Transcript reviewproblemschapter4

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
NB[ 5 sin ( 40deg )  15( 2.5 cos ( 40deg )
 0
NB  8.938lb
a
Fx  0
T cos ( 10deg )  8.938  0
Namas Chandra
Introduction to Mechanical engineering
T  9.08lb
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
2 NB( 40)  180000( 31)  22000( 38)  50000( 40)  0
a
Fy  0
NB  105200 lb
NA  2( 105200)  180000  22000  50000  0
NA  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:
Mx  0
Cx( 0.6)  800( 0.3)  400( 0.3)  0
Cx  600 N
My  0
Bx( 0.8)  600( 0.4)  400( 0.4)  0
Bx  500 N
Fx  0
A z  600  500  800  400  600  0
A z  700 N
Fx  0
Bx  0
Namas Chandra
Introduction to Mechanical engineering
Fy  0
By  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
2i  3j  6k
TBC  700

 2
2
2
 2  ( 3)  6 
TDE  TDE 


2
2
2
(

3
)

(

6
)

2


3i  6j  2k
FA  A x i  A y  j  A z k
Namas Chandra
Introduction to Mechanical engineering
TBC  ( 200i  300j  600k)  lb
TDE 
3
7
 TDE i 
6
7
 TDE j 
2
7
 TDE k
W  ( W  k)lb
Hibbler
Chapter 2-6
EML 3004C
Problem 4-70 (continued)
Force Summation
F  0
( 200i  300j  600k) 
  3  T  i  6  T  j  2  T  k    A  i  A  j  A  k  0
  7 DE 7 DE
  x
y
z 
7 DE  

 200  3  T  A  i   300  6 T  A  j   600  2 T  A  W   0




z
y
z
7 DE
7 DE
7 DE

 
 

Equilizing i, j, k components

Fx  0
200 

Fy  0
300 

Fz  0
600 
3
T
 A x  0
7 DE
Namas Chandra
Introduction to Mechanical engineering
6
T
 A y  0
7 DE
2
T
 A z  W  0
7 DE
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   ( 8j)  ( W  k)  0

DE
DE
DE 
7
7
7


 1800  12 T  8W  i 


DE
7


 18 T  600  k  0


DE
7


Equating i and k components

M x  0
1800 

M z  0
18
7
12
T
 8W  0
7 DE
TDE  600  0
Solving Equations
TDE  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
Fx.  0
4000 sin     Fa  Fb  0
Fy.  0
Ma  0
NA  NB  4000 cos     0
NB( 8)  4000 2 sin     4000 4 cos     0
Assuming Slipping occurs. Therefore:
FA  0.4NA
FB  0.4NB
Substituting
NA  1484.6 lb
NB  2228.3 lb
  21.8deg
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:
Fx  0
N  ( 600  500)  0
N  1100 lb
Fy  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
Fy  0
Case II: Only block A moves
Fx.  0
Fy.  0
N  600 cos ( 20 deg )  P cos ( 70 deg )  0
P sin ( 70 deg )  600 sin ( 20 deg )  0.3N  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:
Fx.  0
N  P sin    W  cos     0
Fy.  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 
P 
d
d
W  cos    sin     sin    cos    
cos     cos    sin     sin  
W  sin     
cos     
P 
W  sin       sin     
cos
   
2
W  sin       sin      0
W  sin       0
sin       0
    0
P 
W  sin     
cos     
Namas Chandra
Introduction to Mechanical engineering
  0
P  W  sin     
Hibbler
Chapter 2-13