Torque / Static Equilibrium Question

AI Thread Summary
The discussion revolves around a physics problem involving a uniform ladder leaning against a frictionless wall, requiring the calculation of the minimum angle to prevent slipping. The user initially attempted to solve the problem using torque equations but arrived at an incorrect answer. Key points include the need for clarity in defining variables such as torque, normal force, and friction force, as well as the importance of considering all forces acting on the ladder. Ultimately, the user successfully identified their mistake and found the correct solution without further assistance. The conversation highlights the complexities of static equilibrium and the role of friction in such scenarios.
ohlhauc1
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I need help with the following question.

1. A uniform ladder of mass M and length L leans at an angle A against a frictionless wall. If the coefficient of static friction between the ladder and the ground is Q, what is the minimum angle at which the ladder will not slip?

Answer so far:
let T stand for torque

Tladder + TNwall = 0
rFcosA + rFsinA = 0
LFcosA = -LFsinA
mgcosA / sinA = Ffriction
mg = FfrictiontanA
(1 / QcosA) = tanA
A = tan^-1(1 / QcosA)

*The real answer is supposed to be A = tan^-1(1 / 2Q)

I was wondering if you could tell me what I did wrong, and what I should do to get the right answer. Thanks!
 
Last edited:
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Can cosA equal Q, which is actually mu?
 
ohlhauc1 said:
I need help with the following question.

1. A uniform ladder of mass M and length L leans at an angle A against a frictionless wall. If the coefficient of static friction between the ladder and the ground is Q, what is the minimum angle at which the ladder will not slip?

Answer so far:
let T stand for torque

Tladder + TNwall = 0
rFcosA + rFsinA = 0
LFcosA = -LFsinA
mgcosA / sinA = Ffriction
mg = FfrictiontanA
(1 / QcosA) = tanA
A = tan^-1(1 / QcosA)

*The real answer is supposed to be A = tan^-1(1 / 2Q)

I was wondering if you could tell me what I did wrong, and what I should do to get the right answer. Thanks!

Maybe its just the notation I can't follow, but I'm not seeing some things I think I should see. What is the normal force of the floor on the ladder? What is the friction force in terms of the normal force? What is the normal force of the wall on the ladder. Is that your r? I can't see you having a net zero torque equation with less than three torques.

Please define your variables.
 
I put my pivot point at the base of the ladder so the normal force of the floor on the ladder would be 0. The friction force would be equal in magnitude to the normal force the wall exerts on the top of the ladder. r is the length of the ladder (L) because it represents the length of the lever arm a.k.a. ladder.

T represents torque
Q represents mu, the coefficient of friction
A represents theta, the angle
M represents the mass of the ladder
F represents force

Does this help?
 
Got It!

I was able to get the answer, so you do not have to reply. I found my mistake! :)
 
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