A 5.0-m long,12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder from the wall.The coefficient of static friction between the floor and the ladder is 0.28.What distance,measured along the ladder from the bottom,can a person climb before the ladder starts to slip?
A) 4.0 m
B) 3.7 m
C) 1.7 m
D) 1.3 m
E) 3.3 m

Question

A 5.0-m long,12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder  from the wall.The coefficient of static friction between the floor and the ladder is 0.28.What distance,measured along the ladder from the bottom,can a  person climb before the ladder starts to slip?
A) 4.0 m 
B) 3.7 m 
C) 1.7 m 
D) 1.3 m 
E) 3.3 m

Quizplus · Accepted Answer

The ladder is in equilibrium when the forces acting on it are balanced. The forces acting on the ladder are its weight and the normal force from the ground, as well as the friction force and the horizontal component of the tension force from the climber.

The maximum friction force that can act on the ladder is given by the coefficient of static friction times the normal force. In this case, the normal force is equal to the weight of the ladder plus the weight of the climber:

N = (12 kg + 70 kg) × 9.81 m/s² = 882.9 N

The maximum friction force is therefore:

f friction,max = μN = 0.28 × 882.9 N = 246.8 N

For the ladder to remain in equilibrium, the horizontal component of the tension force from the climber must be less than or equal to the maximum friction force:

T cosθ ≤ f friction,max

where θ is the angle the ladder makes with the ground. In this case, θ can be found from the right triangle formed by the ladder, the wall, and the ground:

tanθ = 5.0 m / 11.83 m
θ = 22.2°

Therefore, the maximum allowed tension force is:

T ≤ f friction,max / cosθ = 246.8 N / cos22.2° = 267.9 N

The distance along the ladder that the climber can climb is given by the point at which the tension force is equal to the maximum allowed value:

T = f friction,max

T cosθ = μN
T cosθ = μ(m ladder + m climber)g
T = μ(m ladder + m climber)g / cosθ

Setting this equal to the maximum allowed tension force and solving for the distance x, we get:

x = (T cosθ / μg) - l
x = [(μ(m ladder + m climber)g / cosθ) cosθ / μg] - 5.0 m
x = (m ladder + m climber)g / μ - 5.0 m
x = [(12 kg + 70 kg) × 9.81 m/s² / 0.28] - 5.0 m
x ≈ 1.7 m

Therefore, the answer is C) 1.7 m.

A 50-M Long,12-Kg Uniform Ladder Rests Against a Smooth Vertical Wall