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1. I am confused with the following question as there are two rough surfaces:

A uniform ladder AB of length 8m leans against a rough wall at an angle of 45 to horizontal, The co-effiecient of friction between the ground and ladder is 1/2 and wall and ladder is 1/3. If someone climbing the ladder weighs half as much as it, how far can they climb before the ladder falls?

So far I have, with h being height climbed, N being reaction force from wall, R being reaction force from the ground, F being friction on the ground and G being the friction on the wall, with A being on the ground and B being on the wall:

Taking moments around A:
(mg x hcos45) + (4cos45 x 2mg) = (8sin45 x N) + (8cos45 x G)

and as G=1/3N

the right side can become (8cos45 x N/3) + (8sin45 x N) therefore you now only have two unknowns. I also know that R + G = 3mg and F=N, but I can't work out where to go from here as because there are two rough surfaces I don't know how to get rid of the unknowns and get a number out in place of N which you can therefore use to solve for h.

Many thanks!
2. (Original post by mayjb)
...
Okay it's a bit hard to follow so I had to do this question myself, so read through this to make sure you've done it correctly so far.

We want to find the distance

Let

Then taking moments we have:

...which simplifies down to:

Then you know that and which you can use to express in terms of and find the distance from there.

To verify, you can also take and see that you get the same value of

P.S. Haven't done M2 in months so if I made a mistake someone will point it out
3. Take moments around A and B and also resolve forces horizontally and vertically. This should give you enough equations to find your unknowns
4. (Original post by RDKGames)
Okay it's a bit hard to follow so I had to do this question myself, so read through this to make sure you've done it correctly so far.

We want to find the distance

Then taking moments we have:

Then you know that and which you can use to express in terms of and find the distance from there.

To verify, you can also take and see that you get the same value of

P.S. Haven't done M2 in months so if I made a mistake someone will point it out
I haven't done the question but looking at your diagram, there should be some s and s in your working.

I think you've tried to take moments like M1, where the line of action of the force is perpendicular to the point you're taking moments about. But M2 includes angles.

EDIT: I think I'm wrong here. See below.
5. (Original post by notnek)
I haven't done the question but looking at your diagram, there should be some s and s in your working.

I think you've tried to take moments like M1, where the line of action of the force is perpendicular to the point you're taking moments about. But M2 includes angles.
I have indeed taken angles into account, but simplified it all down for the response as 45 degrees is nice to work with.
6. (Original post by RDKGames)
I have indeed taken angles into account, but simplified it all down for the response as 45 degrees is nice to work with.
Oh okay - sorry. I guessed that because you said that you might have forgotten M2 that you made this error. I didn't actually check your working.
7. (Original post by RDKGames)
I have indeed taken angles into account, but simplified it all down for the response as 45 degrees is nice to work with.
I'm not sure that really helps the OP - you should put angles in.

I would also suggest a comment about 'In equilibrium ....' to explain why you are taking moments.
8. (Original post by notnek)
Oh okay - sorry. I guessed that because you said that you might have forgotten M2 that you made this error. I didn't actually check your working.
No worries - I added it in just for clarity then.
9. (Original post by RDKGames)
No worries - I added it in just for clarity then.
I haven't worked through this, but please could your weights have a g attached.
10. Let if you wish then. and are constants and since they always come together in these types of questions I just denoted their product as a single variable just so the equations are cluttered with less variables.
11. (Original post by RDKGames)
Let if you wish then. and are constants and since they always come together in these types of questions I just denoted their product as a single variable just so the equations are cluttered with less variables.
Initially you hadn't stated that M = mg. M would therefore have been taken as being the mass and you'd have been losing accuracy marks for missing g out. That's how Edexcel treat it anyway. Strictly speaking, your equations are dimensionally incorrect which might incur a bigger penalty with other boards, I don't know.
12. I got the answer as 11.2m. Does anyone know if that is correct?
13. (Original post by AspiringUnderdog)
I got the answer as 11.2m. Does anyone know if that is correct?
But the ladder is only 8m in length...
14. (Original post by RDKGames)
But the ladder is only 8m in length...
fml
15. (Original post by RDKGames)
But the ladder is only 8m in length...
Okay I accidentally took friction against the wall as going down. I now got 5.71...m
You get anything like that?
16. (Original post by AspiringUnderdog)
Okay I accidentally took friction against the wall as going down. I now got 5.71...m
You get anything like that?
Yep that's what I got.
17. (Original post by RDKGames)
Yep that's what I got.
Hooray!

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