A 14, 650-N crane pivots around a friction-free axle at its base and is supporte

Question

A 14, 650-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 11, 300-N pallet of bricks by a 2.2-m very light cord, find the following. a) the tension in the cable (b) the horizontal and vertical components of the force that the axle exerts on the crane horizontal component N vertical component N

Explanation / Answer

F1=14650 N

F2=11300 N

a)

let,

T be the tension in the cable,

at equilibrium the net torque must be zero,

F1*7*cos(55) + F2*16*cos(55) + T*cos(60)*(16-3)*cos(55) - T*sin(60)*(16-3)*sin(55)=0

14650*7*cos(55) + 11300*16*cos(55) + T*cos(60)*(16-3)*cos(55) - T*sin(60)*(16-3)*sin(55)=0

===> T=29581.7 N

b)

horizontal componenet, FH=T*sin(60)

=29581.7*sin(60)

=25618.5 N

and

verical component, Fv=F1+F2+T*cos(60)

=14650+11300+29581*cos(60)

=40740.5 N

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