Solving Evil Torque Problem: Get Help Here!

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Homework Help Overview

The discussion revolves around a torque problem involving a plank attached to a wall and the effects of gravity on its center of mass. The original poster expresses confusion regarding the calculated torque compared to a provided solution in a textbook.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate torque using the formula torque = r x F, questioning the discrepancy between their result and the textbook's answer. Participants explore the significance of the distance (r) and the point of application of the gravitational force.

Discussion Status

Participants are actively engaging with the problem, with some suggesting that the point of application of the gravitational force affects the torque calculation. There is a productive exchange of ideas regarding the center of mass and its implications for the torque value.

Contextual Notes

There is an assumption that the mass of the plank is uniformly distributed, which influences the location of the center of mass. The original poster is seeking clarification on the correct interpretation of the torque calculation.

tonix
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Hi

I have a problem for which the given solution makes no sense to me.

A 200 kg plank of length 2m is attached by a hinge at one of its ends to a wall. It is held in place horizontally by a string. The string breaks and I am supposed to find the torque at that point. (See attached image).

I have that torque = r x F which gives me a torque vector with a magnitude of 9.8 m/s^2 * 2m * 200kg = 3920 Nm.
The solution the book gives, however, is half of that, ie 1960Nm.

Could somebody either reassure me that the book is wrong or clarify where I made a mistake?

Thank you,
tonix
 

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Can u see why your answer is twice more??Is it F,or is it "r"??

Daniel.
 
Well, I don't know why it is twice more. That is the problem. It seems that r should be 2m for sure. I suppose it could be F since the gravity is only acting on half the mass?
 
Nope,gravity affects all mass...Think about the point of application of the gravity force and its whereabouts wrt the rotation point of the rod...

Daniel.
 
ahh... so the overall point of application of the gravity force is in the middle of the rod and therefore r is only 1m instead of 2m?
 
It's actually in the center of mass,but,assuming the mass is distributed uniformly throughout the rod,it would be exactly at the middle.
Yes,that distance is only of 1 meter.


Daniel.
 
That's what I meant to say. Thanks!
 

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