# Homework Help: Minimum angle the ladder can make with the floor without slipping

1. Dec 7, 2014

### sncobra

1. The problem statement, all variables and given/known data

A 3.0-m-long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.32.

What is the minimum angle the ladder can make with the floor without slipping?

2. Relevant equations

With the axis of rotation at the bottom of the ladder:
ΣT = 0 <- torque
ΣF = 0
T = LF
T = LFsinφ
ƒ = μN

3. The attempt at a solution
X to the right is positive, Y up is positive

Nw = Normal of wall
Nf = normal of floor
fs = static friction force
Fg = gravity

ΣFx = 0
Nw - fs=0
Nw = μNf

ΣFy = 0
Nf-Fg = 0
Nf = mg

So,
Nw=μmg

Now I'm still having trouble with torques so bear with me. Since these two forces go through the axis then they produce no torque correct?

TNf = 0
Tfs = 0

and

TFg = (Fg)(L/2) = mg(L/2)
TNw = (Nw)(L)

Am I on the right path? I think you sum the two torques but I don't have the mass so that wouldn't work.
Also, how do I know if the torque is negative or positive? and which way do I make torque positive? CW or CCW?

Appreciate any and all advice :)

2. Dec 7, 2014

### ehild

Welcome to PF!

Correct so far.

It is right that the forces have zero torques if their line of action goes through the axis. But you calculate the torques incorrectly. Torque is force multiplied with the lever arm, the distance of the line of action from the axis. https://sites.google.com/site/torqu...xamples-of-torque-in-everyday-life/torque.gif
Remember, the torque makes the ladder rotate. Imagine that you exert force in the same direction as the force in question. In what direction will the ladder rotate? Do the normal force and the weight rotate in the same way?

3. Dec 7, 2014

### Simon Bridge

Yep - the sum of the torques must be zero if the ladder is static.
Leave the mass of the ladder in as a variable "m".

4. Dec 7, 2014

### TSny

Hello and welcome to PF!

Yes.

These aren't correct. For example, L/2 is not the "lever arm" for Fg because the direction from the axis to the center of mass is not perpendicular to the direction of the force.

Once you set the sum of the torques to zero, maybe some things will cancel out of the equation.

Usually, CCW is chosen as positive. But it won't make any difference here as long as you take CW and CCW torques to have opposite signs. A torque is CCW if it tends to rotate the ladder CCW around your chosen axis.

5. Dec 7, 2014

### sncobra

Okay I wasn't sure I did the lever arm right thanks all.

So would the lever arm for Fg be d1 in the picture? and then d2 for Nw?

d1 = (L/2)cos60°

I'm still not sure where to go for the torque of Nw

And if I chose CCW to be positive, then Nw would be negative and Fg would be positive? Is that correct?

6. Dec 7, 2014

### TSny

Yes. (Of course, in your problem the angle is not necessarily 60o.)

I'm not sure what you're having a problem with here. Is if finding the distance d2 in the diagram?

Yes, the torque from Nw would be negative and the torque from Fg would be positive.

7. Dec 7, 2014

### sncobra

What do you mean by this? I remember my professor saying something about the angle between the lever arm and force.

Yeah and I'm not sure what the angle you would use here would be either. I'm going to use a right triangle correct?

Thanks, you definitely cleared up the direction for me and are helping quite a bit :)

8. Dec 7, 2014

### ehild

The lever arm is perpendicular to the force. You need the length of the level arm.

Use the angle θ the ladder makes with the horizontal . Your goal is to find the angle when the ladder starts to slip, by solving the equations you wrote up. See the link in my post #2.