Finding the horizontal force and normal force on a 2.0 kg block on a 60° incline

[KT]

This problem is probably very simple, but it is giving me a lot of trouble. I'm hoping someone could explain it to me. The problem is, as follows:

A block of mass 2.0 kg is held in equilibrium on an incline of angle theta=60 degrees by the horizontal force F.
a)Determine the value of F, the magnitude of F.
b)Determine the normal force exerted by the incline on the block (ignore friction).

The answers are:
a)33.9 N
b)39.2 N

I don't know how to arrive at these answers. Thank you to everyone to takes the time to read this.

 

[arildno]

Hint:
the tangent vector along the inclined plane is:
$$\vec{t}=\cos\theta\vec{i}+\sin\theta\vec{j}$$
The normal vector is given by:
$$\vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{j}$$
Hence, we may write:
$$\vec{i}=\cos\theta\vec{t}-\sin\theta\vec{n}$$
$$\vec{j}=\cos\theta\vec{n}+\sin\theta\vec{t}$$

 

[KT]

Still having trouble

Hi!
Thanks for the reply. I tried to use the information you gave me and I'm still not understanding. This is sad because I am majoring in physics. I'll explain how I'm trying to solve the problem and maybe someone can tell me where I'm making the mistake.
Ok, I broke it down into the components mgsin(theta) and mgcos(theta). I'm trying to solve for the horizontal force, so after drawing the force diagram I am left with
Fx=F-mgsin(theta)=0
Fy=n-mgcos=0
I know I am making the mistake somewhere around here, but I don't have a clue where, because I don't see what other forces(excluding friction as it's irrelevant here) come into play.

 

[Doc Al]

Fx=F-mgsin(theta)=0
Fy=n-mgcos=0
I know I am making the mistake somewhere around here, but I don't have a clue where, because I don't see what other forces(excluding friction as it's irrelevant here) come into play.

You are treating the normal force as if it has no horizontal component. The normal force is perpendicular to the surface, so it has both vertical and horizontal components.

While it's perfectly OK to use vertical and horizontal components to analyze this problem, I'll bet you'll find things easier by taking components parallel and perpendicular to the surface of the incline. Try it.

 

[arildno]

Note that we have:
$$\vec{W}=-mg\vec{j}=-mg(\cos\theta\vec{n}+\sin\theta\vec{t})$$
$$\vec{F}=F\vec{i}=F(-\sin\theta\vec{n}+\cos\theta\vec{t})$$