Question

A 10 meter long ladder leans against the wall as shown. If the ladder weighs 200 N and there is just enough frictional force to allow a 800 N person to climb to the top safely, what is ømin? Note: µFloor=0.675.

Answer 1

As the ladder is in equilibrium, net force and net torque acting on it must be zero.

let theta is the minimum angle required.

Apply, Fnety = 0

N - 200 - 800 = 0

N = 1000 N

Force exrted by wall on the laader, F_wall = N*mue_s

= 1000*0.675

= 675 N

Apply, Net torque acting on the ladder about bottom,

Tnet = 0

200*5*sin(90 - theta) - 800*10*sin(90-theta) + 675*10*sin(theta) = 0

1000*cos(theta) - 8000*cos(theta) +6750*sin(theta) = 0

6750*sin(theta) = 7000*cos(theta)

tan(theta) = 7000/6750

tan(theta) = 1.037

theta = tan^-1(1.037)

= 46.04 degrees