Question

An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is 0.55, and the ladder makes a 50.0° angle with the ground. How far up the ladder can an 700 N person climb before the ladder begins to slip?
______ m

Answer 1

Given length of the ladder L = 8.0 m

weight (force ) of ladder is 240 N and the weight of the person is about 700 N

The coefficient of static friction between the ladder and the ground is 0.55

ladder makes an angle of 55 degrees with the ground

let the distance the person can move up the ladder is x , then

if the ladder does not slips , then the system is in equilibrium then , the force exerted by the wall is F_w = Frictional force

   F_w = F_f = mue_s*Fy
vertical forces are

   Fy =    700+240 N = 940 N

and equating the net torque to zero we get the dist

we know that the torque T = rXF = r*F sin theta

taking the pivot point at the ladder is on the ground then the torques (positive ) are

   700N*x + 240*4 cos40 (counter clockwise direction )

and negative torque is

   Fw *8 sin50 (clockwise direction )

but the net torque is zero so

   700*x + 240*4 cos50 - Fw *8 sin50 = 0

   700*x + 240*4 cos50 - 940*0.55 *8 sin50 = 0

solving for x, x =3.64 m , up the ladder