# Block on an incline

1. Apr 12, 2009

### davidmigl

1. The problem statement, all variables and given/known data
See the figure. This is part of a larger problem involving electromagnetism. I understand the electromagnetism but am having trouble with the basic mechanics!

I have a block of mass M on a frictionless incline at angle phi. There is a magnetic force acting horizontally on the block. I need to find the component of the normal force that balances this out.

3. The attempt at a solution
The component of the normal force parallel to the incline is mg*cos(Pi/2-phi)=mgsin(phi). The angle between this component is phi. Therefore, the horizontal component should be mgsin(phi)cos(phi).

However, my book says this should be mgtan(phi). I do not know how they got that. Perhaps setting the component of the magnetic force parallel to the incline equal to the component of the normal force parallel to the incline we get Fcos(phi)=mgsin(phi) so F=mgtan(phi). But if I try to find the component of the normal force in the opposite direction of the magnetic force, I get F=mgsin(phi)cos(phi). What gives?

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2. Apr 12, 2009

### LowlyPion

I can't see your image, but you have gravity acting || to the incline and that component you have as m*g*sinφ . Your horizontal force has a component up the incline || and opposite the m*g component, and that makes you balanced condition F - the horizontally directed F then m*g*tanφ as you have figured.

When you are looking at the forces normal to the incline then don't you have two components there, that are in balance not with each other but with the force normal to the incline? You have your m*g*cosφ and you have your F*sinφ.

3. Apr 12, 2009

### davidmigl

Thank you for your reply. I see how you get $$mgtan(\phi)$$. Where was I going wrong when I got $$mgsin(\phi)cos(\phi)$$?

I.e. $$mgcos(\phi)$$ is at an angle $$\pi/2-\phi$$ with the horizontal. So it's component in the horizontal direction should be $$mgcos(\phi)cos(\pi/2-\phi)=mgcos(\phi)sin(\phi)$$. But that can't be right, since $$sin(\phi)cos(\phi)\neq tan(\phi)$$ in general.