# Homework Help: Plz help, normal force

1. Nov 5, 2007

### Kingrick

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

A small block with mass m=0.100kg can slide along the frictionless loop-the-loop, with loop radius R=0.15m, an initial velocity along the track of vi= 2.0 m/s at point P, at height h=2.0R above the bottom loop.

ok now, so first part to the question is,

determine the normal force that the track exerts on the block when it is at the point Q. Include a FBD in your answer and cleraly identify the laws of physics that are used to solve the problem.

2. Relevant equations

f=ma
pe=ke

3. The attempt at a solution

ok, for my attempt at solving this I'll tell you what i was thinking. I look at this and think it must be inertia holding the block in, and i started thinking there must be something to do with energy in the question, but when i started with energy I realized that finding Vf would not help me know what the normal force at point Q is.

So after examining the question further I realized I'm not sure how to solve it, because if F=ma, and the block is decelerating (which a=9.8 i believe) then Vf gets me no where. But i know that the wall must exert a force on the block to keep it in the circle.

any help would be appreciated on how to solve normal forces in situations like this.

thank you.

2. Nov 5, 2007

### Feldoh

The normal force is the force of the loop-the-loop pushing on the block would you agree. So lets start here: What direction is the normal force when the block is at the top of the loop, and what other forces are acting on the block and in which direction(s)?

3. Nov 5, 2007

### Kingrick

well, the normal force should be downwards, and gravity would be acting on it. and i think that would be it.

4. Nov 5, 2007

### Astronuc

Staff Emeritus
Where is point Q? P is apparently at the top of the loop, h = 2R = D.

There is a force (centripetal) that always points along the radius of the loop, i.e. it is normal to the circumference of the loop, and it changes orientation with respect to vertical. The force due to gravity is always pointing downward.

For the resultant force, one simply adds the force vectors.