# B Torque on a rod?

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1. Jul 21, 2017

### tummbacoco

So I've learned that the torque on an object is just: (perpendicular force)(radius) and that has worked well for things like seesaws but it doesn't take into account the torque of the object itself!

In the picture the rod has a total mass of 5kg, and by definition the torque on the right is greater than that on the left. T=Fr

Now my question is a what distance to the left of the fulcrum (triangle supporting the rod) would I have to put a 3kg weight to balance the rod?? I've tried calculating the net torque Tnet= Iα but have yet to solve this problem. Since the problem is just practice I'm looking for an explanation more so than an answer. Thanks!

2. Jul 21, 2017

### andrewkirk

Assuming the rod is horizontal, you can calculate the torque from the weight of the part of the rod each side of the fulcrum, as an integral of the torques of small segments, from the fulcrum to the end of the rod on that side.

Torque from weight of a side of length $L = \int_0^L m g l\,\frac{dl}L$

3. Jul 22, 2017

### Nidum

Find out what the 'Centre of Gravity' of a body is . This will help you a lot when doing this sort of problem .

4. Jul 24, 2017

### Dr.D

Your phrase "torque of the object itself" really does not make any sense. The object, presumably the rod, does not cause any torque; it is just a chunk of mass. The force of gravity on this mass causes a torque, but that is due to gravity, not due to the rod.