Question

A chair of weight 75.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 42.0 N directed at an angle of 38.0 ∘ 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 normal force that floor exerts on chair is 100.83N

Step-by-step explanation:

Given:

Weight of chair, W = 75N

Applied force, F = 42N

Angle, θ = 38°

Normal force, n = ?

We know,

Vertical component of the force = F sinθ

= 42 X sin38°

= 42 X 0.615

= 25.83N

Total normal force acting on the chair = 75N + 25.83N

= 100.83N

Therefore, the magnitude of normal force that floor exerts on chair is 100.83N