A 16,250 -N crane pivots around a friction-free axle at its base and is supporte

Question

A 16,250-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25

A 16,250-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25 degree angle with the crane (see figure). The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane. When the crane is raised to 55 degree above the horizontal holding an 10,150-N pallet of bricks by a 2.2-m very light cord, find the following.

Explanation / Answer

(a) the tension T in the cable
Net torque about the axle = W*dcos? + w'*Lcos? - T*(L - a)sin? = 0

?T = (Wd + w'L)cos?/[(L - a)sin?] = 3.43 x 104 N

equating forces in the vertical direction, we get,
2.9*10^4*cos(25+(90-55)) +16,250+10,150 = Fy

so,Fy = 44766.25 N <-----vertical component

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similarly along horizontal direction,
Fx = 29000*sin60= 25114.7N <-----horizontal component

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