What is it that makes this problem different?

[Sefrez]

Normally when finding the normal force on an object on a incline, I find the magnitudes of gravity in the perpendicular and parallel directions to the incline and have that multiplied by mass. Here is a diagram of that:
http://www.freeimagehosting.net/38f87

Which means that the magnitude of n is m*g*cos theta.

On another kind of problem were I am asked to find the speed that a mass can move around a circle of an inclined plane, the normal force is different:
http://www.freeimagehosting.net/edfa7

In this case n = m*g / cos theta.

Why is this?

 

[Doc Al]

Normally when finding the normal force on an object on a incline, I find the magnitudes of gravity in the perpendicular and parallel directions to the incline and have that multiplied by mass. Here is a diagram of that:
http://www.freeimagehosting.net/38f87

Which means that the magnitude of n is m*g*cos theta.

What allows you to deduce this is the fact that the acceleration normal to the surface is zero.

On another kind of problem were I am asked to find the speed that a mass can move around a circle of an inclined plane, the normal force is different:
http://www.freeimagehosting.net/edfa7

In this case n = m*g / cos theta.

Why is this?

Here the acceleration normal to the surface is not zero. But since the acceleration is horizontal, you can say that the vertical component of the acceleration is zero.

 

[Sefrez]

Oh, I think I understand what you are saying. So the normal force is greater than m*g*cos theta because the centripetal force also has some force in the normal direction? In other words, it is the sum of these two?

N = m*[(v^2/r)*sin theta + g*cos theta]

 

[Doc Al]

Oh, I think I understand what you are saying. So the normal force is greater than m*g*cos theta because the centripetal force also has some force in the normal direction? In other words, it is the sum of these two?

N = m*[(v^2/r)*sin theta + g*cos theta]

It's better to think that the acceleration has a component in the normal direction and just apply Newton's 2nd law in that direction:
ƩF = ma
N - mg*cosθ = m(v^2/r)*sinθ

Which is entirely equivalent to what you wrote. It's best not to think of 'centripetal force' as a separate force. Here, the 'centripetal force' is provided by a component of the normal force.

Even simpler is to look at the vertical direction:
ƩF = ma = 0
N*cosθ - mg = 0
Thus: N = mg/cosθ

 

[Sefrez]

I see what you mean. The last that you mentioned is how I originally saw it. I guess its simplicity is what cough me off guard. I always over complicate things. For some reason it just seems easier, if that makes any sense.

Thanks.

 

[Redbelly98]

Weird, one of the students I tutor asked me this same question the other day. We were looking at a centripetal acceleration problem, and he proceeded to write N=mg·cosθ because he was used to always seeing it that way.

 

[Sefrez]

Weird, one of the students I tutor asked me this same question the other day. We were looking at a centripetal acceleration problem, and he proceeded to write N=mg·cosθ because he was used to always seeing it that way.

Yep. It's these books (or at least mine). It lacks a lot of intuition and goes straight to the math. I have now gotten to the point that I blearily look at the math examples. I just read the material for the concepts and figure out the math that is intuitive to me.

I imagine it is better this way as you are less likely to become fixated to formulas. That is something I dispise, just grabbing formulas, at which seemed to get the best of me in this example. It suppresses critical thinking/abilities.