Applied Maths Ladder slipping Question

TheSetMiner

A ladder of length 2L rests against a smooth wall and a rough floor of coefficient of friction 1/2. Find the angle x at which the ladder slips.

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Basically I think that you have to set this up by comparing horizontal and vertical forces and by then taking the moments about the floor but the answer Ii got in doing this was approx 26 degrees which seems a little nonsensical given that the whole is smooth. Would anyone like to attempt this question themselves and tell me how they do? It would be interesting to compare answers and maybe see where I have gone wrong.


Thanks fro any help!

Mr Pseudonym

Is this an Ordinary-level question or are you in fifth year?

I presume the angle X is between the horizontal and the ladder.

There are four forces: reaction on the vertical (N), weight, and reaction (N') and friction on the horizontal.

Up = Down; Left = Right: N' = W; F = N.

CW-moments = ACW-moments (about the point in contact with the ground): N(2l)cosX = W(l)sinX ... N = WtanX/2.

At the point of slipping, F = coefficient of friction x N': WtanX/2 = (0.5)W ... tanX = 1 ... X = 45 degrees.


Caveat: having done my LC a few years ago, I am very rusty!

TheSetMiner

Mr Pseudonym wrote: »

Is this an Ordinary-level question or are you in fifth year?

I presume the angle X is between the horizontal and the ladder.

There are four forces: reaction on the vertical (N), weight, and reaction (N') and friction on the horizontal.

Up = Down; Left = Right: N' = W; F = N.

CW-moments = ACW-moments (about the point in contact with the ground): N(2l)cosX = W(l)sinX ... N = WtanX/2.

At the point of slipping, F = coefficient of friction x N': WtanX/2 = (0.5)W ... tanX = 1 ... X = 45 degrees.


Caveat: having done my LC a few years ago, I am very rusty!

nice one thanks man!

TheSetMiner

Can you explain how you get the expressions for the clockwise and anti-clockwise moments? Is it a matter of taking the perpendicular component vectors to the ladder? The example I have in my book has confused me as it seems to have a rather convoluted expression for the clockwise moment which involves both cos and sin in the expression.

Appreciate if you can clear this up

Mr Pseudonym

I'm not entirely sure what you don't understand, so I'll briefly explain it in full.

Here's a hastily drawn diagram: Breathtaking App Maths Diagram.png.

Split forces into components parallel and perpendicular to the ladder. Take moments about x, i.e. point where ladder touches the ground. The components that are parallel to the ladder act through x, and so can be ignored. There are two relevant components, each perpendicular to the ladder (important, because the perpendicular distance from each of their lines-of-action to x is the same as each's distance along the ladder), and each acting in opposite directions. Moment = F x perpendicular distance.

Hope that explains everything.

TheSetMiner

i get it now thank you man