Co-efficient of friction of a uniform ladder leaned on rough wall

[nvjnj]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

R(upwards) : R+G=W.........(1)
R(away from wall) : S=F........(2)
Moments(about foot) : 4S+2G=W......(3)
Moments(about top of ladder) : W+4F=2R...(4)
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
(Answer μ=(√5)-1 )

 

[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.

 

[ehild]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

R(upwards) : R+G=W.........(1)
R(away from wall) : S=F........(2)
Moments(about foot) : 4S+2G=W......(3)
Moments(about top of ladder) : W+4F=2R...(4)
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
(Answer μ=(√5)-1 )

Hi nvjnj, welcome to PF.

Your thread would fit better to the sub-forum "Introductory Physics"

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

 

[nvjnj]

Quote by ehild

Hi nvjnj, welcome to PF.

Your thread would fit better to the sub-forum "Introductory Physics"

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

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.

 

[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.

 

[ehild]

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.

You are welcome.

ehild