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> Torque Of Cable Supporting Pivoted Boom And Suspen
Anna Werner
Posted: Mar 12 2012, 12:33 PM


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I have a homework problem, due this coming Wednesday, that has me stumped. It is worded as follows: "A mass M = 153 kg is suspended from the end of a uniform boom . . . The boom (mass = 93.0 kg, length = 3.60 m) is at an angle [theta]=52.0 deg from the vertical, and is supported at its mid-point by a horizontal cable and by a pivot at its base. Calculate the tension in the horizontal cable."

Here's what I've been trying:
1) Find torque of boom: (L/2)(boom mass)(9.8 m/(s^2)), L=length of boom
2) Find torque of suspended weight: L(mass of weight)(accel. gravity)
3) Set up equation for torque (T) of horizontal cable:
T(L)(sin alpha), alpha=angle between cable and boom, found by 180-(90+given angle)
4) Add all torque values and set to equal zero. Find T.

I am not sure where I have gone wrong here. Any assistance would be appreciated.

Regards,
Anna

This post has been edited by Anna Werner on Mar 12 2012, 12:35 PM
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