Torque on a Beam Due to Gravity

[cs44167]

Homework Statement

A rigid beam of mass 7.78 kg and length 1.44 m is fixed at point P, around which the beam rotated (P is at the far leftmost point of the beam). What is torque on the beam due to gravity only?

Relevant Equations

Torque = r*F*sin (theta)
w = mg

I tried using r * f * sin theta and calculated this:

1.14 m * 9.80 m/s/s * 7.78 kg = 109.7 N*m

this was wrong; I needed three significant figures so I did 1.10E2 N*m which was also wrong.

Since the torque is due to gravity; would it be -1.10E2 N*m since it’s angle is -270° which is -1?

 

[berkeman]

Homework Statement:: A rigid beam of mass 7.78 kg and length 1.44 m is fixed at point P, around which the beam rotated (P is at the far leftmost point of the beam). What is torque on the beam due to gravity only?
Homework Equations:: Torque = r*F*sin (theta)
w = mg

I tried using r * f * sin theta and calculated this:

1.14 m * 9.80 m/s/s * 7.78 kg = 109.7 N*m

this was wrong; I needed three significant figures so I did 1.10E2 N*m which was also wrong.

Since the torque is due to gravity; would it be -1.10E2 N*m since it’s angle is -270° which is -1?

Where did 1.14m come from in your calculation? A typo of 1.44m? And the force acts at the center of mass for the beam, not at the end of it...

 

[cs44167]

Where did 1.14m come from in your calculation? A typo of 1.44m? And the force acts at the center of mass for the beam, not at the end of it...

Yes, the 1.14 was a typo - it should’ve been 1.44 which I had in my calculations.

What does it mean for the force to act at the center? Would that divide the radius into halves?

 

[berkeman]

Yes. It's hard to be sure without seeing the figure that has the problem, but if it's a simple horizontal cantilever beam supported only at the left side, and the beam is uniform density, then its COM is halfway out.

 

[berkeman]

Is it like this?

https://media.cheggcdn.com/study/c0c/c0c63894-f03d-4021-9d86-9b6660f0bd54/DC-785V1.png

 

[haruspex]

What does it mean for the force to act at the center?

What you calculated is for the case where the beam is weightless except for a 7.78kg mass at its free end. Clearly that will create more torque than a uniform beam would.
In principle, one should consider the beam made of small elements of length dx, find the torque due to each, and integrate along the beam.
But you can see what the answer will be just by taking these elements in pairs: a small element length dx at distance x from the fixed end, plus an element length dx at x from the free end will exert the same torque as two such elements in the middle.