Friction of conveyor belt on suitcase

1. Jun 18, 2008

habibclan

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

A baggage handler drops your 10 kg suitcase onto a conveyer belt running at 2 m/s. The materials are such that us= 0.5 and uk=0.3 [coefficients of static and kinetic friction]. How far is your suitcase dragged before it is riding smoothly on the belt?

2. Relevant equations

fs= usmg
fk= ukmg

3. The attempt at a solution

I don't know how to set up the problem because I don't know how to define the force applied by the baggage handler. Any help would be greatly appreciated!! Thanks in advance! =)

2. Jun 18, 2008

dirk_mec1

Draw a picture and identify the friction force then use an energy balance to find the travelled distance.

$$\frac{1}{2}mv^2 = \mu_{total}mg \Delta x$$

Last edited: Jun 18, 2008
3. Jun 18, 2008

habibclan

I really have no idea. I was able to do all the other questions in this chapter, just not this one. I would say that the friction is in the direction of the motion of the conveyer belt because due to friction, the luggage is able to stick to the conveyer belt. Can you please guide me as to how to set up the problem!

4. Jun 18, 2008

habibclan

How did you get the friction equation with delta x in it? I've never seen that before.

5. Jun 18, 2008

dirk_mec1

Yes you have it's just:

$$E_{kinetic} = F \cdot \Delta s$$

with s the distance

6. Jun 18, 2008

habibclan

Oh okay! This is the work equation. Thanks a lot =). But logically, what happens? When the guy puts the suitcase onto the belt, from this equation, it means that the kinetic energy of the belt is converted into energy for friction?

7. Jun 18, 2008

dirk_mec1

No the guys drops the suitcase that's something else then putting it on the belt.

Imagine yourself a suitcase dropping on a conveyer belt the suitcase 'slides' a distance and then just moves with the belt, get the picture?

8. Jun 18, 2008

habibclan

That I get. But I feel so stupid coz I can't understand how Efric=Kinetic energy. Is it that the kinetic energy of the belt converts itself into friction energy to hang onto the luggage? Sorry for bugging you so much :\$.

9. Jun 18, 2008

tiny-tim

Hi habibclan!

Think of the belt as an infinite source of energy …

… rather like the earth is an "infinite" source of mass … when you bounce a ball off a wall, you tend to assume that the wall is fixed to the earth and the earth doesn't move!

The belt will go on moving at the same speed no matter how kmany cases are put on it … the speed is determined by (a) the engine running it, and (b) the person who wrote the question.

Forget the KE of the belt.

When you do FBDs, you do them for only one body at a time.

(You could do an FBD for the belt, but you'd have to pult the belt motor into it, and you've no idea what that's doing! )

In this case do an FBD, or conservation equation, or work-energy equation, for the luggage only.

You know it starts with zero horizontal velocity, and there's only one horizontal force on it, which is constant until it reaches the velocity of the belt.

You can either do a Newton's second law equation … which has the disadvantage of giving you a differential equation to solve , or, preferably in this case, use the Work-energy equation that dirk_mec1 suggested, which tells you that the KE of the luggage increases because of the work done on it by the friction force.

10. Jun 18, 2008

habibclan

I get it now! THanks so much! So basically this is one of the rare cases where thermal energy is converted into kinetic energy. I was reading my textbook today and it said that usually its KE--> thermal, but I guess this is an exception to that case!

11. Jun 18, 2008

tiny-tim

No no no … the energy comes from the motor which drives the belt.

Some of that energy moves the luggage, and some is converted into thermal energy.

Friction is not thermal … the useful work done by friction is converted into KE … the wasted work becomes thermal energy.

12. Jun 18, 2008

habibclan

Thanks a lot!