Anyone up for an A level maths question?

[Mykingdomforanickname]

A uniform ladder of length 4m and mass 30kg rests against a smooth vertical wall. The ladder makes a 70 degree angle with the rough horizontal ground and is in limiting equilibrium. Find the coefficient of friction between the ladder and the ground.

Can anyone confirm whether the normal reaction force is 147? Can anyone explain how to calculate the friction force?

Any help much appreciated!

[JustRichmal]

Call the point where the ladder touches the floor, a point half way along the ladder and the point where the ladder touches the wall A, B and C respectively. Assume the mass of the ladder is all at point B.

Resolve vertically. So the downward force at B of 30g is matched with an upward force at A of 30g.

Resolve horizontally. At A there is a horizontal force of friction, F, towards the wall, matched by a force at C, out from the wall.

You can then take moments about point C. These are assumed to be perpendicular to the ladder. The clockwise and anti-clockwise moments need to balance.

So 230gcos70+4Fsin70=430gcos70.

So Fsin70=230gcos70

F=230gcos70/sin70

It is in limiting equilibrium and I do not have a symbol for mu on my keyboard so I'm calling the coefficient of friction m.

So F=m30g

We now have 2 equations for F, so can out them equal to each other.

So m30g=2*30g/tan70
m=2/tan70

This is what I am remembering from dd's maths 2 years ago, so it could be horribly wrong. I can remember saying to resolve vertically, resolve horizontally, take moments. I would wait for Noblegiraffe or some other good maths person to confirm this.

I know dd used the CGP books for A level maths, and they were really clear and made it seem simple!

[DadDadDad]

@JustRichmal - Where you've got F = 230g/tan70, I'm getting F = (1/2)30g/tan70. (I took moments about A because that way I could ignore two forces). I'll see if I can work out where you or I have gone wrong. (I agree with your approach).

[DadDadDad]

So Fsin70=230gcos70*

That line in Just's working should be 4Fsin70 = 2*30gcos70.

That leads to m = 0.5/tan70 or 0.5tan20 = 0.182

and F is around 55N (assuming g = 10).

[DadDadDad]

Bold fail with use of !

I was correcting Just's line which says: So Fsin70=230gcos70

there should be a 4 in front of the F.

[JustRichmal]

DadDadDad Oops, yes, I did not carry the 4 down. I should have put in more steps, rather than just skip through it in my head. Thank you for correcting, but I'm pleased I got the physics basically right.

Also, it did occur to me, when I had posted, that taking the moments about A would have been easier. (otherwise, why bother resolving horizontally) It is good to do these questions, as making mistakes is how to learn. I am teaching dd Further maths this year, so I need to get up to speed on all this.

[JustRichmal]

At least my use of * did not fail.

[Mykingdomforanickname]

Thank you both very much!