Can someone help me with this physics problem?

Question

A 135.0 kg box is pushed by a horizontal force F at constant speed up a frictionless ramp which makes an angle of 23.0 deg with the horizontal. What is the magnitude of the Normal force between the ramp and the box?

**I found the magnitude of the applied force=561.6 N. For some reason, I can't get the right answer on what the normal force is. If someone could help me that would be great!

Answer 1

If only to find for the normal force. Fn. then

start with the sum of all forces that act on the block and the ramp, then putting the sum of those forces = to ma , so

P + Fn+ Fg = ma , a =0 coz the block is not accelerated, so

P+Fn+Fg = 0

P = the force that push the block along the x axis

Fn = the normal force which the ramp exerting on the block on the y axis

Fg = gravitational force acting on the block which opposes

to Fn, and since Fn and Fg is the only forces along the y axis

which act on the block

therefore the force P along the x axis will do no work so P = 0

then

0 + Fn + Fg = 0

but since Fg = mg sin 23 and mg sin 23 is on negative y axis (if u can draw the diagram u no what i mean,)

therefore = - mg sin 23 and then,

0+Fn + (-mg sin 23 ) = 0

Fn = mg sin 23

Fn = (135 kg )(9.8 m/s^2) sin 23 degree

Fn = 516.937 newtons (or 516. 94)

Answer 2

Weight of the body will be acting vertically downwards. Resolve this force in the direction of the normal at the point of contact. This component will oppose the Normal force. Next resolve the externally applied horizontal force F in the direction of the normal force which is equal to F sin 23 degree but this component will go against the normal force. It is therefore clear that Normal force will be equal to

FN = 130g cos 23 degree + F sin 23 degree.

Now if you know the value of horizontal Force F you can calculate the required force.