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4 July, 03:05

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.

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  1. 4 July, 05:16
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    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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