COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 27.
Have
where
M C = TAD d
JJJG
d = Perpendicular distance from C to line AD
with
M C = rA/C TAD
and
rA/C = ( 3.1 m ) j + (1.2 m ) k
JJJG
AD
TAD = TAD
AD
( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k
TAD = ( 369 N )
( 2.4 m )2 + ( − 3.1 m )2 + ( −1.2 m )2
= ( 216 N ) i − ( 279 N ) j − (108 N ) k
Then
i
j
k
MC = 0
3.1 1.2 N ⋅ m
216 − 279 −108
= ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k
and
MC =
( 259.2 N ⋅ m )2 + ( −669.6 N ⋅ m )2
= 718.02 N ⋅ m
∴ 718.02 N ⋅ m = ( 369 N ) d
or d = 1.946 m W
.
COSMOS Chapter 3 Solution 27
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