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# m1 help

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1. can someone pls explain this
how can it be the greatest possible value of this distance like it doesnt change.. and how do we even get the greatest possible value
im so confused
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2. (Original post by pondsteps)
can someone pls explain this
how can it be the greatest possible value of this distance like it doesnt change.. and how do we even get the greatest possible value
im so confused
Although it doesn't change, you don't know where it is. It could be anywhere on the rod. Question is asking you what's the greatest value it could take such that the rod remains in equilibrium.
3. (Original post by ghostwalker)
Although it doesn't change, you don't know where it is. It could be anywhere on the rod. Question is asking you what's the greatest value it could take such that the rod remains in equilibrium.
so why does x have max value when the rod is tilting??? what makes it max and how did we know
4. (Original post by pondsteps)
so why does x have max value when the rod is tilting??? what makes it max and how did we know
Well for small values of x, the weight is close to A, and the rod is stable.
If x was 3l, and the weight was at C, then I think it's clear the rod will tilt.

So, some point in between (when x is a maximum) the rod will be on the verge of tilting.
At that point the reaction at A will be zero. The rod will just be balancing on the other support. Taking moments about that will give you x.
5. (Original post by ghostwalker)
Well for small values of x, the weight is close to A, and the rod is stable.
If x was 3l, and the weight was at C, then I think it's clear the rod will tilt.

So, some point in between (when x is a maximum) the rod will be on the verge of tilting.
At that point the reaction at A will be zero. The rod will just be balancing on the other support. Taking moments about that will give you x.
btw will there still be a normal force at A if its about to tip???
6. (Original post by pondsteps)
btw will there still be a normal force at A if its about to tip???
No; when it's about to tip, there is no reaction / normal force at A.
7. Do you have the answers for these:
I got 2.25l for the first one and 1.75l for the second.
8. (Original post by RKM21)
Do you have the answers for these:
I got 2.25l for the first one and 1.75l for the second.
First one is wrong, second one is right.

Post your working if you can't find your mistake. This is the Jan 2014 IAL paper Q6 so you can check your answers there.
9. (Original post by notnek)
First one is wrong, second one is right.

Post your working if you can't find your mistake. This is the Jan 2014 IAL paper Q6 so you can check your answers there.
no its 2015 jan ial
10. (Original post by pondsteps)
no its 2015 jan ial
Oops. Thanks for the correction.
11. Aaah yes. Forgot to minus it from 2l.

Thanks,
12. (Original post by ghostwalker)
No; when it's about to tip, there is no reaction / normal force at A.
does at the point of tipping mean that it's still horizontal and at equilibrium?
13. (Original post by pondsteps)
does at the point of tipping mean that it's still horizontal and at equilibrium?
Yes.

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