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    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!
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    (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.

    Name:  IMG_0864.jpg
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    We want to find the distance a

    Let M=mg

    Then taking moments we have:

    M_o(A): M\cdot a\cos(45) + 2M\cdot 4\cos(45) = \frac{1}{3}S\cdot 8\cos(45)+S\cdot 8 \sin(45)

    ...which simplifies down to:

    M_o(A): \frac{1}{2}aM+4M=\frac{16}{3}S

    Then you know that 3M=R+\frac{1}{3}S and S=\frac{1}{2}R which you can use to express S in terms of M and find the distance a from there.

    To verify, you can also take M_o(B) and see that you get the same value of a

    P.S. Haven't done M2 in months so if I made a mistake someone will point it out
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    Take moments around A and B and also resolve forces horizontally and vertically. This should give you enough equations to find your unknowns
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    (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.

    Name:  IMG_0864.jpg
Views: 40
Size:  502.8 KB

    We want to find the distance a

    Then taking moments we have:

    M_o(A): \frac{1}{2}aM+4M=\frac{16}{3}S

    Then you know that 3M=R+\frac{1}{3}S and S=\frac{1}{2}R which you can use to express S in terms of M and find the distance a from there.

    To verify, you can also take M_o(B) and see that you get the same value of a

    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 (\cos 45)s and (\sin 45)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.
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    (Original post by notnek)
    I haven't done the question but looking at your diagram, there should be some (\cos 45)s and (\sin 45)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.
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    (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.
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    (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.
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    (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.
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    (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.
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    Let M=mg if you wish then. m and g 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.
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    (Original post by RDKGames)
    Let M=mg if you wish then. m and g 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.
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    I got the answer as 11.2m. Does anyone know if that is correct?
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    (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...
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    (Original post by RDKGames)
    But the ladder is only 8m in length...
    fml
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    (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?
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    (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.
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    (Original post by RDKGames)
    Yep that's what I got.
    Hooray!
 
 
 
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