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.

The magnitude of normal force that floor exerts on chair is 100.83N

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

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