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A chair of weight 105 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 43.0 ∘ 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.

Sagot :

sqdancefan

Answer:

  133.6 N

Explanation:

The vertical component of the force pushing on the chair is ...

  Fp = (42.0 N)sin(43°) ≈ 28.6 N

There is no acceleration of the chair in the direction normal to the floor, so the net force on it is zero. The floor counters the downward pushing force, together with the weight of the chair, with a force of ...

  105 N +28.6 N = 133.6 N

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