Does force normal change when you push an object at an angle?

[Ace.]

Homework Statement

You exert a force of 131 N [30° below horizontal] onto a 12.6 kg lawn mower, What is the force of gravity and normal force on the lawn mower?

Homework Equations

F_g= m × g
F_{A (y component)} = Sin30° × F_A

The Attempt at a Solution

The force of gravity is equal to the force of normal,
∴ F_N = m × g

= 12.6 kg × 9.8 m/s²​

= 123.48 [up]​

But, the normal force gets added on by the force opposing the y component of the applied force right?

So, F_{A (y component)}Sin30° × F_A

= 65.5 N [down]​

So by Newtons 3rd law there would be a force acting up on the lawn mower of 65.5 N[up].I'm just wondering if these two forces are added to each other for the total normal force which is 123.48 N + 65.5 N = 188.98 N [up]?

 

[Delphi51]

Yes, all correct. The vertical component of the push adds to the weight to make the normal force pressing the mower down on the ground.

 

[Ace.]

Thanks, but I have a question. Why doesn't the force of gravity on the lawn mower increase when you apply your force on it? Aren't you essentially making the lawn mower heavier?

 

[Delphi51]

The force of gravity depends on the MASS of the mower. You probably know it is
Fg = GMm/d² where M is the mass of the Earth and m the mass of the mower.