here the equation has been solved in terms of the veriables
#Hence we directly consider the final result
av=(2*g)/3 #m/s**2

#Result
print'The value of av is',round(av,2),"m/s**2"

The value of av is 6.53 m/s**2


## Example 16.16-21, Page No 349¶

In [23]:
import math
import numpy as np

#initilization of variables
W=16.1 #lb
v=9 #ft/s
# as phi=30 degrees,
sinphi=(2)**-1
cosphi=(3**0.5)*2**-1
r=0.5 #ft
g=32.2 #ft/s**2
OG=4.5 #ft

#Calculations
#Using equations of motion
an=v**2/OG #ft/s**2
#Solving for alpha we get
N=(W*g**-1)*an+W*cosphi #lb
#Using equations of motion
A=np.array([[1,-r],[-1,-r*r]])
B=np.array([[W*sinphi],[0]])
C=np.linalg.solve(A,B)
F=C[0] #lb
at=C[1] #ft/s**2

#Result
print'The value of N and F are',round(N,1),"lb and",round(F,2),"lb respectively."

The value of N and F are 22.9 lb and 2.68 lb respectively.


## Example 16.16-25, Page No 353¶

In [26]:
import math
import numpy as np

#Initilization of variables
W=50 #lb
P=10 #lb
t=5 #s
vo=0 #ft/s
g=32.2 #ft/s**2

#Calculations
#Using equations of motion
ax=(P*g)/W #ft/s**2
#Solving by matrix method for A and B
F=np.array([[1,1],[-4,4]])
Q=np.array([[W],[P]])
R=np.linalg.solve(F,Q)
#Velocity calculations
v=vo+ax*t #ft/s
A=R[0] #lb
B=R[1] #lb

#Result
print'The velocity of the door after 5s is',round(v,1),"ft/s",'and A=',round(A,1),"lb",'and B=',round(B,1),"lb"

The velocity of the door after 5s is 32.2 ft/s and A= 23.8 lb and B= 26.3 lb


## Example 16.16-26, Page No 354¶

In [28]:
import math
import numpy as np
#Initilization of variables
AB=2 #m
m=2 #kg
F=20 #N
g=9.8 #m/s**2

#Calculations
#Using equation of motion
a=F/m #m/s**2
#Solving by matrix method for Na and Nb
A=np.array([[1,-1],[4*5**-1,4*5**-1]])
B=np.array([[m*g],[F*(3*5**-1)]])
C=np.linalg.solve(A,B)

#Result
print'The value of a is',round(a),"m/s**2",'and the reactions are: Na=',round(C[0],1),"N up",'and Nb=',round(C[1],1),"N up"

The value of a is 10.0 m/s**2 and the reactions are: Na= 17.3 N up and Nb= -2.3 N up


## Example 16.16-27, Page No 354¶

In [29]:
import math

#Initilization of variables
vo=0 #ft/s

#Calculations
s=(0.011*5280*2)/(2*0.004)

#Result
print'It travels',round(s),"ft along the level before coming to rest"
#Answer in the textbook is incorrect by 20ft

It travels 14520.0 ft along the level before coming to rest


## Example 16.16-28, Page No 355¶

In [30]:
import math

#Initilization of variables
u=0.3 #coefficient of friction
m=70 #kg
g=9.8 #m/s**2

#Calculations
#CASE 1
#Using equations of motion
Na=m*g #N
ah=(u*Na)/m #m/s**2
#CASE 2
#Applying sum of moments equal to zero
F=(Na*0.3)/1.2 #N
a_h=F/m #m/s**2

#Result
#Intutive insights can be attained after we get these results
print'The value of Na is',round(Na),"N"
print'and that of acceleration are:'
print'1st value=',round(ah,2),"m/s**2",'and 2nd value is',round(a_h,2),"m/s**2"
print'and the value of F is',round(F),"N"

The value of Na is 686.0 N
and that of acceleration are:
1st value= 2.94 m/s**2 and 2nd value is 2.45 m/s**2
and the value of F is 172.0 N


## Example 16.16-29, Page No 356¶

In [32]:
import math

#Initilization of variables
m=60 #kg
me=660 #kg
a=6 #m/s**2
g=9.8 #m/s**2

#Calculations
#Using equations of motion
P=m*a+m*g #N
R=P/g #kg
#Increase in mass
I=R-m #kg
#Tension
T=me*a+me*g #N

#Result
print'The value of P is',round(P),"n"
print'The apparent icrease in weight is',round(I,1),"kg",'and the tension in the cable is',round(T),"N"
#Answer in the textbook is off by 28 #N in Tension

The value of P is 948.0 n
The apparent icrease in weight is 36.7 kg and the tension in the cable is 10428.0 N


## Example 16.16-30, Page No 356¶

In [30]:
import math
import numpy as np

#Initilization of variables
u=0.2 #coefficient of friction
ma=1.2 #kg
mb=2 #kg
g=9.8 #m/s**2

#Calculations
Nb=mb*g #N
F=u*Nb #N
#Using equations of motion
#Solving for T and a
A=np.array([[-1,-ma],[1,-mb]])
B=np.array([[-ma*g],[F]])
C=np.linalg.solve(A,B)
T=C[0] #N
a=C[1] #m/s**2
#Taking the sum of the moments
x_m=-(F*0.15+T*0.15)/Nb #m
x=x_m*1000 #mm

#Result
print'The acceleration of block A is',round(a,2),"m/s**2"
print'and Nb acts at a distance of',round(x,1),"mm.(Negative sign indictaes that the side assumed is incorrect)"

The acceleration of block A is 2.45 m/s**2
and Nb acts at a distance of -97.5 mm.(Negative sign indictaes that the side assumed is incorrect)


## Example 16.16-31, Page No 357¶

In [31]:
import math
import numpy as np

#Initilization of variables
a=2.5 #m/s**2
mA=3 #kg
mB=7 #kg
g=9.8 #m/s**2

#Calculations
F=(mA+mB)*a #N
#Using equations of motion
Py=mB*g #N
#Solving for Px and H
A=np.array([[1,1],[-0.0375,0.0375]])
B=np.array([[mB*a],[Py*0.05]])
C=np.linalg.solve(A,B)
Px=C[0] #N
H=C[1] #N

#Result
print'The value of H is',round(H,1),"N"

The value of H is 54.5 N


## Example 16.16-32, Page No 358¶

In [32]:
import math

#Initilization of variables
m=20 #kg
g=9.8 #m/s**2
vo=3 #m/s
v=0 #m/s
s=4 #m

#Calculations
#Using equations of motion
Na=m*g #N
F=(Na*0.075)/0.125 #N
a=F/m #m/s**2
#Displacement
d=-(v**2-vo**2)/(2*a) #m
displ=s-d #m
v_f=sqrt(2*a*displ) #m/s

#Result
print'The final velocity is',round(v_f,2),"m/s to the left."

The final velocity is 6.17 m/s to the left.


## Example 16.16-33, Page No 358¶

In [33]:
import math
import numpy as np

#Initilization of variables
mA=30 #kg
mB=45 #kg
u_ab=3**-1 #coefficient of friction between two blocks
u_bp=10**-1 #coefficient of friction between block and horizontal plane
g=9.8 #m/s**2

#Calculations
#By inspection
Na=mA*g #N
Nb=Na+mB*g #N
a=(u_ab*Na-u_bp*Nb)/mB #m/s**2
P=(mA*a+u_ab*Na) #N
#For block A
#Solving for P,F and a
A=np.array([[1,-1,-mA],[-0.05,-0.075,0],[0,1,-mB]])
B=np.array([[0],[-Na*0.050],[Nb*u_bp]])
C=np.linalg.solve(A,B)
P_new=C[0] #N

#Result
#As p < p_new
print'The maximum value of P is',round(P),"N"

The maximum value of P is 114.0 N


## Example 16.16-34, Page No 359¶

In [40]:
import math

#Initilization of variables
Vo=1.5 #m/s
V=0 #m/s
g=9.8 #m/s**2

#Calculations
a=(g*0.2)/0.75 #m/s**2
t=-(V-Vo)/a #s

#Result
print'The maximum acceleration is',round(a,2),"m/s**2",'and minimum time is',round(t,2),"s"

The maximum acceleration is 2.61 m/s**2 and minimum time is 0.57 s


## Example 16.16-36, Page No 361¶

In [41]:
import math
import numpy as np

#Initilization of variables
vo=0 #mi/h
v=60 #mi/h
t=13.8 #s
W=3385 #lb
xb=46 #in
xf=66 #in
xv=31 #in
g=32.2 #ft/s**2

#Calculations
a=(((v*88*60)/3600)-vo)/t #ft/s**2
#Summing horizontal forces
F=(W/g)*a #lb
#Solving for Rf and Rr
A=np.array([[1,1],[-xf,xb]])
B=np.array([[W],[-F*xv]])
C=np.linalg.solve(A,B)
Rr=C[0] #lb
Rf=C[1] #lb

#Result
print'The value of reactions are Rf=',round(Rf),"lb",'and Rr=',round(Rr),"lb"

The value of reactions are Rf= 1809.0 lb and Rr= 1576.0 lb


## Example 16.16-43, Page No 366¶

In [43]:
import math
import numpy as np

#Initilization of variables
W=161 #lb
F=16.1 #lb
t=2 #s
g=32.2 #ft/s**2

#Calculations
#Using equations of motion
#Solving for T and alpha
A=np.array([[r*12**-1,-0.5*(W*g**-1)*(r*12**-1)**2],[-1,-F*g**-1]])
B=np.array([[0],[-F]])
C=np.linalg.solve(A,B)

#Result

#The ans is incorrect in textbook.

The angular speed is 7.58 rad/s


## Example 16.16-47, Page No 369¶

In [45]:
import math

#Initilization fo variables
r=2000 #ft
g=32.2 #ft/s**2
d=4.71 #ft
v=176 #ft/s

#Calculations
e=(d*v**2)/(g*r) #ft

#Result
print'The superelevation is',round(e,2),"ft"
#Watch the unit in the final answer

The superelevation is 2.27 ft


## Example 16.16-48, Page No 369¶

In [49]:
import math

#Initilization of variables
a=5 #ft/s**2
C=50 #lb-ft
W=161 #lb
g=32.2 #ft/s**2

#Calculations
T=0.5*(W/g)*1**2*a+C #lb
Ox=-T*(2/sqrt(a)) #lb
Oy=T*(1/sqrt(a))+W #lb
Wa=T/(1-(a/g)) #lb

#Result
print'The values are: T=',round(T,1),"lb",', Wa=',round(Wa),"b",',Ox=',round(Ox,1),"lb",'and Oy=+',round(Oy),"lb"

The values are: T= 62.5 lb , Wa= 74.0 b ,Ox= -55.9 lb and Oy=+ 189.0 lb


## Example 16.16-49, Page No 370¶

In [48]:
import math

#Initilization of variables
m=100 #kg
mr=20 #kg
l1=300 #mm
l2=600 #mm
g=9.8 #m/s**2

#Calculations
r_bar=(mr*l1+m*750)/120 #mm
I=(3**-1)*mr*(l2*1000**-1)**2+(2*5**-1)*m*(l1*2000**-1)**2+m*(0.75)**2 #kg.m**2
On=(m+mr)*(r_bar*1000**-1)*w**2 #N
Ot=((m+mr)*(r_bar*1000**-1)*alpha)-(m+mr)*g #N

#Result
print'The angular acceleration is',round(alpha,1),"rad/s**2",'and On=',round(On),"N",'and Ot=',round(Ot,1),"N"

#Due to decimal accuracy there is discrepancy in answers with the textbook

The angular acceleration is 13.3 rad/s**2 and On= 5184.0 N and Ot= -96.3 N


## Example 16.16-50, Page No 370¶

In [59]:
import math

#Initilization of variables
W=40 #lb
r=2 #in
g=32.2 #ft/s**2

#Calculations
#Using equations of motion
On=(W*g**-1)*(1*6**-1)*w**2 #lb
Ot=(W*g**-1)*(1*6**-1)*alpha
Io=(0.5*(W*g**-1)*0.5**2)*2+((W*g**-1)*(1*6**-1)**2)*2

#Result
print'The reaction components are On=',round(On,1),"lb to the right",'and Ot=',round(Ot,2),"lb up"
print'The value of Io is',round(Io,2),"lb-ft"

The reaction components are On= 20.7 lb to the right and Ot= 0.41 lb up
The value of Io is 0.38 lb-ft


## Example 16.16-51, Page No 371¶

In [49]:
import math
import numpy as np

#Initilizatin of variables
W=6 #lb
l=8 #ft
v=10 #ft/s
g=32.2 #ft/s**2
# as theta1=60 degrees & theta2=30 degrees
costheta1=2**-1
costheta2=(3**0.5)*2**-1

#Calculations
Fe=(W*v**2)*(g*l*0.5)**-1 #lb
#Using equations of motion
#Solving for C and T
A=np.array([[costheta1,-costheta2],[costheta2,costheta1]])
B=np.array([[-Fe],[W]])
P=np.linalg.solve(A,B) #lb
C=P[0] #lb
T=P[1] #lb

#Result
print'The value of C is',round(C,2),"lb",'and T is',round(T,2),"lb"

The value of C is 2.87 lb and T is 7.03 lb


## Example 16.16-52, Page No 372¶

In [50]:
import math

#Initilization of variables
W=32.2 #lb
T=120 #lb
m=1 #slug
r=6*12**-1 #ft

#Calculations

#Result

The angular speed permissible is 13.9 rad/s


## Example 16.16-53, Page No 372¶

In [51]:
import math
import numpy as np

#Initilization of variables
m=30 #kg
k=0.45 #m
g=9.8 #m/s**2

#Using equations of motion
#Solving for T1,T2 and alpha
A=np.array([[1,0,-m],[0,-1,-45],[-0.6,0.3,-m*k**2]])
B=np.array([[50*g],[-150*g],[0]])
C=np.linalg.solve(A,B)

#Result

# The answer for T2 and alpha is off by 4 & 0.1 units

The values are: T1= 607.0 N ,T2= 1294.0 N and alpha= 3.9 rad/s**2


## Example 16.16-54, Page No 373¶

In [65]:
import math

#Initilization of variables
Wc=28 #lb
v=16 #ft/s
Ib=12 #ft-lb-s**2
u=0.4 #coefficient of friction
t=2 #s
g=32.2 #ft/s**2

#Calculations
T=Wc+(Wc*g**-1)*8 #lb
F=((Ib*alpha)+(T*1.25))/t #lb
N=F/u #lb
P=(N*8+F*3)/40 #lb
#Summing forces horizontally and vertically
Dx=151-P #lb
Dy=-F #lb

#Result
print'The reactions at D are: Dx=',round(Dx),"lb to right",'and Dy=',round(Dy,1),"lb down"

The reactions at D are: Dx= 116.0 lb to right and Dy= -60.2 lb down


## Example 16.16-55, Page No 374¶

In [79]:
import math
import numpy as np

#Initilization of variables
m=8 #kg
n=90 #rpm
g=9.8 #m/s**2

#Calculations
Fg=m*g #N
#using equations of motion
By=m*g #N
#Solving for Bx and C
A=np.array([[1,1],[-0.3,0.9]])
B=np.array([[m*0.3*w**2],[By*0.3]])
C=np.linalg.solve(A,B) #N

#Result
print'The solution is: Bx=',round(C[0]),"N",',By=',round(By,1),"N",'and C=',round(C[1],1),"N"

The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N


## Example 16.16-56, Page No 375¶

In [66]:
import math

#Initilization of variables
m=8 #kg
n=90 #rpm
g=9.8 #m/s**2
r=0.3 #m

#calculations
#Using equations of motion
C=(m*g*0.3+m*r*w**2*r)/1.2 #N
Bx=-C+m*r*w**2 #N
By=m*g #N

#Result
print'The solution is: Bx=',round(Bx),"N",',By=',round(By,1),"N",'and C=',round(C,1),"N"

The solution is: Bx= 140.0 N ,By= 78.4 N and C= 72.9 N


## Example 16.16-57, Page No 375¶

In [68]:
import math

#Initilization of variables
Na=294 #N
Nb=735 #N

#Calculations
a=(10**-1*Nb-3**-1*Na)/45 #m/s**2
P=(3**-1*Na)-30*a #N

#Result
print'The solution is P=',round(P),"N",'and a=',round(a,3),"m/s**2"

The solution is P= 114.0 N and a= -0.544 m/s**2


## Example 16.16-58, Page No 376¶

In [69]:
import math

#Initilization of variables
W=50 #lb
g=32.2

#Calculations
#Using equations of motion
a=(10/(W/g)) #ft/s**2
B=((2.5*(W/g)*a)+4*W-1.5*10)/8 #lb
A=50-B #lb

#Result
print'The solution is A=',round(A,1),"lb",',B=',round(B,1),"lb",'and a=',round(a,2),"ft/s**2"

The solution is A= 23.8 lb ,B= 26.3 lb and a= 6.44 ft/s**2


## Example 16.16-60, Page No 377¶

In [71]:
import math

#Initilization of variables
g=9.8 #m/s**2
r1=0.3 #m
m1=20 #kg
m2=100 #kg
r2=0.75 #m

#Calculations

#Result

The angular acceleration is 13.4 rad/s**2


## Example 16.16-61, Page No 378¶

In [78]:
import math

#Initilization of variables
r=15*12**-1 #ft
W=600 #lb
# as theta=25 degrees,
sintheta=0.422

#calculations
ax=(r*W*sintheta)*((r**-1)*14.5+r*18.6)**-1 #ft/s**2
F=(W*sintheta)-(18.6*ax) #lb

#Result
print'The solution is F=',round(F,1),"lb",'and ax=',round(ax,2),"ft/s**2"

The solution is F= 84.3 lb and ax= 9.08 ft/s**2


## Example 16.16-62, Page No 378¶

In [79]:
import math

#Initilization of variables
m=7 #kg
g=9.8 #m/s**2
r=0.5 #m
I=0.875 #kg.m**2

#Calculations
#Solving for alpha and T

The soultion is alpha= 13.1 rad/s**2 and T= 22.9 N