Answer to Question #143353 in Physics for Zech

Question

A Ladder of 6.0 meters in length weighs 150 N rests on horizontal ground and leans at an angle of 65 degrees with the horizontal against a smooth vertical wall. How far up the ladder may a 700 N person go before the ladder slips? The coefficient of friction between ladder and ground is 0.40.

Answer 1

The diagram:

Assume that the system is in equilibrium when the person is at the highest point $l$ meters from the top of the ladder (measured along the ladder).

The force of friction:

$$f=\mu N_F.$$

The equilibrium of forces:

$$x:N_W-f=0,\\ y: N_F-W-F=0.$$

Substitute the force of friction and find the normal forces:

$$N_F=W+F,\\ N_W=\mu N_F=\mu(W+F).$$

Now, it's time to write the torque around the lower end:

$$W\text{ cos}\theta\cdot\frac L2+F\text{ cos}\theta\cdot(L-l)-N_WL\text{ sin }\theta=0,$$

substitute the normal force from the wall:

$$W\text{ cos}\theta\cdot\frac L2+F\text{ cos}\theta\cdot(L-l)-\\-\mu(W+F)L\text{ sin }\theta=0,\\ l=0.4\text{ m}.$$

The person may climb 5.6 m before the ladder starts slipping.