Question

A uniform ladder of weight W leans without slipping

against a wall making an angle with the floor. There is friction between the ladder and the floor, but friction between the ladder and the wall is negligible.

The magnitude of the normal force exerted by
the floor on the ladder is

(A) W
(B) Wsin(theta)
(C) Wcos (theta)
(D)W/2 sin(theta)

Answer 1

OPTION D.

The magnitude of the normal force exerted by the floor on the ladder is equal to the weight of the ladder times the cosine of the angle of inclination with the floor, resulting in Wcos(theta).

Step-by-step explanation:

The student's question revolves around identifying the magnitude of the normal force exerted by the floor on a ladder that leans against a wall. We know that the weight of the ladder acts downward because of gravity, and there is friction between the ladder and the floor, but negligible friction between the ladder and the wall. To find the normal force, we must analyze the forces in the vertical direction.

Since friction between the ladder and the wall is negligible, the normal force exerted by the wall is in the horizontal direction and does not affect the vertical forces. Consequently, in the vertical direction, we have two main forces: the weight of the ladder (W) and the normal force from the floor (N). The ladder is in equilibrium, which means the vertical components of all forces must sum to zero. Hence, the normal force is equal in magnitude but opposite in direction to the total vertical component of the weight.

The correct formula to identify the magnitude of the normal force is N = W × cos(θ), where θ is the angle of inclination with the floor. This equation comes from breaking down the weight of the ladder into two components: one parallel to the floor (which is counteracted by friction) and one perpendicular (normal) to the floor (which is counteracted by the normal force). Therefore, the magnitude of the normal force exerted by the floor on the ladder is Wcos(θ).

Answer 2

The magnitude of the normal force exerted by the floor on the ladder is (C) W cos(theta).

Step-by-step explanation:

The magnitude of the normal force exerted by the floor on the ladder is W cos(theta).

In this case, the weight of the ladder W can be broken down into two components:

1. The component parallel to the floor, which is W cos(theta).
2. The component perpendicular to the floor, which is W sin(theta).

Since the ladder is in equilibrium and not slipping, the normal force from the floor must balance the component of the weight parallel to the floor, which is W cos(theta).

Therefore, the magnitude of the normal force exerted by the floor on the ladder is (C) W cos(theta).