[1] Q1. (Problem F2.24, page 63 ) Determine the resultant force at A
Q1. (Problem F2.24, page 63 ) Determine the resultant force at A
Formulation ! ! ! ! ! ! R A = Σ F = FAB + FAC = 600 uAB + 490 uAB Coordinates A(2,0,6)
B(0,4,2)
C(-4,3,4)
Position Vectors " 0 − 2 & " −2 & ! ! ! $ $ $ $ ! rAB = # 4 − 0 ' = # 4 ' = −2 i + 4 j − 4 k $ 2 − 6 $ $ −4 $ ( % ( % " −4 − 2 & " −6 & ! ! ! $ $ $ $ ! rAC = # 3 − 0 ' = # 3 ' = −6 i + 3 j − 2 k $ 4 − 6 $ $ −2 $ % ( % ( Unit Direction Vectors " −2 & " ! $ $ ! rAB 1 $ uAB = = # 4 '=# rAB 6 $ −4 $ $ % ( % " ! −6 &$ "$ ! rAC 1 $ uAC = = # 3 '=# rAC 7 $ −2 $ $ % ( %
rAC = (−2)2 + (4)2 + (−4)2 = 6
rAC = (−6)2 + (3)2 + (−2)2 = 7
& ! ! ! $ ' = −0.3333 i + 0.6667 j − 0.6667 k $ ( −0.8575 &$ ! ! ! 0.4286 ' = −0.8575 i + 0.4286 j − 0.2857 k −0.2857 $ (
−0.3333 0.6667 −0.6667
Cartesian Forces " ! $ ! FAB = FB uAB = 600 lb # $ % " ! $ ! FAC = FC uAC = 490 lb # $ %
& " $ $ '=# $ $ ( % −0.8571 &$ "$ 0.4286 ' = # −0.2857 $ $ ( %
−0.3333 0.6667 −0.6667
& ! ! ! $ ' lb = −200 i + 400 j − 400 k lb $ ( −420 &$ ! ! ! 210 ' lb = −420 i + 210 j −140 k lb −140 $ (
−200 400 −400
(
)
(
)
Solution " −200 & " −420 & " −620 & ! ! ! ! ! ! $ $ $ $ $ $ R A = FAB + FAC = # 400 ' + # 210 ' = # 610 ' lb = −620 i + 610 j − 540 k lb $ −400 $ $ −140 $ $ −540 $ % ( % ( % (
)
(
R A = (−620)2 + (610)2 + (−540)2 = 1024 lb (Magnitude) * o " −620 & " −0.6056 & ! , 127.3 $ $ $ $ ! R 1 , uR = A = # 610 ' = # 0.5958 ' → , 53.4o A R A 1024 $ −540 $ $ −0.5275 $ ,, 121.8o % ( % ( +
/ / (Direction) / // .
[1]
Q2. (Problem 3.42, page 101) The ball D has a mass of 20 kg. If a force F =100N is applied horizontally to the ring at A, determine the dimension d so that the force in cable AC is zero. Y TB
FBD
F
θB
A
X
W
Geometry
Known Forces
Tan θB = (1.5+d)/2
W = 20 kg (9.8 m/s2) = 196 N
Tan θC = d/2
TC = 0
F = 100 N
Equilibrium Equations ΣFx = 0 − TBCos θB +100 = 0
(1)
ΣFy = 0
(2)
TBSin θB – 196 = 0
Solution (20) Divide Eq (2) by Eq (1) TBSin θB 196 = −TBCos θB –100 1.5 + d 2 d = 2(1.96) −1.5 = 2.42 m
Tan θB = 1.96 =
◃ Answer (10)
[2]
Q3. (Problem F4.14, p 144) (
!
!
!
!
Determine the magnitude of the moment of the force F = (300 i – 200 j +150 k) N about the OA axis.
Formulation ! ! ! ! ! MOA = MO • uOA = rOB × FB • uOA or... ! ! ! ! ! MOA = MA • uOA = rAB × FB • uOA Coordinates A(0.3, 0.4, 0) B(0.3, 0.4, -0.2) # 0.3 ' # 0 ! ! % % " % ! ! Position Vectors rOB = $ 0.4 ( m = 0.3 i + 0.4 j − 0.2 k m or rAB = $ 0 % −0.2 % % −0.2 & ) & # 300 ' ! ! ! % % " Force Vector FB = $ −200 ( N = 300 i − 200 j +150 k N % 150 % & ) # 0.3 ' # 0.6 ' ! % % % % ! rOA 1 Unit Direction Vector uOA = = $ 0.4 ( = $ 0.8 ( rOA (0.3)2 + (0.4)2 + (0)2 % 0 % % 0 % & ) & ) # 0.3 ' # 300 ' # 0.6 ' ! ! % % % % % % ! Triple Scalar Product rOB × FB • uOA = $ 0.4 ( × $ −200 ( • $ 0.8 ( % −0.2 % % 150 % % 0 % & ) & ) & ) # 0 ' # 300 ' # 0.6 ' ! ! % % % % % % ! rAB × FB • uOA = $ 0 ( × $ −200 ( • $ 0.8 ( % −0.2 % % 150 % % 0 % & ) & ) & )
(
(
)
' % (m % )
)
Solve by separate vector products (× → •) . . . or by determinant # 0.3 ' # 300 ' # 20 ' # 0.6 ' ! % % % % % % % % ! rOB × FB = $ 0.4 ( × $ −200 ( = $ −105 ( → • $ 0.8 ( = −72 N.m ◃ Answer % −0.2 % % 150 % % −180 % % 0 % & ) & ) & ) & ) # 0 ' # 300 ' # −40 ' # 0.6 ' ! % % % % % % % % ! rAB × FB = $ 0 ( × $ −200 ( = $ −60 ( → • $ 0.8 ( = −72 N.m ◃ Answer % −0.2 % % 150 % % 0 % % 0 % & ) & ) & ) & )
[3]
Formulation ! ! ! ! ! ! R A = Σ F = FAB + FAC = 600 uAB + 490 uAB Coordinates A(2,0,6)
B(0,4,2)
C(-4,3,4)
Position Vectors " 0 − 2 & " −2 & ! ! ! $ $ $ $ ! rAB = # 4 − 0 ' = # 4 ' = −2 i + 4 j − 4 k $ 2 − 6 $ $ −4 $ ( % ( % " −4 − 2 & " −6 & ! ! ! $ $ $ $ ! rAC = # 3 − 0 ' = # 3 ' = −6 i + 3 j − 2 k $ 4 − 6 $ $ −2 $ % ( % ( Unit Direction Vectors " −2 & " ! $ $ ! rAB 1 $ uAB = = # 4 '=# rAB 6 $ −4 $ $ % ( % " ! −6 &$ "$ ! rAC 1 $ uAC = = # 3 '=# rAC 7 $ −2 $ $ % ( %
rAC = (−2)2 + (4)2 + (−4)2 = 6
rAC = (−6)2 + (3)2 + (−2)2 = 7
& ! ! ! $ ' = −0.3333 i + 0.6667 j − 0.6667 k $ ( −0.8575 &$ ! ! ! 0.4286 ' = −0.8575 i + 0.4286 j − 0.2857 k −0.2857 $ (
−0.3333 0.6667 −0.6667
Cartesian Forces " ! $ ! FAB = FB uAB = 600 lb # $ % " ! $ ! FAC = FC uAC = 490 lb # $ %
& " $ $ '=# $ $ ( % −0.8571 &$ "$ 0.4286 ' = # −0.2857 $ $ ( %
−0.3333 0.6667 −0.6667
& ! ! ! $ ' lb = −200 i + 400 j − 400 k lb $ ( −420 &$ ! ! ! 210 ' lb = −420 i + 210 j −140 k lb −140 $ (
−200 400 −400
(
)
(
)
Solution " −200 & " −420 & " −620 & ! ! ! ! ! ! $ $ $ $ $ $ R A = FAB + FAC = # 400 ' + # 210 ' = # 610 ' lb = −620 i + 610 j − 540 k lb $ −400 $ $ −140 $ $ −540 $ % ( % ( % (
)
(
R A = (−620)2 + (610)2 + (−540)2 = 1024 lb (Magnitude) * o " −620 & " −0.6056 & ! , 127.3 $ $ $ $ ! R 1 , uR = A = # 610 ' = # 0.5958 ' → , 53.4o A R A 1024 $ −540 $ $ −0.5275 $ ,, 121.8o % ( % ( +
/ / (Direction) / // .
[1]
Q2. (Problem 3.42, page 101) The ball D has a mass of 20 kg. If a force F =100N is applied horizontally to the ring at A, determine the dimension d so that the force in cable AC is zero. Y TB
FBD
F
θB
A
X
W
Geometry
Known Forces
Tan θB = (1.5+d)/2
W = 20 kg (9.8 m/s2) = 196 N
Tan θC = d/2
TC = 0
F = 100 N
Equilibrium Equations ΣFx = 0 − TBCos θB +100 = 0
(1)
ΣFy = 0
(2)
TBSin θB – 196 = 0
Solution (20) Divide Eq (2) by Eq (1) TBSin θB 196 = −TBCos θB –100 1.5 + d 2 d = 2(1.96) −1.5 = 2.42 m
Tan θB = 1.96 =
◃ Answer (10)
[2]
Q3. (Problem F4.14, p 144) (
!
!
!
!
Determine the magnitude of the moment of the force F = (300 i – 200 j +150 k) N about the OA axis.
Formulation ! ! ! ! ! MOA = MO • uOA = rOB × FB • uOA or... ! ! ! ! ! MOA = MA • uOA = rAB × FB • uOA Coordinates A(0.3, 0.4, 0) B(0.3, 0.4, -0.2) # 0.3 ' # 0 ! ! % % " % ! ! Position Vectors rOB = $ 0.4 ( m = 0.3 i + 0.4 j − 0.2 k m or rAB = $ 0 % −0.2 % % −0.2 & ) & # 300 ' ! ! ! % % " Force Vector FB = $ −200 ( N = 300 i − 200 j +150 k N % 150 % & ) # 0.3 ' # 0.6 ' ! % % % % ! rOA 1 Unit Direction Vector uOA = = $ 0.4 ( = $ 0.8 ( rOA (0.3)2 + (0.4)2 + (0)2 % 0 % % 0 % & ) & ) # 0.3 ' # 300 ' # 0.6 ' ! ! % % % % % % ! Triple Scalar Product rOB × FB • uOA = $ 0.4 ( × $ −200 ( • $ 0.8 ( % −0.2 % % 150 % % 0 % & ) & ) & ) # 0 ' # 300 ' # 0.6 ' ! ! % % % % % % ! rAB × FB • uOA = $ 0 ( × $ −200 ( • $ 0.8 ( % −0.2 % % 150 % % 0 % & ) & ) & )
(
(
)
' % (m % )
)
Solve by separate vector products (× → •) . . . or by determinant # 0.3 ' # 300 ' # 20 ' # 0.6 ' ! % % % % % % % % ! rOB × FB = $ 0.4 ( × $ −200 ( = $ −105 ( → • $ 0.8 ( = −72 N.m ◃ Answer % −0.2 % % 150 % % −180 % % 0 % & ) & ) & ) & ) # 0 ' # 300 ' # −40 ' # 0.6 ' ! % % % % % % % % ! rAB × FB = $ 0 ( × $ −200 ( = $ −60 ( → • $ 0.8 ( = −72 N.m ◃ Answer % −0.2 % % 150 % % 0 % % 0 % & ) & ) & ) & )
[3]
