EML 3004C
Problem 2-12 (page 26)
The cable exerts a force of 600 N on the frame.
Resolve this force into components acting (a) along
the x and y axes and (b) along the u and v axes.
What is the magnitude of each component?
Solution:
a)
Fx  600cos( 75de g)  N
Fx  155.291N
Fy  600s in( 75de g) N
Fy  579.555N
Fu  600cos( 45de g)  N
Fu  424.264N
Fv  600s in( 45de g) N
Fv  424.264N
b)
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-1
EML 3004C
Problem 2-16 (page 26)
If the resultant FR of the two forces acting on the jet
aircraft is to be directed along the positive x axis and have
a magnitude of 10 kN, determine the angle  of the cable
attached to the truck at B such that the force FB in this
cable in a minimum. What is the magnitude of force in
each cable when this occurs?
Solution:
  90de g  20de g
  70de g
3
FB  3.42 10 N
3
Fc  9.397 10 N
FB  10s in( 20de g)  10 N
Fc  10cos( 20de g)  10  N
Namas Chandra
Introduction to Mechanical engineering
3
3
Hibbler
Chapter 2-2
EML 3004C
Problem 2-29 (page 35)
Three forces act on the bracket.
Determine the magnitude and orientation
 of F2 so that the resultant force is
directed along the positive u axis and has
a magnitude of 50 lb.
Solution:
5
Sum mation of
( 1)


50cos( 25de g)  80  Fs  cos 25de g   
 52
Force s in the
13
X direction.
F2  cos 25de g     54.68
Sum mation of
Force s in the
Namas
Chandra
Hibbler
Y direction.
Introduction to Mechanical engineering
Chapter 2-3
EML 3004C
Problem 2-29 continued
5
Sum m at ion of
50cos( 25de g)  80  Fs  cos 25de g    
 52
Force s in the
13
X dir ection.
F2  cos 25de g     54.68
Sum m at ion of
5
 52
Force s in the 50 s in( 25de g)  Fs  cos 25de g    
13
Y dir ection.
F2  s in 25de g     69.13
( 2)
Eq 1 and 2 yields :
tan 25de g     1.2642
25de g    128.34
  103de g
Substituting into Eqn 1 or 2 yie lds:
F2  88.1 lb
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-4
EML 3004C
Problem 2-42 (page 47)
The pipe is subjected to the force F which
has components acting along the x, y, z
axes as shown. If the magnitude of F is 12
kN, and α = 120 deg and γ = 45 deg,
determine the magnitudes of its three
components.
Solution:
2  42
2
2
2
cos   cos   cos   1
2
2
2
cos 120de g  cos   cos 45de g  1
Fr om the Figure , cos = + 0.5
cos  0.5 plus or m inus
    0 de g
Fx  F cos 
Fx  12 cos 120de g
Fx  6  k N
Fy  F cos 
Fy  12 cos 60de g
Fx  6  k N
Fz  F cos 
Fz  12 cos 45de g
Fx  8.49 k N
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-5
EML 3004C
Problem 2-48 (page 55)
Express the position vector r in
Cartesian vector form; then determine
its magnitude and coordinate direction
angles.
Solution:
Pos ition vector:
M agnitude :
Namas Chandra
Introduction to Mechanical engineering
R = (4 - 0) i +[ - 4 - ( - 2)] j + ( 6 - 3) k
= { 4i - 2j + 3k } m
r 
2
2
2
4  ( 2)  3
r  5.39 m
Hibbler
Chapter 2-6
EML 3004C
Problem 2-48 continued
Coor dinat e dire ction angles :
R
ur 
r
ur 
4i  2j  3k
5.385
ur  0.74281i 0.3714j 0.5571k
cos   0.7428
  42 de g
cos   0.3714
  112 de g
cos   0.5517
  56.1 de g
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-7
EML 3004C
Problem 2-59 (page 57)
Express each of the two
forces in Cartesian vector
form and then determine the
magnitude and coordinate
direction angles of the
resultant force.
Solution:
2  59
rab  ( 0  4)  i  ( 8  8)  j  [ 0  ( 12) ]  k
rab  ( 4i  0j  12k)ft
rAB 
( 4)
2
2
 0
 ( 12)
2
rab
F1v  F1 
rAB
rAB  12.649ft
 4i  0j  12k 

12.649


F1v  12
F1v  ( 3.79i 11.38k)  l b
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-8
EML 3004C
Problem 2.59 continued 1
rac  ( 5  4)i  ( 8  8) j  [ 4  ( 12) ]k
rac  ( 9i  16j  16k)ft
rAC 
2
2
( 9)  ( 16)  ( 16)
2
rAC  24.352ft
9i  16j  16k 

F1v  18

24.352


rac
F2v  F2 
rAC
F2v  ( 6.65i 11.8j 11.8k)  lb
FR  F1  F2
FR  ( 3.79i 11.38k)  ( 6.65i 11.82j 11.82k)
FR  ( 10.44i  11.82j 23.21k)  lb
Fr 
 10.442  11.822  23.212
Namas Chandra
Introduction to Mechanical engineering
FR  28.067lb

FR  28.1 lb
Hibbler
Chapter 2-9
EML 3004C
Problem 2-59 continued 2
Coor dinate dir ection angles :
FR
ur 
Fr
ur 
10.44i 11.82j 23.21k
28.067
ur  0.37i 0.42j 0.82k
cos   0.37
  112 de g
cos   0.42
  115 de g
cos   0.82
  34.2 de g
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-10
EML 3004C
Problem 2-65 (page 59)
The cylindrical vessel is supported by three cables
which are concurrent at point D. Express each force
which the cables exert on the vessel as a Cartesian
vector, and determine the magnitude and coordinate
direction angles of the resultant force.
Solution:
2  65
ra  ( 0  0.75)  i  ( 0  0)  j  ( 3  0)  k
ra  ( 0.75i 0j  3k)m
rA 
2
2
( 0.75)  0  ( 3)
2
ra
FAv  FA
rA
rA  3.0923m
 0.75i 0j  3k 

3.0923


FAv  6
FAv  ( 1.461i 5.82k)  k N
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-11
EML 3004C
Problem 2-65 continued 1
rc  [ 0  ( 0.75s in 45de g) ]i  [ 0  ( 0.75cos 45de g) ] j  ( 3  0)k
rc  ( 0.5303i 0.5303j 3k)m
rC 
2
2
( 0.5303)  ( 0.5303)  ( 3)
2
rC  3.0923
 0.5303i 0.5303j 3k 

3.0923


rc
FC v  FC 
rC
FC v  5
FC v  ( 0.857i 0.857j 4.85k)  k N
rb  [ 0  ( 0.75s in 30de g) ]i  [ 0  ( 0.75cos 30de g) ] j  ( 3  0)k
rb  ( 0.375i 0.6495j 3k)m
rB 
2
2
( 0.375)  ( 0.6495)  ( 3)
rc
FBv  FB 
rC
2
rB  3.0923
 0.375i 0.6495j 3k 

3.0923


FBv  8
FBv  ( 0.970i 1.68j 7.76k)  k N
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-12
EML 3004C
Problem 2-65 continued 2
Res ultant For ce Vector :
FR  FA  FB  FC
FR  ( 1.45i 5.8208k)  ( 0.97i 1.68j 7.76k)  ( 0.85i 0.85j 4.85k)
FR  ( 0.3724i 0.8228j 18.4326k)  k N
Fr 
0.37242  ( 0.8228) 2  18.43262
Namas Chandra
Introduction to Mechanical engineering
FR  18.4547lb

FR  18.5 lb
Hibbler
Chapter 2-13
EML 3004C
Problem 2-65 continued 3
Coor dinate dire ction angle s:
FR
ur 
Fr
ur 
0.3724i 0.8228j 18.4326k
18.4547
ur  0.02018i  0.04458j  0.9988k
cos   0.02018
  88.8 de g
cos   0.04458
  92.6 de g
cos   0.9988
  2.81 de g
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-14
EML 3004C
Problem 2-78 (page 65)
Determine the magnitudes of the projected
components of the force F={-80i + 30j + 20k} lb in
the direction of the cables AB and AC.
Solution:
uAC 
uAB 
( 4  0)i  ( 3  0)j  ( 0  8)k
2
2
( 4  0)  ( 3  0)  ( 0  8)
uAC 
2
( 0  5)i  [ 0  ( 4) ]j  ( 8  0)k
2
2
( 0  5)  [ 0  ( 4) ]  ( 8  0)
Namas Chandra
Introduction to Mechanical engineering
2
uAC 
4i  3j  8k
89
5i  4j  8k
105
Hibbler
Chapter 2-15
EML 3004C
Problem 2-78 continued
FAC  F uAC
4i  3j  8k 

FAC  ( 80i  30j  20k)  

89


FAC 
( 80)  ( 4)  30 ( 3)  20 ( 8)
 lb
89
FAC  26.5lb
FBA  F uBA
5i  4j  8k 

FBA  ( 80i  30j  20k)  

105


FBA 
( 80)  ( 5)  30 ( 4)  20 ( 8)
 lb
105
FAC  26.5lb
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-16
EML 3004C
Problem 2-95 (page 68)
Express each of the three forces acting on the
column in Cartesian vector form and
determine the magnitude of the resultant force.
Solution:
F1  ( 140s in( 30de g) i  140cos( 30de g) j
F1  ( 70i  121j)  lb
F2  ( 180j)  lb
F3  ( 125cos( 45de g) i  125s in( 45de g) j)
FR  F1  F2  F3
F3  ( 88.4i 88.4j)  lb
FR  ( 70i  121.1j)  ( 180j)  ( 88.4i 88.4j)
FR  ( 18.38i 389.63j)  lb
M agnitude :
FR 
2
( 18.38)  ( 398.63)
2
FR  390 lb
Namas Chandra
Introduction to Mechanical engineering
Hibbler
Chapter 2-17