A weight (with a mass of 42 kg) is suspended from a point near the right-hand en

Question

A weight (with a mass of 42 kg) is suspended from a point near the right-hand end of a uni- form boom with a mass of 43 kg . To support the uniform boom a cable runs from this same point to a wall (the left-hand vertical coordi- nate in the figure) and by a pivot on the same wall at an elevation of 6.5 m. Calculate the tension T in the cable. The acceleration of gravity is 9.8 m/s2 . In the figure: the vertical height between the boom and the cable is 3.5 m. The horizontal distance between the wall and where the boom and cable meet is 6 m. The entire length of the boom is 10 m long.

Explanation / Answer

first, angle wire makes with the boom

tan@ = 3.5/6 @ = 30.256 deg

Then

torques acting on boom...

torque from weight of boom + torque from sign = torque from tension of cable

Mg * 1/2 L + mg*L = tension * 6 * sin30.256

43 * 9.8 * 5 + 42 * 9.8 * 10 = tension * 3.0232

tension = 2058 Newtons

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