Newton's 2nd Law Question (2-Dimensions)

[chroncile]

Homework Statement

A proud homeowner is out mowing their lawn. If the force of friction on the mower is 50.0 N and it is accelerating forward at 0.15 m/s², what is the magnitude of the applied force if the owner is pushing on the handle of the mower at an angle of 70 below the horizontal? How big is the normal force acting on the lawnmower? The mass of the lawnmower is 11 kg.

Homework Equations

Fnet = Fapp + Ff
Fnet = ma
Fg = mg

The Attempt at a Solution

Fnet = Fapp + Ff
0.15 * 11 = Fapp_x + (-50.0)
1.65 + 50.0 = Fapp_x
Fapp_x = 51.65 N

Fapp_y = 51.65 * tan(70) = 142 N

Fapp = SquareRoot (51.65² + 142²)
Fapp = 151 N

0 = Fapp_y + Fn + Fg
0 = 142 + Fn + 11*9.8
Fn = -250 N

Correct answers are:

Fapp = 151 N
Fn = 156 N

How would you get the normal force?

 

[PhanthomJay]

Homework Statement

A proud homeowner is out mowing their lawn. If the force of friction on the mower is 50.0 N and it is accelerating forward at 0.15 m/s², what is the magnitude of the applied force if the owner is pushing on the handle of the mower at an angle of 70 below the horizontal? How big is the normal force acting on the lawnmower? The mass of the lawnmower is 11 kg.

Homework Equations

Fnet = Fapp + Ff
Fnet = ma
Fg = mg

The Attempt at a Solution

Fnet = Fapp + Ff
0.15 * 11 = Fapp_x + (-50.0)
1.65 + 50.0 = Fapp_x
Fapp_x = 51.65 N

Fapp_y = 51.65 * tan(70) = 142 N

Fapp = SquareRoot (51.65² + 142²)
Fapp = 151 N

0 = Fapp_y + Fn + Fg
0 = 142 + Fn + 11*9.8
Fn = -250 N

Correct answers are:

Fapp = 151 N
Fn = 156 N

How would you get the normal force?

Your way. There seems to be an error in the problem solution.