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Mme_Bonii
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#1
A uniform rod AC, of weight W and length 3l, rests horizontally on two supports, one at A and one at B, where AB= 2l. A particle of weight 2W is placed on the rod at a distance x from A. The rod remains horizontal and in equilibrium.
(a) find the greatest possible value of x.
The magnitude of the reaction of the support at A is R. Due to a weakness in the support at A, the greatest possible value of R is 2W.
(b) find the least possible value of x.
I tried understanding how it's solved from the ms, but it was fruitless. Can anyone explain to me how and why it's solved that way?
(a) find the greatest possible value of x.
The magnitude of the reaction of the support at A is R. Due to a weakness in the support at A, the greatest possible value of R is 2W.
(b) find the least possible value of x.
I tried understanding how it's solved from the ms, but it was fruitless. Can anyone explain to me how and why it's solved that way?
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username1765117
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#2
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#2
If you hadn’t already ALWAYS draw a diagram for this kind of stuff if you haven’t done so already. I personally can’t be bothered rn to draw and attach a photo of one so I’m visualising but to understand you definitely should and also I might muck up by not drawing one.
I can’t see the mark-scheme so idk what it says there but this is how I’d approach it.
a) So the greatest possible value of x is how far along the rod the weight can be without the whole thing tilting. The maximum point is when it’s in limiting equilibrium, so anti-clockwise moment = clockwise moment and the whole thing will be about to tilt about B so B bears all of the weight which is 3W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(3W)
2xW = 4.5Wl
x = 2.25l
b) Ok so I just said reaction at A is 0 but that was when x was as far along as possible. Now we’re finding x so the weights as close to A as possible, without the reaction at A becoming more than 2W which would cause it to break. So we need to find when it’s exactly 2W, and therefore reaction at B is exactly W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(W)
2xW = 1.5Wl
x = 0.75l
I can’t see the mark-scheme so idk what it says there but this is how I’d approach it.
a) So the greatest possible value of x is how far along the rod the weight can be without the whole thing tilting. The maximum point is when it’s in limiting equilibrium, so anti-clockwise moment = clockwise moment and the whole thing will be about to tilt about B so B bears all of the weight which is 3W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(3W)
2xW = 4.5Wl
x = 2.25l
b) Ok so I just said reaction at A is 0 but that was when x was as far along as possible. Now we’re finding x so the weights as close to A as possible, without the reaction at A becoming more than 2W which would cause it to break. So we need to find when it’s exactly 2W, and therefore reaction at B is exactly W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(W)
2xW = 1.5Wl
x = 0.75l
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Mme_Bonii
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#3
(Original post by TheTree0fDeath)
If you hadn’t already ALWAYS draw a diagram for this kind of stuff if you haven’t done so already. I personally can’t be bothered rn to draw and attach a photo of one so I’m visualising but to understand you definitely should and also I might muck up by not drawing one.
I can’t see the mark-scheme so idk what it says there but this is how I’d approach it.
a) So the greatest possible value of x is how far along the rod the weight can be without the whole thing tilting. The maximum point is when it’s in limiting equilibrium, so anti-clockwise moment = clockwise moment and the whole thing will be about to tilt about B so B bears all of the weight which is 3W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(3W)
2xW = 4.5Wl
x = 2.25l
b) Ok so I just said reaction at A is 0 but that was when x was as far along as possible. Now we’re finding x so the weights as close to A as possible, without the reaction at A becoming more than 2W which would cause it to break. So we need to find when it’s exactly 2W, and therefore reaction at B is exactly W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(W)
2xW = 1.5Wl
x = 0.75l
If you hadn’t already ALWAYS draw a diagram for this kind of stuff if you haven’t done so already. I personally can’t be bothered rn to draw and attach a photo of one so I’m visualising but to understand you definitely should and also I might muck up by not drawing one.
I can’t see the mark-scheme so idk what it says there but this is how I’d approach it.
a) So the greatest possible value of x is how far along the rod the weight can be without the whole thing tilting. The maximum point is when it’s in limiting equilibrium, so anti-clockwise moment = clockwise moment and the whole thing will be about to tilt about B so B bears all of the weight which is 3W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(3W)
2xW = 4.5Wl
x = 2.25l
b) Ok so I just said reaction at A is 0 but that was when x was as far along as possible. Now we’re finding x so the weights as close to A as possible, without the reaction at A becoming more than 2W which would cause it to break. So we need to find when it’s exactly 2W, and therefore reaction at B is exactly W.
So taking moments about A: 1.5l(W)+ x(2W) = 2l(W)
2xW = 1.5Wl
x = 0.75l
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username1765117
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#4
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#4
(Original post by Mme_Bonii)
would you mind explaining what is limiting equilibrium?
would you mind explaining what is limiting equilibrium?
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