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A 1500-N uniform boom at ? = 68.0

A 1500-N uniform boom at ? = 68.0

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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 ago
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