Answer to Question #144059 in Physics for Laiza

Question

A ladder of 6.0 meters in length weight 150 N rest 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

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

The force of friction:

$$f=\mu N_F$$

The equilibrium of forces:

$$N_W-f=0\\N_F-W-F=0$$

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:

$$W\cos{\theta}(0.5L)+F\cos{\theta}(L-l)-\\-\mu(W+F)L\sin{\theta}=0\\l=0.4\ m$$

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