A crate with a mass of 191.5 kg is suspended from the end of a uniform boom with

Question

A crate with a mass of 191.5 kg is suspended from the end of a uniform boom with a mass of 76.9 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable. A crate with a mass of 191.5 kg is suspended from the end of a uniform boom with a mass of 76.9 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable.

Explanation / Answer

let tension in the cable be T .

angle made by the cable with -ve x axis=arctan((9-7)/(11-1))=11.31 degrees

component of tension along -ve x axis=T*cos(11.31)=0.98*T

component of tension along +ve y axis=T*sin(11.31)=0.196*T

weight of the mass =191.5*9.8=1876.7 N is acting in -ve y axis direction.

weight of the boom is acting at a horizontal distance from the base(marked as X)=(11-1)2=5 m

weight of the boom=76.9*9.8=753.62 N, acting along -ve y axis.

balancing torque about point X:

0.98*T*(7-1)+0.196*T*(11-1)=1876.7*(11-1)+753.62*5

==>7.84*T=22535.1

==>T=2874.375 N

hence tension in the cable is 2874.375 N

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