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M2 ladders question

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!

Reply 1

RDKGames

Study Forum Helper

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

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 :smile:

(edited 7 years ago)

Reply 2

Darth_Narwhale

Take moments around A and B and also resolve forces horizontally and vertically. This should give you enough equations to find your unknowns

Reply 3

Notnek

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

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 :smile:

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.

(edited 7 years ago)

Reply 4

RDKGames

Study Forum Helper

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.

Reply 5

Notnek

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.

Reply 6

Muttley79

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.

Reply 7

RDKGames

Study Forum Helper

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. :smile:

Reply 8

tiny hobbit

Original post by RDKGames

No worries - I added it in just for clarity then. :smile:

I haven't worked through this, but please could your weights have a g attached.

Reply 9

RDKGames

Study Forum Helper

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.

(edited 7 years ago)

Reply 10

tiny hobbit

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.