What Are the Equations to Solve a Ladder's Minimum Angle Problem with Friction?

[wolfgarre]

Need help solving/setting up this problem for Physics:

A ladder of length L and weight W rests against a wall. The coefficient of static friction between the ladder and the floor and between the ladder and the wall is mu=0.5 What is the minimum angle the ladder can make with the floor (theta) without slipping?

What i know:
sum of torques = 0
sum of forces in y dir = 0
sum of forces in x dir = 0
forces in y dir = normal force from floor, friction with wall (mu times normal force from wall),and weight (W)
forces in x dir = friction with floor (mu times normal force from floor), normal force from wall

I have set the torque point at the ladder's contact with the floor and know there will be five equations to deal with.
If anyone could help me find these equations or help me set the problem up i'd appreciate it.

Jeff

 

[NateTG]

The torque from a force is r sin α where r is the length of the level arm to the point where the force is applied, and α is the angle that the force makes with the arm.

 

[M98Ranger]

To solve this problem, we will use the equations for rotational equilibrium and the equations for static friction. Let's start by setting up the problem and defining our variables:

- L: length of the ladder
- W: weight of the ladder
- mu: coefficient of static friction
- theta: angle between the ladder and the floor

We can start by drawing a free body diagram of the ladder, which will help us visualize the forces acting on it. The ladder is in equilibrium, so the sum of all forces and torques acting on it must equal zero.

First, let's consider the forces acting in the y-direction (perpendicular to the floor). These include the normal force from the floor, the normal force from the wall, and the weight of the ladder. We can write an equation for the sum of forces in the y-direction:

ΣFy = Nf + Nw - W = 0

where Nf is the normal force from the floor, Nw is the normal force from the wall, and W is the weight of the ladder. We can also use the equation for static friction to relate the normal force and the coefficient of friction:

Nf = mu*Nw

Next, let's consider the forces acting in the x-direction (parallel to the floor). These include the friction force from the floor and the normal force from the wall. We can write an equation for the sum of forces in the x-direction:

ΣFx = f - Nw = 0

where f is the friction force from the floor. We can also use the equation for static friction to relate the friction force and the normal force:

f = mu*Nw

Now, let's consider the torques acting on the ladder. We will choose the point of contact between the ladder and the floor as our pivot point. The forces acting on the ladder that will create a torque are the weight of the ladder and the friction force from the wall. We can write an equation for the sum of torques:

Στ = -W*L*sin(theta) + f*L*cos(theta) = 0

where L is the length of the ladder and theta is the angle between the ladder and the floor. We can also substitute in our expression for f from the equation for the sum of forces in the x-direction:

Στ = -W*L*sin(theta) + mu*Nw*L*cos(theta) = 0

Now, we have four equations and