1. Dec 24, 2008

### MM156

1. The problem statement, all variables and given/known data
A uniform ladder 5 m long (and 12 kg) is leaning against the wall. The height from the ground to the point at which the ladder touches the wall is 4 m. The wall is frictionless but the ground is not. A painter (55kg) climbs 70% of the way up the ladder when it begins to slip. What is the minimum coefficient of friction?

2. Relevant equations
T=-T
T=F dcos@
F friction = F normal * Mu (coefficient)

3. The attempt at a solution
I figured all of the dimension and distances of the triangle. It turns out that the painter traveled 3.5 m along the ladder and is 2.8 m off the ground. The center of mass (when extended to the floor) is 2.1 m away from the point the ladder touches the ground. There is a force coming from the ground Fg which I broke into components Fy and Fx. I know Fx = Fw and that mg + M(painter)g + Fy = 0. I'm having difficulty in determining which of the forces from the ground is the friction force and the normal force. Would Fy be the normal force and Fg the frictional force?

2. Dec 25, 2008

### Staff: Mentor

Good.
Recheck that result.
Good.
Fg is the total force of the ground on the ladder. Fy is the normal force; Fx is the frictional force.

3. Dec 25, 2008

### MM156

Correction: It is 2.1 m from the point on the ground to the painter.
I see now. The Fg and its components were confusing me.
So, I can just add up the mg and M(painter)g to find Fy.
Then use torque and set the point on the ground as the pivot point. mg*2 + Mg*2.1= Fw*3
I didn't do the calculations yet but since Fwall = Fx...it also equals friction.
Ffriction = Fnormal*Mu
Ffriction = Fy*Mu
Mu = Ffriction/Fy

4. Dec 26, 2008

### FedEx

We would be considering two two normals. One form the wall and other from the ground. We would also consider the Mg of ladder and that of the man. Consider the frictional force. Equate them.

Also consider rotational equilibrium

5. Dec 26, 2008

### FedEx

Ffriction = Fnormal*Mu

There is a subtle relation between fy, fx and mu.

6. Dec 26, 2008

### Staff: Mentor

OK. (That's the horizontal distance.)
Right.
Right.
You'll have to redo this. The first and last terms have the wrong distance. (Looks like you mixed up vertical and horizontal distances.)
OK.

7. Dec 27, 2008

### MM156

Wouldn't the horizontal distance from the point on the ground to the wall be 3m because it's a 345 triangle? Or maybe it's 4 m because it has to be perpendicular with the radius when it's extended.

8. Dec 27, 2008

### Staff: Mentor

Exactly.

9. Oct 28, 2009

### SASGood

Think of an imaginary wheel around the pivot point and torques acting perpendicular to the wheel. Total clockwise equals total anticlockwise. seehttp://www.cutescience.com/pp/physics/torque/135Torque_files/v3_document.htm [Broken]

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