Question

To get up on the roof, a person (mass 65.0 kg) places a 5.40 m aluminum ladder (mass 13.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom?

Answer 1

let theta be the angle made by the ladder with horizantal at the bottom

costheta = x/L = 2/5.4

theta = 68.3

IN equilibrium net torque = 0

Ftop*L*sintheta - Wperson*l*costheta - Wladder*L/2*costheta = 0

(Ftop*5.4*sin68.3) - (65*9.8*3*cos68.3) - (13*9.8*5.4/2*cos68.3) = 0

Ftop = 166.1 N   <<<<=============answer

at the bottom

along vertical

Fv - Wperson - Wladder = 0

Fv = (65+13)*9.8 = 764.4 N

along horizantal

Fh - Ftop = 0

Fh = Ftop = 166.2 N

force Fbottom = sqrt(Fh^2+Fv^2)

Fbottom = 782.26 N <<<<=============answer