EML 3004C
Problem 3-6 (page 84)
A force of 30 lb is applied to the handle of the
wrench. Determine the moment of this force about
point 0.
Solution:
3-6
Mo   [ 3 0 si n( 3 0 d eg)  ( 0 .5)  3 0 co s( 3 0 d eg)  ( 7) ]  l b i n
Mo  1 74 .36 5
l b i n
Mo  1 4.5 3l b ft
Mo   1 4.5 3l
 b ft
Cl ock wi se
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-1
EML 3004C
Problem 3-17 (page 86)
Determine the moment of the force F at
point A about point P. Express the result
as a Cartesian vector.
Solution:
Posit ion V ec tor:
rp   [ 5  ( 3) ]i  [ 0  ( 6) ] j  ( 4  2)k
F orc e Vec t or:
F   2 0 
( 2  5)i  ( 8  0) j  [ 0  ( 4) ]k


2
2
2
 ( 2  5)  ( 8  0)  [ 0  ( 4) ] 
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-2
EML 3004C
Problem 3-17 continued
Moment about point P:
M p   rp  F
j
k 
 i
Mp    8
6
6 


 6 .36 0 1 6.9 60 8 .48 0
M p   [ 6( 8 .48 0)  1 6.9 60
( 6) ]i  [ 8( 8 .48 0)  ( 6 .36 0) ( 6) ] j  [ 8( 1 6.9 60)  ( 6 .36 0) ( 6) ]k
M p   ( 1 53i
  2 9.7 j  1 74k
 )l b ft
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-3
EML 3004C
Problem 3-22 (page 87)
Using Cartesian vectors, determine the moment
of each of the two forces acting on the pipe
assembly about point A. Add these moments
and calculate the magnitude and coordinate
direction angles of the resultant moment.
Solution:
Posit ion Vec tor:
r1  ( 0  0)i  ( 1 .2  0) j  ( 0 .8  0 .8)k
r1  ( 1 .2 j)  m
r2  ( 0 .4  0)i  ( 1 .2  0) j  ( 0 .8  0 .8)k
r2  ( 0 .4i  1 .2j)  m
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-4
EML 3004C
Problem 3-22 continued
Moment about point A:
M1  r1  F1
i j k 
M1   0 1 .2 0 


3
0
2
0

3
0


M1  [ 1 .2( 3 0)  2 0( 0) ]i  [ 0( 3 0)  ( 3 0) ( 0) ] j  [ 0( 2 0)  ( 3 0) ( 1 .2) ]k
M 1  ( 3 6 i  3 6k) N m
M 2  r2  F2
 i j k 
M 2   0 .4 1 .2 0 


 6 0 1 0 3 5
M2  [ 1 .2( 3 5)  1 0( 0) ]i  [ 0 .4( 3 5)  ( 6 0) ( 0) ] j  [ 0 .4( 1 0)  ( 6 0) ( 1 .2) ]k
Namas Chandra
Introduction to Mechanical engineering
M 2  (42i  14 j  68k) Nm
Hibbler
Chapter 2-5
EML 3004C
Problem 3-22 continued
R es ult ant moment about point A:
MR  
A
 MA
MR   M1  M2
A
1
M R   ( 3 6 i  3 6 k)  ( 4 2 i  1 4 j  6 8 k)
A
M R   ( 7 8i  1 4j 1 04k) N m
A
Magnit ude :
2
2
M R   ( 7 8)  ( 1 4)  ( 1 04)
2
A
M R   1 31N m
A
C oordinate direc tion angles :
co s    
7 8
co s    
1 4
co s    
1 04
Namas Chandra
Introduction to Mechanical engineering
1 31
1 31
1 31
   1 27d eg
   9 6.1d eg
   1 43d eg
Hibbler
Chapter 2-6
EML 3004C
Problem 3-38 (page 95)
Determine the moment that the force F
exerts about the y axis of the shaft. Solve
the problem using a Cartesian vector
approach and using a scalar approach.
Express the result as a Cartesian vector.
Solution:
Forc e Vect or:
F   1 6co s( 3 0) i  1 6si n( 3 0) k
F   ( 1 3.8 56 i 8k) N
Posit ion V ec tor:
r   0 .2co s( 4 5) i  0 .05 j 0 .2si n( 4 5) k
Namas Chandra
Introduction to Mechanical engineering
r   ( 0 .14 14 i 0 .05 j 0 .14 14
k)m
Hibbler
Chapter 2-7
EML 3004C
Problem 3-38 continued
Magnit ude of t he moment along y axis :
M y   j ( r  F)
1
0
 0


My   0 .14 14 0 .05 0 .14 14



1
3.8
56
0
8


M y   0  1[ 0 .14 148
( )  ( 1 3.8 56
) ( 0 .14 14
)]  0
My   0 .82 8N m
I n Cart es ian v ec t or f orm:
M y   ( 0 .82 8j
) N m
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-8
EML 3004C
Problem 3-38 continued
Sc alar Analy s is:
My  
 My
1
M y   [ 1 6 co s( 3 0 d eg)  0 .2 si n( 4 5 d eg)  1 6 si n( 3 0 d eg)  ( 0 .2 co s( 4 5 d eg) ) ]  N m
M y   0 .82 8N
 m
I n Cart es ian v ec tor f orm:
M y   ( 0 .82 8j
) N m
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-9
EML 3004C
Problem 3-50 (page 103)
The cord passing over the two small pegs
A and b of the board is subjected to a
tension of 10 lb. Determine the minimum
tension P and the orientation  of the cord
passing over the pegs C and D, so that the
resultant couple moment produced by the
two cords is 20 lb*in., clockwise.
Solution:
I n order t o y ield a max imum c ount erc loc kwise c ouple moment a minim um
f orc e P whic h ac t s perpendicular t o C D in needed. Thus. .. .
   4 5d eg
MR  
M
2 0   P( 3 0d eg)  1 0co s( 1 5d eg)  3 0
1
P   8 .99l b
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-10
EML 3004C
Problem 3-56 (page 103)
Determine the couple moment.
Express the result as a Cartesian Vector.
Solution:
F orc e Vect or:
F   7 0



2
2
2
[ 0  ( 6) ]  [ 2 2  ( 1 0) ]  ( 0  4) 
[ 0  ( 6) ]i  [ 2 2  ( 1 0) ] j  ( 0  4)k
F   ( 3 0i  6 0j  2 0k) N
Posit ion Vec tor:
r   ( 6  5)i  ( 1 0  8) j  [ 4  ( 3) ]k
Namas Chandra
Introduction to Mechanical engineering
r   ( 1 1i  1 8j  7k)m
Hibbler
Chapter 2-11
EML 3004C
Problem 3-56 continued
C ouple Mom ent Vec tor:
Mc   r  F
j
k 
 i
M c    1 1 1 8 7 


3
0

6
0

2
0


M c   [ 1 8( 2 0)  ( 6 0) ( 7) ]i  [ 1 1( 2 0)  ( 3 0) ( 7) ] j  [ 1 1( 6 0)  ( 3 0) ( 1 8) ]k
M c   ( 7 80 i 1 0j  1 20 0k) N m
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-12
EML 3004C
Problem 3-65 (page 118)
Replace the force and
couple moment system by a
equivalent force and couple
moment acting at point P.
Solution:
Forc e Summat ion:
FRx  
 Fx
3
FRx   2 co s( 4 5d eg)  3 si n( 3 0d eg)
FRx   0 .08 5791 0  N
FRy   2 si n( 4 5d eg)  3 co s( 3 0d eg)
FRy   1 .18 39
1 0 N
1
FRy  

Fy
3
1
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-13
EML 3004C
Problem 3-65 continued
2
2
FR  FRx  FRy
 FRy 
  atan

FRx
 
2
2
3
FR  ( 0.08579
)  ( 1.1839)  10  N
3
FR  1.187 10 N
  85.855deg
Moment Summation:
MR 
p
 Mp
MR  3cos( 30deg) 4  3sin( 30deg)  4  2cos( 45deg)  3
p
1
MR  21.6kN
 m
counterclockwise
p
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-14
EML 3004C
Problem 3-82 (page 121)
The frame is subjected to the coplanar system of
loads. Replace this system by an equivalent
resultant force and couple moment acting at point B.
Solution:
Forc e Sum mat ion:
FRx  
 Fx
FRx   4 0l b
FRx   4 0 l b
t o the lef t
1
FRy  
 Fy
FRy   ( 8 0  6 0  7 5)  l b
FRy  2 15l b
FRy   2 15l
 b
d ow n
1
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-15
EML 3004C
Problem 3-82 continued
2
2
FR   FRx  FRy
 FRy 
   at an

 FRx 
2
2
FR   ( 4 0)  ( 2 15)  N
FR  2 18 .68 9N
  7 9.4 61d eg
Moment Summation:
MR  
p
 Mp
M R   8 0 3  7 5 6  6 0 9  4 0 4
p
1
M R   1 39 0 l b ft
p
Namas Chandra
Introduction to Mechanical engineering
M R   1 .39k
 ip ft
c loc kwis e
p
Hibbler
Chapter 2-16
EML 3004C
Problem 2-100 (page 124)
The main beam along the wing of an airplane is swept
back at a angle of 26deg. From load calculations it is
determined that the beam is subjected to couple
moment Mx=17 kip*ft and My=25 kip*ft. Determine
the resultant couple moments created about the x’ and
y’ axes. The axes all like in the same horizontal
plane.
Solution:
M
xn
  Mx
xn
 My
xn
M xn   1 70 00co s( 2 5d eg)  2 50 00s i n( 2 5d eg)
M xn   4 84 2l
 b ft
My n   My
yn
M xn   4 .84k
 ip ft
 Mx
yn
M y n   2 50 00co s( 2 5d eg)  1 70 00s i n( 2 5d eg)
M y n   2 98 42
 l b ft
Namas Chandra
Introduction to Mechanical engineering
M y n   2 9.8 2k
 ip ft
Hibbler
Chapter 2-17