# Simple Ladder leading against a vertical wall become complicated.

1. Dec 11, 2007

### dr3vil704

1. The problem statement, all variables and given/known data
A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is μs = 0.500, determine the smallest angle the ladder can make with the floor without slipping.

2. Relevant equations
I believe the set up of the problem is:
(1/2 X mg x Cos$$\theta$$ = ( N x L Sin$$\theta$$ ) +( Ff X Cos$$\theta$$ )

3. The attempt at a solution
(Here Maybe the problem number)

I bring Ff X cos$$\theta$$ to the other side, and factor out the mg and divived the Sin. to get Tan$$\theta$$ by it self.
So the result is:
tan$$\theta$$= (1/2 X L X mg - $$\mu$$mg) / (mg X L)
Tan$$\theta$$ = (1/2 L - $$\mu$$) / L

Okay, then the problem is that I'm suppose to get a numeric answer, to $$\theta$$ which is 36.8 degree.
If my attempt was on the right track, then How could i find the Numeric value of this if I only know what $$\mu$$ is?

2. Dec 11, 2007

### cryptoguy

I'm not sure if you did this, but you are using torque so you should set up the pivot point at either the floor or the wall to cancel out one of the Friction forces so you are left with only one F and mg. Then you make them equal (Tcw = Tccw).

3. Dec 12, 2007

### Staff: Mentor

Several problems: (1) Your torque equation should have an L in every term, which then cancels; (2) It looks like you assumed that friction equals $\mu mg$, but mg is not the normal force. Also, I don't quite see how you went from your first equation to your final version.

In addition to the torque equation, take advantage of the other conditions for equilibrium.

4. Dec 12, 2007

### dr3vil704

yes, I did put the pivot point on the floor, which cancel out the normal and friction between floor and ladder. So the frictional force that is left is between the wall and floor. And my friction and normal is not incline so it just $$\mu$$mg. The first equation is Tcw=Tccw

5. Dec 12, 2007

### Staff: Mentor

Don't forget the friction force from the wall.
Yes, but with errors.

6. Dec 12, 2007

### dr3vil704

Oh, I see the mistake now. I guess I forgot to put the distance with the force
So the equation should be
(1/2 X mg X L X Cos$$\theta$$) = (N X Lsing$$\theta$$) +(Ff X cos$$\theta$$X L)

But I do no get why the Ff agaisnt the wall is not $$\mu$$N

7. Dec 12, 2007

### Staff: Mentor

Who says it isn't?

8. Dec 13, 2007

### rl.bhat

For equilibrium we require force balance and torque balance.
For force balance Sigms Fx = 0 i.e f1 - N2 = 0 where f1 frictional force at floor and N2 is the normal reaction on the wall. Sigma Fy = 0 i.e. mg - N1 - f2 = 0.
For torque balance mg*L/2*cos(theta) = f2*cos(theta) + N2*sin(theta).Substitute the value of mg. Then (N1+ f2)*L/2*cos(theta) =f2*cos(theta) + N2*sin(theta). Since Coeff. of friction is 0.5, N1 = 2f1 and N2 = 2f2 or f1 = 2f2. Substituting these values in the torque balance equation and simplifying you get tan(theta) = 3/4 or theta = 36.8 degree.