EML 3004C
Problem 4-4 (page 138)
Draw the free body diagram of the beam which
supports the 80-kg load and is supported by the pin
at A and a cable which wraps around the pulley at D.
Explain the significance of each force on the
diagram.
Solution:
80(9.81)N is the effect of the cable
(the weight of the object) on the beam.
T is the effect of the cable on the beam.
Ax and Ay are the effect of the pin
Support on the beam.
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-1
EML 3004C
Problem 4-8 (page 139)
Draw the free-body diagram of the winch,
which consists of a drum radius 4 in. It is
pin-connected at it center C, and at its
outer rim is a ratchet gear having a mean
radius of 6 in. The pawl AB serves as a
two-force member (short link) and holds
the drum for rotating.
Solution:
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-2
EML 3004C
Problem 4-23 (page 155)
The uniform rod AB has a weight
of 15 lb. Determine the force in the
cable when the rod is in the
position shown.
Solution:



 M 
 0
a
 F x  0
N B [ 5 sin ( 40 deg )  15 ( 2.5 cos ( 40 deg )
T  cos ( 10 deg )  8.938  0
Namas Chandra
Introduction to Mechanical engineering
 0
N B  8.938 lb
T  9.08 lb
Hibbler
Chapter 2-3
EML 3004C
Problem 4-32 (page 156)
Determine the resultant normal force acting on each
set of the wheels of the airplane. There is a set of
wheels in the front, A, and a set of wheels under
each wing, B. Both wings have total weight of 50
kip and center of gravity at Gw, the fuselage has a
weight of 180 kip and center of gravity at Gf, and
both engines(one on each side) have a weight 22 kip
and center of gravity at Ge.
Solution:



 M 
 0
a
2  N B ( 40 )  180000 ( 31 )  22000 ( 38 )  50000 ( 40 )  0
N B  105200  lb
 F y  0
N A  2 ( 105200 )  180000  22000  50000  0
N A  41600  lb
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-4
EML 3004C
Problem 4-57 (page 172)
The triangular plate is supported by a balland-socket joint at B and rollers at A and
C. Determine the x, y, z components of
each reaction at these supports due to the
loading shown.
Solution:
 M x  0
C x( 0.6 )  800 ( 0.3 )  400 ( 0.3 )  0
C x  600  N
 M y  0
 B x( 0.8 )  600 ( 0.4 )  400 ( 0.4 )  0
B x  500  N
 F x  0
A z  600  500  800  400  600  0
A z  700  N
 F x  0
B x  0
Namas Chandra
Introduction to Mechanical engineering
 F y  0
B y  0
Hibbler
Chapter 2-5
EML 3004C
Problem 4-70 (page 174)
Cable BC and DE can support a max. tension
fo 700 lb before it breaks. Determine the
greatest weight W that can be suspended from
the end of the boom. Also, determine the x, y,
z components of reaction at the ball-andsocket joint A.
Solution:
Assuming failure at cable BC
T BC  700
 2i  3j  6k

 2
2
2
 2  (  3)  6 

 3i  6j  2k

2
2
2
(

3
)

(

6
)

2


T DE  T DE 

F A  A x i  A y  j  A z k
Namas Chandra
Introduction to Mechanical engineering
T BC  ( 200i  300j  600 k )  lb
T DE 
3
7
 T DE  i 
6
7
 T DE  j 
2
7
 T DE  k
W  (  W  k ) lb
Hibbler
Chapter 2-6
EML 3004C
Problem 4-70 (continued)
Force Summation
 F  0
( 200i  300j  600 k ) 
   3  T  i  6  T  j  2  T  k    A  i  A  j  A  k  0
 x

DE
DE
DE  
y
z 
7
7
 7

 200  3  T
 i    300  6 T
 j   600  2 T


A

A




DE
z
DE
y
DE  A z  W   0
7
7
7

 



Equilizing i, j, k components

F x  0
200 

F y  0
 300 

F z  0
600 
3
7
T DE  A x  0
6
7
Namas Chandra
Introduction to Mechanical engineering
2
7
T DE  A y  0
T DE  A z  W  0
Hibbler
Chapter 2-7
EML 3004C
Problem 4-70 (continued)
Moment Summation

M a  0
( 3j ) 
  3 T  i  6 T  j  2 T  k   ( 8 j)  (  W  k )  0

DE
DE
DE 
7
7
7


 1800  12 T
 i   18 T
 k  0

8W

600




DE
DE
7
7

 

Equating i and k components

M x  0

M z  0
1800 
12
7
18
7
T DE  8W  0
T DE  600  0
Solving Equations
T DE  233.33  lb
233.33  700  lb
so assumption OK
W  275  lb
A z   392  lb
A x   100  lb
Namas Chandra
Introduction to Mechanical engineering
A y  500  lb
Hibbler
Chapter 2-8
EML 3004C
Problem 4-78 (page 189)
The car has a weight of 4000 lb and a center
of gravity at G. If it pulls off the side of a
road, determine the greatest angle of tilt, ,
it can have without slipping or tipping over.
The coeffiecient of static friction between
its wheels and the ground is 0.4.
Solution:
Equilibrium
 F x.  0
4000 sin     F a  F b  0
 F y.  0
 M a  0
N A  N B  4000 cos     0
N B ( 8 )  4000  2 sin     4000  4  cos     0
Assuming Slipping occurs. Therefore:
F A  0.4 N A
F B  0.4 N B
Substituting
N A  1484.6  lb
N B  2228.3  lb
  21.8 deg
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-9
EML 3004C
Problem 4-85 (page 190)
The two stone blocks have weights of
Wa=600lb and Wb=500lb. Determine the
smallest horizontal force P that must be applied
to block A in order to move it. The coeffeicient
of the static friction between the blocks is 0.3
and between the floor and each block is 0.4.
Solution:
Case I: Block A and Block B move together:
 F x  0
N  ( 600  500 )  0
N  1100  lb
 F y  0
0  0.5 ( 1100 )  P
P  550  lb
Namas
Chandra
Case
II: Only block A moves
Introduction to Mechanical engineering
Hibbler
Chapter 2-10
EML 3004C
Problem 4-85 continued
 F y  0
Case II: Only block A moves
 F x.  0
N  600 cos ( 20  deg )  P  cos ( 70  deg )  0
 F y.  0
P  sin ( 70  deg )  600  sin ( 20  deg )  0.3 N  0
Solving
P  447.2  lb
N  716.8  lb
Choose the smalles P among the two cases:
P  447  lb
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-11
EML 3004C
Problem 4-110 (page 194)
Determine the angle  at which the applied force
P should act on the log so that the magnitude of P
is as small as possible for pulling the log up the
incline. What is the corresponding value of P?
The log weighs W and the slope  is known.
Express the answer in terms of the angle of
kinetic friction,  = atan ().
Solution:
 F x.  0
N  P  sin    W  cos     0
 F y.  0
P  cos    W  sin     tan      W  cos     P  sin     0
P 
N  W  cos     P  sin  
W  sin     tan    cos    
cos    tan     sin  
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-12
EML 3004C
Problem 4-110 (continued)
P 
W  cos
cos
P 
W  sin     
cos
d
d
   sin     sin    cos    
    cos    sin     sin  
P 

 
W  sin       sin     
cos
2

 
W  sin       sin    
sin       0
P 
W  sin     
cos

 
Namas Chandra
Introduction to Mechanical engineering
 0
W  sin       0
    0
  0
P  W  sin     
Hibbler
Chapter 2-13