#### physicsman2

1. The problem statement, all variables and given/known data
A uniform ladder of mass m and length l leans at an angle theta against a frictionless wall. If the coefficient of static friction between the ladder and the ground is mu, determine a formula for the minimum angle at which the ladder will not slip.

2. Relevant equations

3. The attempt at a solution
I know how to create the free body diagram and use the rules for static equilibrium in the x and y direction, but I don't know why the torque equation is set up as: W(lsin(theta))-(1/2)(lmgcos(theta)) = 0 where W is the Force exerted by the wall and mg is the Force due to gravity acting on the ladder at the center of its mass. Mainly, where do the sin and cos come from when the the wall force is parallel and gravity perpendicular to the ground. I know it has something to do with the ladder being at an angle, but I can't wrap my mind around where they exactly came from. Also, it's a frictionless wall which makes me believe that the wall force should not have any components. Thanks in advance.

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#### Delphi51

Homework Helper
Hi Physicsman. The theta you are using is between the ladder and the ground, right? The wording suggests to me that it should be the angle between the wall and the ladder . . .

Are you finding the torque about the point where the ladder touches the ground? That should be the easiest route because the force is complicated at that point and we can just ignore it this way. The first torque is due to the force of the wall Fw which must be perpendicular to the wall because it is frictionless so the torque is Fw*sin(θ)*L (the component of Fw perpendicular to the ladder times distance from pivot).

#### physicsman2

I found the diagram online: https://egp.rutgers.edu/gifs/CJ/9-7.gif

I originally did make the point where the ladder touches the ground the pivot point, but for some reason I don't understand why the force has a component when it's just the same force exerted. Wait, is it because you need to use the perpendicular force of the Fw to the ladder instead of the wall since you're working with torque? Then what about the mgcos(theta), where does cos come from?

Homework Helper

#### physicsman2

But why is it cos and not sin for the perpendicular Fg? I don't understand why in this case the angle is sin(pi/2 - theta) and not just sin theta like in the perpendicular Fw.

Edit: I think I see why now, I thought theta would be the angle between the ladder and mg but I think theta would be the angle between the ladder and the perpendicular force. Am I correct and if so can someone explain why it's so in this case?

Last edited:

#### Delphi51

Homework Helper
Maybe a diagram will help.

I have trouble "seeing" which angle is θ so I marked the angles complementary to θ (90-θ) with a solid dot.

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