Question

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F = 40.0 N that is directed at an angle of 37.0∘ below the horizontal, and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton’s laws to calculate the normal force that the floor exerts on the chair.

Answer 1

Answer: Normal force, N = 141.64 Newton

Explanation:

All the forces acting on the system and described in free body diagram  are:

1) gravitational pull in downward direction  

2) Normal force in upward direction

3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:

  i) F Cos 37° along the horizontal plane  in forward direction and

  ii) F Sin 37° along the vertical plane in downward direction

Applying the Newton's second law, net forces in the vertical plane are:

Net force, f = N - (mg + F Sin 37°)

As there is no acceleration in the vertical plane hence, net force f = 0.

So,

N - (mg + F Sin 37°) = 0

Adding (mg + F Sin 37°) both the sides in above equation, we get

N = mg + F Sin 37°

N = 12 $\times$ 9.8 + 40 $\times$ 0.601      because (Sin 37° = 0.601)

N = 117.6 + 24.04

N = 141.64 Newton

Answer 2

A) The clearly labeled free body diagram has been attached.

B) The normal force that the floor exerts on the chair is; N = 141.7 N

We are given;

Mass of chair; m = 12 kg

Applied force; F = 40 N

Angle of force below the horizontal; θ = 37°

A) I have drawn a clearly labeled free body diagram and attached it.

B) To get the normal force(N) that the floor exerts, using Newton's laws, we have;

N = F sinθ + mg

This is because sum of upward forces must be equal to sum of downward forces in equilibrium.

Plugging in the relevant values;

N = 40(sin 37) + (12 × 9.8)

N = 24.1 + 117.6

N = 141.7 N

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