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m1 help

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 :frown:
im so confused
m1.PNG39.7KB
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 :frown:
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.
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
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.
(edited 7 years ago)
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???
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.
Reply 6
Avatar for RKM21
RKM21
3
Do you have the answers for these:
I got 2.25l for the first one and 1.75l for the second.
Reply 7
Avatar for Notnek
Notnek
21
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.
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
Reply 9
Avatar for Notnek
Notnek
21
Original post by pondsteps
no its 2015 jan ial

Oops. Thanks for the correction.
Reply 10
Avatar for RKM21
RKM21
3
Aaah yes. Forgot to minus it from 2l.

Thanks,
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?
Original post by pondsteps
does at the point of tipping mean that it's still horizontal and at equilibrium?


Yes.

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