COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 116.
Free-Body Diagram:
Express all forces in terms of rectangular components:
rA = ( 2.4 m ) i
rB = (1.8 m ) j
uuur
AD = − ( 2.4 m ) i + ( 0.3 m ) j + (1.2 m ) k
uuur
BE = − (1.8 m ) i + ( 0.6 m ) j − ( 0.9 m ) k
W = − ( 880 N ) j
Then
uuur
ur
AD
= TAD
T AD = TAD
AD
uuur
ur
BE
T BE = TBE
= TBE
BE
− 2.4i + 0.3j + 1.2k
( − 2.4 )
2
2
+ ( 0.3) + (1.2 )
2
−1.8i + 0.6 j − 0.9k
( −1.8)
2
2
+ ( 0.6 ) + ( − 0.9 )
2
8
1
4
= − TADi + TAD j + TADk
9
9
9
6
2
3
= − TADi + TAD j − TADk
7
7
7
continued COSMOS: Complete Online Solutions Manual Organization System
ΣM C = 0:
or
rA × TAD + rB × TBE + rA × W = 0
i
2.4
8
−
9
j
0
1
9
k
i j k
0 TAD + 1.8 0 0 TBE + ( 2.4 ) i × ( − 880 ) j = 0
4
6 2
3
−
−
9
7 7
7
Equating the coefficients of the unit vectors to zero:
−
j:
9.6
5.4
TAD +
TBE = 0
9
7
2.4
3.6
TAD +
TBE − 2112 = 0
9
7
k:
or TAD = 2160 N
TBE = 2990 N
Force equations:
Cx −
8
6
( 2160.0 N ) − ( 2986.7 N ) = 0,
9
7
Cy +
1
2
( 2160.0 N ) + ( 2986.7 N ) − 880 N = 0, or
9
7
Cz +
4
3
( 2160.0 N ) − ( 2986.7 N ) = 0,
9
7
or
or
C x = 4480.0 N
C y = − 213.34 N
C z = 320.01 N
C = ( 4480 N ) i − ( 213 N ) j + ( 320 N ) k
COSMOS Chapter 4 Solution 116
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