Question

A chair of weight 80.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 38.0 N directed at an angle of 40.0 ∘ below the horizontal and the chair slides along the floor. Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.

Answer 1

The magnitude of the normal force that the floor exerts on the chair is calculated by considering the vertical components of the force applied and the weight of the chair. The normal force is equal to the sum of the chair's weight and the vertical component of the applied force, giving a value of approximately 104.4 N.

Step-by-step explanation:

To calculate the normal force N that the floor exerts on the chair, we need to consider the vertical forces acting on the chair.

The weight of the chair (W) acts downward, and the normal force (N) from the floor acts upward.

The vertical component of the applied force (F) will either add to or subtract from the normal force depending on the direction of the force's vertical component.

Here, since the force is applied at an angle of 40 degrees below the horizontal, its vertical component is directed downward, adding to the chair's weight.

This vertical component is Fvertical = F * sin(Θ), where Θ is the angle below the horizontal.

Thus, the total downward force is W + Fvertical.

The normal force is then equal to this total downward force because the chair does not accelerate vertically, remaining in contact with the floor but not moving through it.

Mathematically, this can be represented as N = W + F * sin(Θ).

Substituting the given values:

• W = 80.0 N
• F = 38.0 N
• Θ = 40.0 degrees

We get N = 80.0 N + 38.0 N * sin(40.0 degrees).

Calculating the sine component and then summing up the weight and the vertical force component, we find:

N = 80.0 N + 38.0 N * 0.6428

= 80.0 N + 24.4264 N

= 104.4264 N.

Therefore, the magnitude of the normal force is approximately 104.4 N.