What is the maximum mass that can be hung from the end of a boom?

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SUMMARY

The discussion focuses on calculating the maximum mass that can be hung from a boom with a mass of 1000 kg, supported by a base weighing 10,000 kg. The key equations involved include torque equilibrium and weight calculations, specifically using the formula for torque net = 0. The tension in the cable, which makes a 46.5-degree angle with the horizontal, is also a critical factor in determining the system's stability. Participants emphasize the importance of correctly identifying the pivot point for torque calculations to prevent overbalancing.

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Homework Statement


In the figure, a boom of mass 1000kg is supporting a mass m. The base has a mass of 10,000kg. (a) What is the maximum mass m that can be hung from the end of the boom? (b) If this amount of mass is haning off the boom, where is the center of gravity of the entire system relative to C.G. (base)? (c) What is the corresponding tension in the cable assuming the cable makes an angle of 46.5 degrees relative to the horizontal? (d) What is the strain in the boom's pin (support)?


Homework Equations



Torque net=0
Force net=0
Weight=mg

The Attempt at a Solution



Weight of boom=(1000kg)(9.80m/s2)=9800N
Torque net=0
0=-W(boom)rsin(theta)-W(mass)rsin(theta)+Tensionrsin(theta)
0=(9800N)(10m)sin(theta)-m(9.80m/s2)rsin(theta)+Trsin(theta)

I don't know what values to use for my angles, but there would still be too many unknowns. so i think that I'm setting up my equation completely wrong, but can't think of anything else to do. once i find the mass then i think i know how to do the other parts of the question.
 

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Welcome to PF!

Hi a.wright! Welcome to PF! :smile:

(have a theta: θ :wink:)
a.wright said:
In the figure, a boom of mass 1000kg is supporting a mass m. The base has a mass of 10,000kg. (a) What is the maximum mass m that can be hung from the end of the boom?

Weight of boom=(1000kg)(9.80m/s2)=9800N
Torque net=0
0=-W(boom)rsin(theta)-W(mass)rsin(theta)+Tensionrsin(theta)
0=(9800N)(10m)sin(theta)-m(9.80m/s2)rsin(theta)+Trsin(theta)

I don't know what values to use for my angles, but there would still be too many unknowns. so i think that I'm setting up my equation completely wrong, but can't think of anything else to do. once i find the mass then i think i know how to do the other parts of the question.

I don't understand your torque equation …

about which point are you taking torques?

and why does the tension come into it? :confused:

Perhaps you're missing the point … if the mass is too heavy, the whole system will overbalance, and your torque equation should be determining when that happens. :wink:
 

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