Gravitational Torque HW: M1,M2,X1,X2,g | Calculate T

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SUMMARY

The discussion centers on calculating the torque exerted by a 500kg steel beam and a 70kg construction worker positioned at the end of the beam, which is 4.0m long. The correct formula for torque, T = Mgxcm, was applied, where Mg represents the total gravitational force and xcm is the moment arm. The initial calculation yielded a torque of 11.9k Nm, while the expected answer was 12.5k Nm, indicating a miscalculation in determining the center of mass. The resolution involved recognizing the need to calculate individual torques rather than relying solely on the center of mass.

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  • Basic knowledge of center of mass and its significance in physics
  • Ability to perform calculations involving mass and distance
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Blistex
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Homework Statement



A 4.0m long, 500kg steel beam extends horizontally from where it has been bolted to the framework of a new building under construction. A 70kg construction worker stands at the far end of the beam. What is the magnitude of the torque about the point where the beam is bolted in place?

M1 = 500kg
M2= 570kg
X1 = 0m
X2 = 4m
g = 9.8 m/s2

Homework Equations



T = Mgxcm

Where Mg is the net gravitational force on the object and Xcm is the moment arm between the rotational axis and the center of mass.

The Attempt at a Solution



I thought this problem would quite simple but I think there might be an error in my reasoning.

I start by calculating the center of mass. {(500kg)(0m) + (570kg)(4m)}/1070kg = 2.13m
So the CM is 2.13 from the fulcrum.

Next I substitute the given values into the equation. T = (500kg + 70kg)(9.8m/s2)(2.13m) = 11902NM or 11.9k Nm.
This should be the torque about the fulcrum.

Thanks for the help :-p the book says 12.5k Nm. So far it's been pretty spot on this far.
 
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Welcome to PF!

Hi Blistex! Welcome to PF! :smile:
Blistex said:
A 4.0m long, 500kg steel beam extends horizontally from where it has been bolted to the framework of a new building under construction. A 70kg construction worker stands at the far end of the beam. What is the magnitude of the torque about the point where the beam is bolted in place?
…
M1 = 500kg
M2= 570kg

erm :redface: …

where did 570 come from? :wink:​

(and anyway, you don't need to find the centre of mass, just find each torque separately, and add! :smile:)
 
Thank Tim!

Given Tgrav = Mgxcm

I summed the individuals particles that had torque on the object. I.e only the worker. which amounted to dsin(x)*m*g = -2744Nm

The other side of the equation asks for the mass relative to the center of rotation, that's where my miscalculation entered. Whereas the center of mass is simply 4m/2, the center of the length of the beam. I had thought the weight of the worker would cause the CM to shift farther towards the worker. I guess that's what relative meant. After plugging in the rest of the variables I was able to solve it.

Thanks for your help.
 

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