A mass M=171 kg is suspended from the end of a uniform boom as shown. The...

Question

Question

A mass M=171 kg is suspended from the end of a uniform boom as shown. The boom (mass=76.0 kg, length=2.40 m) is at an angle =

Answer 1

Answer #1

The beam is in equilibrium, So, kex={fy = {Mo=0 my fyVEN As Emo=0 TX Loso = wx 2 sino + mg x L sine & T = . (W tan o + 2mg ta

Answer 2

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