# Physics

A 4.00-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70.0 kg 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 into place?

I searched on this website and found a similar problem, but had an additional question. Is the torque at the center of gravity equal to the torque about the point where the beam was bolted into place?

Here's the work I've done so far.
T = 2.0m * 500kg * 9.81m/s^2 + 4.0m * 70kg * 9.81m/s^2 = 12,557Nm

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1. A beam which is fixed to a wall by bolting or which is extended from inside the building, the free other end of which is freely hanging in the air is called a cantilever. This may make for easier search if you intend to do so.

The torque caused by weights on the beam is called moment, in structural engineering terms (again, easier for searches).

Your formula for the "torque" is correct, namely ΣFD, where F is an applied force on the beam, and D is the distance from the point at which torque is desired.

Note that F carries a sign. If the force "attempts" to turn the beam in the opposite direction, it would be negative.

Here's the cantilever beam:
|
|=========Mg*L/2========mgL
|

Moment (torque)
= ΣFD
= Mg*L/2 + mgL

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posted by MathMate

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