# Conservatoin Laws

1. Nov 27, 2004

### Goldenlemur

Here is the following problem: A box of mass m is dropped on a converyor belt that has constant velocity u.

a) Let friction between box and belt be nu. How far does the box slide until the box is moving at with same speed of belt?

b) What force F must be applied to the belt to keep it moving at constant speed after the box falls on it, and for how long?

My problem is setting up the equations of moition.....

a) doesn't this force F need to be known for part a?

or do we just consider some other force F' where

m a = F' - nu m g

or is

m a = (dp/dt) = m (dv/dt) + v (dm/dt)

I am close or not can't get anything to work out....

I keep ending up with more unknowns in whatever setup i choose

What am not seeing?

2. Nov 27, 2004

### Staff: Mentor

Are you sure you are not misreading the problem and the friction isn't meant to be the usual $f = \mu N$? (Where $\mu$ is the coefficient of friction and N is the normal force.)

3. Nov 27, 2004

### Goldenlemur

Yes, frictional force suppse to be $f = \mu N$ I thought the normal is mg cause

$0 = N - mg$ ; since the block remains in contact with belt

4. Nov 27, 2004

### Staff: Mentor

Then you should have no problem doing part a, since the friction is the only horizontal force acting on the box.

5. Nov 27, 2004

### Goldenlemur

Yes the frictional force suppose to be cofg. of friction times normal.

i thought normal was equal to the weight of the box because the box remains on the conveyor belt.

6. Nov 27, 2004

### Goldenlemur

Oh the all the stuff that with linear momentum I as doing eailier was meant for part b?

I'm just finding when the box comes to rest with respect the belt and just double intergrate eq of motion to find distance traveled?

7. Nov 27, 2004

### Staff: Mentor

You can integrate if you like, but it's a simple kinematics problem. The force tells you the acceleration.

8. Nov 28, 2004

### Goldenlemur

The part (a) of this question shows that

distance is x = v' t + .5 (u g) t^2

where v' is initial velocity of box and (u g) the acceleration from solving equations of moition....

yet the distance the box should be indepent of time... wouldn't the distnace the box's slide depend on teh magnitude of the belts velocity???

Also, having tourble with part (b):

F = (dp/dt) = (m + M) (dv/dt) + v (d(m + M)/dt)

since belt velocity is constant....

F = v (d(m + M)/dt) where v is the velocity.....

so it additional ask how long should the force be appllied and what is this impluse of this force....?

am I suppose to slove for t from the eq of distance and interage F which just

impluse => is v d(m +M)

the intial momemntum and moemtum while box is sliding should be conserved

9. Nov 29, 2004

### Staff: Mentor

The initial velocity of the box is zero. That is just one of several kinematic equations that applies to uniformly accelerated motion. It is not the best one to solve this problem. What's another?

Why is that?
Absolutely. You need to find the distance it takes to accelerate the box from a speed of zero to the speed of the belt.

I think you are over complicating things. If you found the answer to a, then you know the distance the box had to be accelerated. Now find how long it took. If the box exerts a frictional force on the belt, what force must be applied to the belt to equalize it?

Impulse is $F \Delta t$.

10. Nov 29, 2004

### e(ho0n3

Let me jump in here.

This is similar to the problem of a box sliding on a slippy floor (except the floor is sliding and not the box). If you look at it from the moving belts perspective, then the box is the one that is actually moving and you can find the acceleration on it (friction) and then use kinematics to determine how far the box moves until it's speed is zero (with respect to the belt).

11. Nov 29, 2004

### Staff: Mentor

Nothing wrong with that. Just to be clear: I take the view of the ground based frame in which the belt moves with speed v, and the box begins with a horizontal speed of zero until friction accelerates it to match the speed of the belt. Either way should work.