COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 115.
For equivalence
ΣF : FB + FC + FD = R A
R A = − ( 240 N ) j − (125 N ) k − ( 300 N ) i + (150 N ) k
∴ R A = − ( 300 N ) i − ( 240 N ) j + ( 25 N ) k
Also for equivalence
ΣΜ Α : rB/ A × FB + rC/ A × FC + rD/ A × FD = M A
or M A
i
j
k
i
j
k
i
j
k
= 0 0.12 m
0
+ 0.06 m 0.03 m − 0.075 m + 0.06 m 0.08 m − 0.75 m
0 − 240 N −125 N
−300 N
0
0
0
0
150 N
= − (15 N ⋅ m ) i + ( 22.5 N ⋅ m ) j + ( 9 N ⋅ m ) k + (12 N ⋅ m ) i − ( 9 N ⋅ m ) j
or M A = − ( 3 N ⋅ m ) i + (13.5 N ⋅ m ) j + ( 9 N ⋅ m ) k
.
COSMOS Chapter 3 Solution 115
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