Question

A chair of weight 95.0N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 39.0N directed at an angle of 37.0 degrees below the horizontal and the chair slides along the floor.
1)Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.

Answer 1

Answer:

Answer:

118.4 N

Explanation:

weight of chair, mg = 95 N

Push, F = 39 N

Ф = 37 ° below x axis

Let n be the normal force.

So, by using the diagram and resolve the components of Force F.

n = mg + F SinФ

n = 95 + 39 Sin 37°

n = 95 + 39 x 0.6

n = 118.4 N

Explanation:

Answer 2

Answer:118.47

Explanation:

Given

force (F)= 39 N acting at an angle of 37 below horizontal

Force can be divided in to sin and cos component

Thus Fsin37 will act in Y direction

Total Force in Y direction is Normal reaction , Fsin37 and weight

N-mg-Fsin37=0

$N=95+39\times \frac{3}{5}$

N=118.47 N