# Homework Help: Co-efficient of friction of a uniform ladder leaned on rough wall

1. Jul 17, 2013

### nvjnj

1. The problem statement, all variables and given/known data

A uniform ladder of weight (W) is leaned across a rough wall and on a rough floor. The reaction at the wall is (S) with frictional force (G)(upward direction) while the reaction at the floor is (R) with frictional force (F)(away from the wall direction). Frictional forces are taken to be limiting and the ladder is in equilibrium. The ladder's top is 4m above the ground while its foot is 2m away from the wall. Assuming that the coefficient of friction is same at the wall and the ground, calculate the coefficient of friction.

2. Relevant equations

No equations specified, guessing F=ma, Moments equilibria and F=μR

3. The attempt at a solution

R(upwards) : R+G=W......................................(1)
R(away from wall) : S=F..................................(2)
F=μR............................................................(5)
G=μS............................................................(6)

I may/may not have 6 independent equations to arrive at the solution so best not to consider them, but please do point out the errors, i can't seem to calculate μ with all these equations

Last edited: Jul 17, 2013
2. Jul 17, 2013

### Simon Bridge

Divide all forces into components through the center of mass of the ladder and perpendicular?
You are making people guess what all those letters stand for.

3. Jul 17, 2013

### ehild

Hi nvjnj, welcome to PF.

The problem text says that F, the friction force from the wall points away from the wall. That can not be true, and your equations correspond to the correct orientation.
It is enough one equation for the momenta. If the momentum is zero with respect to a point of the ladder, it will be zero with respect to other points of it.

ehild

4. Jul 17, 2013

### nvjnj

Quote by ehild

You're right, the friction is in the opposite direction, F(towards the wall direction). Can you please help me to evaluate μ?? I'm pretty overwhelmed with the no. of equations. I need to get μ=√5-2.

P.S. I'm not quite sure how to transfer this post to the into. physics area...I'm a rookie in forums, sry.

5. Jul 17, 2013

### nvjnj

Hey guys, thanks for all your effort, i got the answer just now so ignore the last post. I figured I was just going around substituting without a clear aim on getting μ.

Thanks @ehild for pointing out that two momenta equations were unnecessary, helped a great deal.

6. Jul 17, 2013

### ehild

You are welcome.

ehild