Question
In: Physics
A 1500-N uniform boom at ? = 68.0
A 1500-N uniform boom at ? = 68.0
Solutions
Expert Solution
In this fig, boom and its supported cable are 900 to each other ( 68+22= 90)
W' = weight of boom = 1500 N = 1.5 KN
W = weight of object = 1850 N = 1.85 KN
(a) sum of the moments about pivot point is ( from fig)
0 = W' cos 68 *( L/2) + W*L cos 68 - T * ( 3/4) L ( moment = force*perpendicular distance)
T* ( 3/4) *L = L* cos 68* [( 1500/2) +1850] ( by putting values and rearranging the terms)
T = 1298 N
T = 1.29 KN
(B) Sum of forces in horizontal direction:
0 = FH - T cos 22 ( FH = horizontal component of reaction force )
FH = 1.29* cos 22= 1.196 KN
sun of forces in vertical direction
0 = Fv + T *sin 22 - W' - W
Fv ( vertical component of force) = 2.86 KN
genius_generous answered 1 year agoRelated Solutions
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