Question

A chair of mass 30.0 kg is at rest on a horizontal floor. The floor is not frictionless. You push on the chair with a force of 80.0 N directed at an angle of 35.0° to the floor. What is the normal force that the floor exerts on the chair?

Answer 1

First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.

To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N

Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.

Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.

Fn +45.89 -294 = 0
Fn +45.89 = 294
Fn = 248.11 N

You'll get a more accurate answer if you don't round Fn2y when solving for it, it would be something along the lines of 45.88611 etc

Answer 2

Answer: The normal force that the floor exerts on the chair is 360 N

Step-by-step explanation:

Given

Mass of chair, m= 30.0 kg

Force applied on chair, F=80.0 N at an angle of
`\Theta =35.0^(\circ)` with horizontal

Consider X-Y co-ordinate system with upward as positive y-axis and towards right as positive x-axis

Y-component of applied force is

`F_y=-80.0\cos (35.0^(\circ)) N=-65.5N`

Let normal force exerted by floor be
`F_n`

Using Newton's second law along y-axis

`F_n+F_y-mg=0`

=>
`F_n=(65.5+30.0* 9.8)N=360N`

Thus the normal force that the floor exerts on the chair is 360 N