STEP II 2003 Q9 MechanicsWatch

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#1
Here's the paper.

http://www.stepmathematics.org.uk/do...2003paper2.pdf

I don't understand why the frictional force at C must point to B.
It might point to A and make the answer different.
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11 years ago
#2
Well, given the forces acting on it, the rod can't be on the point of moving upwards.
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#3
(Original post by Dystopia)
Well, given the forces acting on it, the rod can't be on the point of moving upwards.
Could you explain the reason?
why can't it be like that? ( see attachment )
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11 years ago
#4
I may be missing something, but I don't see an easy way of showing the friction from the cylinder can't act downwards, even though I think it's "obvious" that it doesn't. I don't know what happens if you actually solve the problem starting with that assumption, you either find a contradiction or that the question setter slipped up, I guess.
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11 years ago
#5
(Original post by DFranklin)
I may be missing something, but I don't see an easy way of showing the friction from the cylinder can't act downwards, even though I think it's "obvious" that it doesn't. I don't know what happens if you actually solve the problem starting with that assumption, you either find a contradiction or that the question setter slipped up, I guess.
Yes, it's one of those things that are so obvious that you struggle to see how it could be proved...

Would this be acceptable?

If the friction from the cylinder acts downwards, then the rod is on the point of slipping 'up', so the friction at A would also have to act to the left. Resolving horizontally (to ground) shows that it cannot be in equilibrium.
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#6
(Original post by Dystopia)
Yes, it's one of those things that are so obvious that you struggle to see how it could be proved...

Would this be acceptable?

If the friction from the cylinder acts downwards, then the rod is on the point of slipping 'up', so the friction at A would also have to act to the left. Resolving horizontally (to ground) shows that it cannot be in equilibrium.
Fair enough.
Thank you.
I didn't realise frictional force is to prevent the motion.
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11 years ago
#7
I'm unsure, to be honest. It's the "friction at A acts towards C, friction at C acts towards A" case I'm struggling to rule out. Common sense says "it doesn't happen", but the "friction is limiting at A and B because they're pushing towards each other frantically for no obvious reason" case doesn't actually seem impossible from a mathematical point of view.

The only "formal" way I see of ruling it out is to show that from the equilibrium position with limiting friction, you can actually reduce the frictional force without losing equilibrium, so it wasn't actually limiting friction after all.

Or I'm just being stupid.
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#8
(Original post by DFranklin)
I'm unsure, to be honest. It's the "friction at A acts towards C, friction at C acts towards A" case I'm struggling to rule out. Common sense says "it doesn't happen", but the "friction is limiting at A and B because they're pushing towards each other frantically for no obvious reason" case doesn't actually seem impossible from a mathematical point of view.

The only "formal" way I see of ruling it out is to show that from the equilibrium position with limiting friction, you can actually reduce the frictional force without losing equilibrium, so it wasn't actually limiting friction after all.

Or I'm just being stupid.
I agree with you.
What I thought was a bit weird.
let's say the rod tends to slip down if smooth everywhere.
The frictional force at A is trying to stop it from slipping down.
Then the frictional force at A might contribute too much.
The rod tends to slip up instead.
The frictional force at B has to stop it.

but at the end, i found it might be a bit silly.
the reason is what you've said.
"you can actually reduce the frictional force without losing equilibrium, so it wasn't actually limiting friction after all."
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11 years ago
#9
In practice, I think you're supposed to think:

Suppose there wasn't any friction with the floor. Then the friction with the cylinder has to act towards B to stop sliding.

And suppose instead there wasn't any friction with the cylinder. Then the friction with the floor has to act towards B to stop sliding.

But the friction is all acting together to stop sliding. So it all must act towards B.

I'm not sure if that's actually rigourous, though.
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#10
Thank you.
i think i understand now.
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