ladder question (forces)

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bexxr
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#1
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#1
Currently looking at ladder questions.

- Uniform beam, with mass m and length 2a, in equilibrium on rough ground and against a smooth wall.
- Horizontal force of magnitude kmg (k is a constant) is applied to the beam at A. This forces acts in a direction that is perpendicular to the wall and towards the wall.

I can model the first bit, however I'm not sure if I am modelling the second part correctly as I am getting a negative number for k.

Here's my drawing. Can someone please tell me if I have the additional force (kmg) drawn in the correct direction, and if not, can someone explain why please?

Thanks!
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mqb2766
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#2
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#2
Can you upload a pic of the original question and what you tried?
Last edited by mqb2766; 1 month ago
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mqb2766
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#3
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Could you upload the first part of the question?
It could be that mu*R is in the opposing direction and kmg is much larger, so friction is acting in the same direction as N? It would be an unusual scenario.
Last edited by mqb2766; 1 month ago
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bexxr
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#4
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mqb2766
Here's the first part of the question and my workings:
Last edited by bexxr; 1 month ago
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mqb2766
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#5
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As its in equilibrium in part a) and you apply a force of kmg towards the wall, the limiting equilibrium means the ladder is about to slip UP the wall. So friction must be acting in the opposite direction (away from the wall), as suggested in the previous post.
Last edited by mqb2766; 1 month ago
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bexxr
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#6
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#6
(Original post by mqb2766)
As its in equilibrium in part a) and you apply a force of kmg towards the wall, the limiting equilibrium means the ladder is about to slip UP the wall. So friction is acting in the opposite direction (away from the wall), as suggested in the previous post.
Oh, of course! I can't believe I didn't realise this (I feel a bit stupid now, haha!). Thank you so much for your help.
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bexxr
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#7
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mqb2766
Here's my revised workings. Would this be correct?
Last edited by bexxr; 1 month ago
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mqb2766
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#8
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#8
Looks good.
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