Solving Evil Torque Problem: Get Help Here!

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The discussion revolves around calculating the torque on a 200 kg plank attached to a wall when a supporting string breaks. The initial calculation of torque was incorrectly based on a distance of 2 meters, leading to a value of 3920 Nm. However, the correct distance to the center of mass, where the gravitational force acts, is only 1 meter, resulting in a torque of 1960 Nm as stated in the book. Participants clarified that the torque formula requires the distance to the center of mass, not the full length of the plank. This understanding resolves the discrepancy in the torque calculations.
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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