Torque / Static Equilibrium Question

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Homework Help Overview

The discussion revolves around a physics problem involving a uniform ladder leaning against a frictionless wall, focusing on the conditions for static equilibrium and the minimum angle to prevent slipping, given the mass of the ladder, its length, and the coefficient of static friction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the application of torque and forces acting on the ladder, questioning the setup of the torque equation and the definitions of various forces involved. There are inquiries about the normal force and friction force relationships, as well as the notation used in the original poster's calculations.

Discussion Status

The discussion includes attempts to clarify the original poster's reasoning and notation. Some participants have provided insights into the forces acting on the ladder and the pivot point chosen for torque calculations. One participant has indicated they resolved their confusion and found their mistake.

Contextual Notes

There is mention of the expected answer being different from the original poster's calculations, prompting questions about the assumptions made regarding forces and torques. The discussion also highlights the need for clear definitions of variables used in the problem.

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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