Question

Problem 2

A person with mass m1 = 74 kg person climbs a distance d from the base of a uniform ladder before the ladder starts to slip. The ladder has a mass m2 = 30 kg and length L = 7.0 m and rests against a smooth vertical wall at an angle θ = 65° with the floor. The coefficient of static friction between the floor and the ladder is 0.40. Use the symbolic notations given before using numerical substitutions.

A) Find the normal force N from the floor and the force of the wall Fw on the ladder when the man is on the ladder.

B) Find the distance d that the person climbs on the ladder before the ladder starts to slip.

Answer 1

A)
Equating the vertical forces,
N = m1 * g + m2 * g
Where m1 is the mass of the person and m2 is the mass of the ladder.
N = (m1 + m2) * g
= (74 + 30) * 9.8
= 1019.2 N

Equating the horizontal forces,
Ff = Fw
Where Ff is the frictinoal force and Fw is the force of the wall on the ladder.
Ff can be written as, Ff = * N
Where is the coefficient of friction.
Fw = Ff = * N
= 0.40 * 1019.2
= 407.68 N

B)
Consider the net torque about the bottom of the ladder.
m2 * g * L/2 * cos + m1 * g * d * cos - Fw * L * sin = 0
Where L is the length of the ladder and Fw is the normal force from the wall on the ladder.

Substituting values,
30 * 9.8 * 7/2 * cos(65) + 74 * 9.8 * d * cos(65) - 407.68 * 7 * sin(65) = 0
434.87 + 306.48 * d - 2586.38 = 0
306.48 d = 2151.51
d = 2151.51 / 306.48
= 7 m