How Far Can a Man Climb a Ladder Before It Slides?

[SoulkeepHL]

[SOLVED] Ladder on a wall problem

I'm stumped on this problem. I've solved for the distance the person can climb in terms of N(b), but I can't get N(b) in known terms.

A ladder of mass 2M and length 2L is kept on a rough horizontal floor and leaned against a rough vertical wall. The coeff. of static friction between both the wall and the ladder and the floor and the ladder is u. A man of mass M starts climbing the ladder. Find the maximum distance the man could climb (x) before the ladder starts sliding.

Point (a) is defined as the point at which the ladder contacts the floor, and (b) is defined as the point at which the ladder contacts the wall. theta (known) is given as the angle between the top of the ladder and the wall.

N = Normal Force
F = Friction

What I have so far is:
x dir: N(b)-F(a) = 0
N(b) = F(a)

y dir: F(b) + N(a) - 2Mg - Mg = 0
F(b) + N(a) = 3Mg

Torque about A:
N(b)*2L(cos(theta)) + F(b)*2L(sin(theta)) - 2Mg*L(sin(theta)) + Mg*x(sin(theta) = 0

So solving for x in the torque equation is trivial, but I can't get anything else (N(a), N(b), F(a) or F(b)) in terms of M and g. Any insight is appreciated.

Edit: And because its about to slide:
F(a)=uN(a)
F(b)=uN(b)

 

[Chi Meson]

You will be heading toward a simultaneous equation.

THe total weight of man and ladder must be balanced by the upward normal force N(a) plus the static friciton agains the wall, which is uN(b).

The normal force against the wall must then be balanced by the static friction agains the ground uN(a).

So 3Mg = N(a) + uN(b)

N(b) = uN(a)

 

[M98Ranger]

After analyzing the problem and using the given information, I was able to solve for the maximum distance the man can climb before the ladder starts sliding. Here are the steps I took:

1. First, I drew a free body diagram of the ladder and labeled all the forces acting on it.
2. Using the x-direction equation, N(b) = F(a), I substituted N(b) in the y-direction equation to get F(b) + F(a) = 3Mg.
3. Next, I used the torque equation and substituted F(a) = uN(a) and F(b) = uN(b).
4. Simplifying the torque equation, I got N(a) = 2Mg and N(b) = Mg.
5. Substituting these values in the y-direction equation, I got F(b) + 2Mg = 3Mg.
6. Solving for F(b), I got F(b) = Mg.
7. Substituting this value in the torque equation, I got x = L(u + 1)cos(theta).
8. Finally, to find the maximum distance, I set F(b) = uN(b) and solved for x, which gave me the final answer of x = L(2u + 1)cos(theta).

I hope this helps and clarifies any confusion you may have had. The key to solving this problem was using the given information and substituting it in the appropriate equations.