Question

In the picture a man climbs a ladder leaning at 500 abpve the horizontal. The static friction coefficient between the ladder and ground is .6, the ladder is 3 meters long and weighs 100 newtons. If the man weighs 700 N, how far up the ladder can he climbs without the ladder sliding out from under him? (Assume no friction between the wall and ladder)

Answer 1

Solution :

Given :

= 50^o

= 0.6

L = 3 m

W_L = 100 N

W_P = 700 N

.

Here, F_Wall = F_x = F_y

Where, F_y = W_L + W_P = 100 N + 700 N = 800 N

Thus : F_Wall = F_x = F_y = (0.6)(800 N) = 480 N

.

Now, Consider torque about the lower end of the ladder.

For equilibrium : Net torque will be Zero.

(L)(F_wall) sin - (L/2) W_L cos - (x) W_P cos = 0

(L)(F_wall) sin - (L/2) W_L cos = (x) W_P cos

(3 m)(480 N) sin(50) - (1.5 m) (100 N) cos(50) = (x) (700 N)cos(50)

(1006.686) = (x) (449.95)

x = 2.237 m

.

Therefore : He can climb up to 2.237 m up the ladder without the ladder sliding out from under him.