# Direction of normal acceleration problems

1. Sep 18, 2014

### yugeci

1. The problem statement, all variables and given/known data

2. Relevant equations

An = mv^2 / r

3. The attempt at a solution

I get the 2D free body diagram, and how the vertical components are resolved, but how is Nc sin θ = mv^2 / r when they both are in the same direction?

Also another related problem is this diagram here:

The equation for the mass at B is,

mg = Fc + N

How is this so when both Mg and Fc are directed downwards (Fc being the centripetal force mv^2 / ρ). Shouldn't it be mg + Fc = N?

I guess I don't understand the direction of Fc due to the normal acceleration. in either of these problems.

2. Sep 18, 2014

### BvU

"when they both are in the same direction?" is not all that clear to me. Who is in which direction, and who else ?

In your second problem, a conceptual dwelling becomes a bit clearer.
Basically, there is no Fc. For an object to execute a circular motion, some force has to play that role.

In the first problem, the horizontal component of the normal force plays that role. Hence also the equality Nc sin θ = mv^2 / r.

In the second problem, the vertical component of the gravitational force plays that role. Not all of it, so there remains some normal force. (**)

A "much better" (ahem) way to write this would be mg - N = the force to cause a circular motion.

Since we generally like short notation, we often use Fc for "the force to cause a circular motion", so now that name is back again. But remember it is shorthand for "the force to cause a circular motion" (or the force that changes the direction of the velocity vector...)

(**) Well, if v is big enough, all of mg is needed to keep the object on the circular trajectory. Increasing v beyond that, the car will loose contact with the hill, since there are no other contribuant forces that might keep it on there.

 Ah, I see it's not a car but an unspecified "mass". Story doesn't change.

3. Sep 18, 2014

### Staff: Mentor

Both what? I only see one force here: Nc. It has a component in the n direction. The rest is Newton's 2nd law.

The way to understand it is to think of the net force at B: ∑F = mg - N. (mg is downward, N is upward) That net force is the Fc, so Fc = mg - N. Note that Fc acts downward.

4. Sep 18, 2014

### BvU

Because it's so important, I throw in another reply:

In the top problem, there are only two forces working on the car: gravity and normal force. The (vector) sum of those is not zero due to the incline.

We have learned F = ma (but actually it is $\vec F = m \vec a$). Where F is the net sum of all forces.

Furthermore, we have learned that for uniform circular motion an acceleration towards the center is needed with a magnitude mv2/r

So if we want the object (car, block, ..) to execute a uniform circular motion, we must ensure that the net sum of forces is pointing the right way (exactly towards the center) and has the magnitude mv2/r.

If you look upon Fc as something that is a resultant instead of one of the forces present, you almost can't go wrong.

5. Sep 18, 2014

### yugeci

I see. I thought of Fc as another independent force. It makes sense now. Thanks for all of the replies.

6. Sep 18, 2014

### Staff: Mentor

This is an important point. Fc is a resultant (or net) force, not a separate force of its own. Fc should never appear in a free body diagram.