TPR Torque question?

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pfaction

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A bear of mass 4M sits at the right end of a weightless plank of length L. The plank can rock on a fulcrum L / 6 from its right end. How far from the right end of the plank must a chimpanzee of mass M stand to balance the plank?

Can't understand this answer; got 29% on the actual set....what the f?
 
A balanced plank means net torque = 0.

Torque = r*F

The force of the bear is 40M and the force of the monkey is 10M.

If the bear is sitting at the very end of the plank that is 1/6L from the fulcrum, then where does the monkey have to sit to balance the plank?

The bear is sitting a distance L/6 from the fulcrum. The monkey, which exerts a force 4x less must sit a distance 4x further than the bear to create the same torque. So the monkey must sit 4L/6 from the fulcrum, which is 5L/6 from the right end of the board since the fulcrum is already 1L/6 from the right side.
 
Thanks for this, btw. I have one more question.

So, TPR gave this problem:
800N man, 200N daughter, 5m see saw. If they're on opposite sides of the 5m see saw, where from the man should the fulcrum be placed?

Simple enough, same logic: man weighs 4x more, so it should be 1/4th the distance: 1m.

Then I get this problem:
A massless rod attached to ceiling by a string. 0.4lb on left side, 1.2lb on right side. Length is L, how far from LEFT side should it be attached so rod is horizontal?

I used the same logic: 3x the weight, so 1/3rd the distance: 2/3L from the left side should be the focus.

But it wasn't. It was 3/4th. They wanted me to use COM...why didn't the first strategy work?
 
Thanks for this, btw. I have one more question.

So, TPR gave this problem:
800N man, 200N daughter, 5m see saw. If they're on opposite sides of the 5m see saw, where from the man should the fulcrum be placed?

Simple enough, same logic: man weighs 4x more, so it should be 1/4th the distance: 1m.

Then I get this problem:
A massless rod attached to ceiling by a string. 0.4lb on left side, 1.2lb on right side. Length is L, how far from LEFT side should it be attached so rod is horizontal?

I used the same logic: 3x the weight, so 1/3rd the distance: 2/3L from the left side should be the focus.

But it wasn't. It was 3/4th. They wanted me to use COM...why didn't the first strategy work?

In your first problem, 1/4 of 5 is not 1, so your logic is not quite right. Sum the moments about your fulcrum and you'll be fine.

Edit. I think where you went wrong is you have to use the proportion of the total weight. The man is 80% of the total weight. 80% of 5 is 4 which is the distance from the girl to the fulcrum. In the rod problem you have 1.6 total with 1.2 on one side. 1.2/1.6=.75. Does this help any?
 
Kind of, but you're using COM, whereas I want to use torque (i have huge problems differentiating when to use torque, fgrav and tension, COM)

I actually ended up making a proportion by making up the following rule:
Make the smaller fraction =1, then the larger will be the multiple of that.

So for the first one, the man is 4x weight, so 1/4 distance, daughter is 4x distance.
1x+4x=5x

Other one:
.4 is 1/3rd 1.2 so it should be 3x away. If I make (1/3)=1, the total is 4m, so 3 over 4L.
So confusing, to be honest. I'm gonna try to logic the monkey problem too. Thanks BTW!
 
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