Zoe M.

asked • 06/12/16

The 2 kg, uniform horizontal rod with a point .25 m from the left side and .75 m from the other side. What is the gravitational torque about the point shown?

The answer is -2.1 N m and I don't know how they got that. 

Arturo O.

Zoe,
 
Where is the point?  (No picture.)  But here is how to solve it:
 
Put the entire weight (W = mg) acting straight down through the center of mass of the rod, in this case its mid-point, then multiply the weight (mg) by the distance from the mid-point to the point in question.  That will be the gravitational torque about that point.
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06/12/16

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Jim S. answered • 06/12/16

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Zoe,
        Lets divide the rod into two segments left and right at .25m and .75m
The center of mass of the segments are at .25/2 and .75/2
Taking moments (torque) about the given point gives ,25mg(,25/2)-.75mg(.75/2)=.6125-5.5125=4.9 N-m which is g/2
 
m=2kg g=9.8 m/sec2 
 
I don't know where the -2.1 N-m came from I think it must be a typo.
 
Hope this helps
 
 
Jim
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Daniel J.

also, i must point out that this 4.9 N-m has a negative sign.
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07/23/19

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