# CHAPTER 6:PLANE KINETICS OF RIGID BODIES¶

## SAMPLE PROBLEM 6/1,PAGE NUMBER:422¶

In [1]:
import math
from scipy.optimize import fsolve
# Given data
W=3220;# lb
v=44;# m/s (30 mi/hr)
s=200;# ft
mu=0.8;# The effective coefficient of friction between the tires and the road
g=32.2;# The acceleration due to gravity in ft/sec**2
d_G=24;# inch
d_BG=60;# inch
d_GA=60;# inch

# Calculation
abar=v**2*(2*s)**-1;# ft/sec**2
theta=math.atan((1*10**-1))*180/math.pi;# degree
W_h=W*math.cos(theta*math.pi/180);# lb
W_v=W*math.sin(theta*math.pi/180);# lb
mabar=(W/g)*abar;# lb
# SigmaF_x = m*abar_x
F=mabar+W_v;# lb
def equations(p):
N_1,N_2=p
return(N_1+N_2-W_h,((d_GA*N_1+(F*d_G)-(N_2*d_BG))))
N_1,N_2=fsolve(equations,(10,10))
FbyN_2=F/N_2;
print"\nThe friction force under the rear driving wheels,F=%3.0f lb \nThe normal force under each pair of wheels,N_1=%4.0f lb & N_2=%4.0f lb"%(F,N_1,N_2);

The friction force under the rear driving wheels,F=804 lb
The normal force under each pair of wheels,N_1=1441 lb & N_2=1763 lb


## SAMPLE PROBLEM 6/2,PAGE NUMBER:423¶

In [2]:
import math
#Given data
m=150;#kg
M=5;#kN
theta=30;#degree
ACbar=1.5;#m
BDbar=1.5;#m
ABbar=1.8;#m
g=9.81;#The acceleration due to gravity in m/s**2

#Calculation
#SigmaM_C=0
A_t=M/ACbar;#kN
#SigmaF_t=m*abar_t
#alpha=14.81-6.54*cos(theta);
A_n=(m*1000**-1)*ACbar*wsquare_30;#kN
A_t=(m*1000**-1)*BDbar*alpha_30;#kN
#SigmaM_A=m*abar*d
B=((A_n*(ABbar-0.6)*math.cos(theta*math.pi/180))+(A_t*0.6))/(ABbar*math.cos(theta*math.pi/180));#kN
print"\nThe force in the link DB,B=%1.2f kN"%B;

The force in the link DB,B=2.14 kN


## SAMPLE PROBLEM 6/3,PAGE NUMBER:432¶

In [3]:
import math
from scipy.optimize import fsolve
# Given data
W_b=644;# lb
r_i=12;# inch
r_o=24;# inch
theta=45;# degree
P=400;# lb
k_o=18;# inch
W=322;# lb
g=32.2;# lb

# Calculation
# Solution 1
# I=k**2*m
k=k_o*12**-1;# ft**2
Ibar=k**2*(W/g);# lb-ft-sec**2
def equations(p):
T,alpha,a=p
# SigmaM_G=Ibar*alpha
# SigmaF_y=m*a_y
# a_t=r*a;
return((((P*(r_o/12))-(T*(r_i/12)))-(Ibar*alpha)),((T-W_b)-((W_b/g)*a)),(a-((r_i/12)*alpha)))
T,alpha,a=fsolve(equations,(100,1,1))
# SigmaF_x=0
O_x=P*math.cos(theta*math.pi/180);# lb
# SigmaF_y=0
O_y=W+T+(P*math.sin(theta*math.pi/180));# lb
O=math.sqrt(O_x**2+O_y**2);# lb
print"\nSolution I:T=%3.0f lb,alpha=%1.2f rad/sec**2,a=%1.2f ft/sec**2,O=%4.0f lb"%(T,alpha,a,O);
# Solution 2
def equations(p):
alpha,a=p
# SigmaM_o=(Ibar*alpha)+(m*abar*d)
# a_t=r*a;
return((((P*(r_o/12))-(W_b*(r_i/12)))-((Ibar*alpha)+((W_b/g)*a*(r_i/12)))),(a-((r_i/12)*alpha)))
alpha,a=fsolve(equations,(1,1))
# SigmaF_y=Sigmam*(a_ybar)
O_y=(W+W_b+(P*math.sin(theta*math.pi/180)))+(((W/g)*(0))+((W_b/g)*alpha));# lb
# SigmaF_x=Sigmam*(a_xbar)
O_x=P*math.sin(theta*math.pi/180);# lb

Solution I:T=717 lb,alpha=3.67 rad/sec**2,a=3.67 ft/sec**2,O=1352 lb


## SAMPLE PROBLEM 6/4,PAGE NUMBER:433¶

In [4]:
import math
# Given data
m=7.5;# kg
rbar=0.250;# m
k_o=0.295;# m
theta_1=0;# degree
theta_2=60;# degree
g=9.81;# The acceleration due to gravity in m/s**2

# Calculation
# SigmaM_o=I_o*alpha;
# alpha=28.2*cos(theta);
# SigmaF_n=m*rbar*omega**2;
O_n=(m*rbar*wsquare)+(m*g*math.sin(theta_2*math.pi/180));# N
# SigmaF_t=m*rbar*alpha;
O_t=(m*g*math.cos(theta_2*math.pi/180))-(m*rbar*28.2*math.cos(theta_2*math.pi/180));# N
O=math.sqrt(O_n**2+O_t**2);# N
q=k_o**2/(rbar);# The distance in m
# SigmaM_Q=0
O_t=(m*g*math.cos(theta_2*math.pi/180)*(q-rbar))/q;# N
print"\nThe total force supported by the bearing,O=%3.1f N \nO_t=%2.2f N"%(O,O_t);

The total force supported by the bearing,O=155.6 N
O_t=10.37 N


## SAMPLE PROBLEM 6/5,PAGE NUMBER:446¶

In [5]:
import math
# Given data
r=6*12**-1;# ft
mu_s=0.15;# The coefficients of static friction
mu_k=0.12;# The coefficients of kinetic friction
theta=20;# degree
g=32.2;# The acceleration due to gravity in ft/sec**2
x=10;# ft

# Calculation
# SigmaF_x=m*abar_x----> mg*math.sin(theta*math.pi/180)-F=m*abar
# SigmaF_x=m*abar_y----> N-mg*math.cos(theta*math.pi/180)=0
# SigmaM_G=Ibar*alpha---> F*r=m*r**2*alpha
abar=(g/2)*math.sin(theta*math.pi/180);# ft/sec**2
# SigmaM_G=Ibar*alpha+m*abar*d----->mgr*sin(theta)=mr**2*(abar/r)+ m*abar*r
# From the above equations,we solve using the coefficients of mg
F=math.sin(theta*math.pi/180)-(math.sin(theta*math.pi/180))/2;# N
N=math.cos(theta*math.pi/180);# N
F_max=mu_s*N;# N
F=mu_k*N;# N
# SigmaF_x=m*abar_x
abar=(math.sin(theta*math.pi/180)-F)*g;# ft/sec**2
t=math.sqrt((2*x)/abar);# sec
print"\nThe angular acceleration of the hoop,alpha=%1.2f ft/sec**2 \nThe time t for the hoop to move a distance of 10 ft down the incline,t=%1.3f sec"%(alpha,t);

The angular acceleration of the hoop,alpha=7.26 ft/sec**2
The time t for the hoop to move a distance of 10 ft down the incline,t=1.646 sec


## SAMPLE PROBLEM 6/6,PAGE NUMBER:447¶

In [6]:
# SAMPLE PROBLEM 6/6
import math
from scipy.optimize import fsolve
# Given data
m=70;# kg
k=0.250;# The radius of gyration in m
mu_s=0.25;# The coefficient of static friction
g=9.81;# The acceleration due to gravity in m/s**2
DCbar=0.30;# m
r_A=0.250;# m
r_Bi=0.150;# m
r_Bo=0.450;# m

# Calculation
a_t=r_A*alpha_0;# m/s**2
abar=r_Bo*alpha;# m/s**2
def equations(p):
F,T=p
N=(m*g);# N
# SigmaF_x=m*abar_x
# SigmaM_G=Ibar*alpha
return(((F-T)-(m*-abar)),(((r_Bo*F)-(r_Bi*T))-(m*k**2*alpha)))
F,T=fsolve(equations,(10,100))
print"\nThe tension in the cable,T=%3.1f N \nThe friction force exerted by the horizontal surface on the spool,F=%2.1f N"%(T,F);
N=(m*g);# N
F_max=mu_s*N;# N
# If the coefficient of static friction had been 0.1
mu_s=0.1;# The coefficient of static friction
F=mu_s*(m*g);# N
# SigmaM_C=Ibar*alpha + m*abar*r
T=((m*(r_A**2)*alpha)+(m*abar*r_Bo))/DCbar;# N
print"\nThe tension in the cable,T=%3.1f N"%T;

The tension in the cable,T=154.6 N
The friction force exerted by the horizontal surface on the spool,F=75.8 N

The tension in the cable,T=154.6 N


## SAMPLE PROBLEM 6/7,PAGE NUMBER:448¶

In [7]:
import math
from scipy.optimize import fsolve
# Given data
W=60;# lb
theta=30*math.pi/180;# degree
F=30;# lb
BGbar=2;# ft
AGbar=2;# ft
l=4;# ft
g=32.2;# The acceleration due to gravity in ft/sec**2

# Calculation
# abar_x=abar*cos(theta)=1.732*alpha;
# abar_y=abar*sin(theta)=1.0*alpha;
def equations(p):
A,B,alpha=p
# SigmaM_G=Ibar*alpha;
# SigmaF_x=m*abar_x;
# SigmaF_y=m*abar_y;
return((((F*(2*math.cos(theta)))-(A*(AGbar*math.sin(theta)))+(B*(BGbar*math.cos(theta))))-((1*12**-1)*(W/g)*l**2*alpha)),((F-B)-((W/g)*(2*math.cos(theta)*alpha))),((A-W)-((W/g)*2*math.sin(theta)*alpha)));
A,B,alpha=fsolve(equations,(10,10,1))
print"\nThe forces on the small end rollers ,A=%2.1f lb and B=%2.2f lb \nThe resulting angular acceleration of the bar,alpha=%1.2f rad/sec**2"%(A,B,alpha);
# Alternative solution
# SigmaM_C=(Ibar*alpha)+(Sigma m*abar*d)
# SigmaF_x=m*abar_x;
abar_y=2*alpha*math.sin(theta);# ft
A=((W/g)*abar_y)+W;# lb
# SigmaF_x=m*abar_x;
abar_x=2*alpha*math.cos(theta);# ft
B=F-((W/g)*abar_x);# lb
print"\nAlternative solution: \nThe forces on the small end rollers ,A=%2.1f lb and B=%2.2f lb \nThe resulting angular acceleration of the bar,alpha=%1.2f rad/sec**2"%(A,B,alpha);

The forces on the small end rollers ,A=68.2 lb and B=15.74 lb
The resulting angular acceleration of the bar,alpha=4.42 rad/sec**2

Alternative solution:
The forces on the small end rollers ,A=68.2 lb and B=15.74 lb
The resulting angular acceleration of the bar,alpha=4.42 rad/sec**2


## SAMPLE PROBLEM 6/9,PAGE NUMBER:464¶

In [8]:
import math
# Given data
F=100;# N
m=40;# kg
k=0.150;# m
theta=15;# degree
r_i=0.100;# m
r_o=0.200;# m
l=3;# The distance in m
g=9.81;# The acceleration due to gravity in m/s**2

# Calculation
W=m*g;# N
l=(r_o+r_i)/r_i;# m
U_12=(F*((r_o+r_i)/r_i)*l)-((W*math.sin(theta*math.pi/180)*l));# J
T_1=0;# J
# T_2=((1/2)*m*vbar**2)+((1/2)*Ibar*omega**2);
# The work-energy equation gives
# Alternatively, the kinetic energy of the wheel may be written
# T=(1/2)*I_C*omega**2
P_100=F*(r_o+r_i)*omega;# W
print"The power input,P=%3.0f W"%P_100;

The power input,P=908 W


## SAMPLE PROBLEM 6/10,PAGE NUMBER:465¶

In [9]:
import math
# Given data
l=4;# ft
W=40;# The weight of the slender bar in N
theta=30;# degree
k=30;# The stiffness of the spring in lb/in
ABbar=24;# inch
BDbar=24;# inch
h=-2;# inch
g=32.2;# The acceleration due to gravity in ft/sec**2

# Calculation
# (a)
# T=[[(1/2)*m*v**2]+((1/2)*I_G*omega**2)];
# T=1.449*omega**2;
T_1=0;# ft-lb
U_12=0;# ft-lb
V_1=0;# ft-lb
V_2=W*((2*math.cos(theta*math.pi/180))-2);# ft-lb
# We now substitute into the energy equation and obtain
# (b)
x=ABbar-18;# ft
V_1=0;# ft-lb
V_3=(1/2)*k*(x**2)/12;# ft-lb
# T=(1/2)*I_A*omega**2;
# T_3=0.828*v_B**2;
U_13=0;# ft-lb
# The final gravitational potential energy is
V_3p=W*h;# ft-lb
v_B=math.sqrt(((T_1+V_1+U_13)-(V_3+V_3p))/0.828);# ft-lb
print"\n(a)The angular velocity of the bar,omega=%1.2f rad/sec \n(b)The velocity with which B strikes the horizontal surface,v_B=%1.2f ft/sec"%(omega,v_B);

(a)The angular velocity of the bar,omega=2.72 rad/sec
(b)The velocity with which B strikes the horizontal surface,v_B=9.83 ft/sec


## SAMPLE PROBLEM 6/11,PAGE NUMBER:466¶

In [10]:
import math
from scipy.optimize import fsolve
# Given data
m=30;# kg
k=0.100;# m
m_OB=10;# kg
m_c=7;# kg
K=30;# kN/m
theta=45;# degree
l=0.375;# m
g=9.81;# m/s**2

# Calculation
# (a)
# T_2=[2*((1/2)*I_G*omega**2]+[(1/2)*m*v**2];
# T_2= 6.83*v_B**2;
T_1=0;# J
l_b=l/math.sqrt(2);# m
V_1=(2*m_OB*g*(l_b/2))+(m_c*g*l_b);# J
V_2=0;# J
U_12=0;# J
v_B=math.sqrt(((T_1+V_1+U_12)-(V_2))/6.83);# m/s
# (b)
T_3=0;# J
U_13=0;# J
def equations(p):
x=p;
return((T_1+V_1+U_13)-(T_3+((-2*m_OB*g*(x/2))-(m_c*g*x)+(K*10**3*x**2/2))));
x=fsolve(equations,(1))
x=x*1000;# mm
print"\n(a)The velocity of the collar as it first strikes the spring,v_B=%1.2f m/s \n(b)The maximum deformation of the spring,x=%2.1f mm"%(v_B,x);

(a)The velocity of the collar as it first strikes the spring,v_B=2.54 m/s
(b)The maximum deformation of the spring,x=60.1 mm


## SAMPLE PROBLEM 6/12,PAGE NUMBER:479¶

In [11]:
import math
# Given data
m_A=3;# kg
m=2;# kg
k=0.060;# The radius of gyration in m
k=1.2;# The spring stiffness in kN/m
F=80;# N
g=9.81;# The acceleration due to gravity in m/s**2

# Calculation
# dT_rack=3a dx
# dT_gear=0.781a dx
# dV_rack=29.4 dx
# dV_gear=9.81 dx
# dV_spring=24 dx
# Canceling dx and solving for a give
a=(80-(29.4+9.81+24))/(3+0.781);
print"\nThe acceleration of rack A,a=%1.2f m/s**2"%a;

The acceleration of rack A,a=4.44 m/s**2


## SAMPLE PROBLEM 6/14,PAGE NUMBER:491¶

In [12]:
import math
import sympy
from scipy.optimize import fsolve
# Given data
# P=1.5*t;
r_i=9*12**-1;# ft
r_o=18*12**-1;# ft
t_1=0;# s
t_2=10;# s
k=10*12**-1;# ft
W=120;# lb
g=32.2;# The acceleration due to gravity in ft/sec**2
v_1=-3;# ft/sec

# Calculation
def equations(p):
F,omega_2=p;
f_1=(((W/g)*v_1)+((1.5*((t_2-t_1)**2)*2**-1)-(F*(t_2-t_1)))-((W/g)*r_o*omega_2));
f_2=((((W/g)*k**2*(v_1/r_o))+(((r_o*F*(t_2-t_1))-(r_i*1.5*(t_2-t_1)**2)*2**-1)))-((W/g)*k**2*omega_2));
return(f_1,f_2)
F,omega_2=fsolve(equations,(10,1))
print"\nThe angular velocity of the wheel,omega_2=%1.2f rad/sec"%omega_2;

The angular velocity of the wheel,omega_2=3.13 rad/sec


## SAMPLE PROBLEM 6/15,PAGE NUMBER:492¶

In [13]:
import math
# Given data
m_E=30;# kg
m_D=40;# kg
v_1=1.2;# m/s
t_1=0;# s
t_2=5;# s
F=380;# N
d=0.375;# m
k_o=0.250;# m
g=9.81;# m/s**2

# Calculation
# [H_O1+(integral(t_2 to t_2))SigmaM_Odt=H_O2]
# Integrating we get
M=((((F*0.750)*t_2)-(((m_E+m_D)*g*d)*t_2))-(((F*0.750)*t_1)-(((m_E+m_D)*g*d)*t_1)));# N.m.s
Ibar=(m_E)*k_o**2;# kg-m**2
H_O1=-((m_E+m_D)*v_1*d)-(Ibar*(v_1/d));# N.m.s
# H_O2=-(m_E+m_D*v_2*d)-(Ibar*(v_2/d));
# H_O2=11.72*omega_2;
# Substituting into the momentum equation gives
omega_2=(H_O1+M)/11.72;# N.m.s
# [G_1+(integral(t_2 to t_2))SigmaFdt=G_2]
m=m_E+m_D;# kg
G_1=m*-(v_1);# (kg.m/s)
G_2=m*(d*omega_2);# (kg.m/s)
# Integrating
# SigmaF=[T*(t_2)+(F*t_2)-(m*g*t_2)]-[T*(t_1)+(F*t_1)-(m*g*t_1)];
T=((G_2-G_1)-(((F*t_2)-(m*g*t_2))-((F*t_1)-(m*g*t_1))))/(t_2-t_1);# N
print"\nThe angular velocity,omega_2=%1.2f rad/s counter clockwise \nThe tension in the cable,T=%3.0f N"%(omega_2,T);

The angular velocity,omega_2=8.53 rad/s counter clockwise
The tension in the cable,T=368 N