1. Jul 4, 2005

### nperk7288

I have two questions I'm stuck on:

Question 1:

1 block rests upon a horizontal surface with an identical second block resting on top of the first block. The coefficient of static friction between the blocks equals the coefficient of static friction between the lower block and the surface. A horizontal force is applied to the top block and when the force reaches 47.0 N the upper block begins to slide. The force is then removed and the blocks are put back in their original positions. What is the magnitude of the horizontal force that should be applied to the lower block so it just begins to slide out from under the upper block?

Question 2:

A damp washcloth is hung over the edge of a table and so part of it (mass = m on) is on the table and part (mass = m off) is hanging over the edge. The coefficient of static friction between table and wash cloth is 0.40. What is the maximum fraction ( m off / (m off + m on)) of the cloth that can hang over the edge without causing the cloth to fall?

Please, any insight would be appreciated...

2. Jul 4, 2005

### Pyrrhus

What have you done? Have you draw your Free Body diagrams?

3. Jul 4, 2005

### nperk7288

Yes, I did Free Body diagrams but I don't know what to do, I don't want a solution, just a start, like what formulas to use first etc.

4. Jul 4, 2005

### nperk7288

I'm really struggling in Physics right now cuz I'm doing an 8-unit full year course in 8 weeks...It's hard to understand everything when we go as quick as we have been.

5. Jul 4, 2005

### Pyrrhus

Well, show me some of the equations you came up with.

6. Jul 4, 2005

### nperk7288

For #1 I had 4 forces acting on the top box, FN pointing in the positive Y direction, FW or the Weight pointing in the negative Y direction, fs (the static friction force) pointing in the negative X direction, and FP = 47.0 N pointing in the positive X direction. On the bottom box I have 6 forces acting. They are ones similar to those mentioned above (a normal force pointing up, a friction forces pointing left, a force from the person pointing right and a force from the box's weight pointing down) plus an additional downward force FB2 from the weight of the top box, and an additional upward force FN pushing against the upper box. What do I do now? and is that correct?

Last edited: Jul 4, 2005
7. Jul 4, 2005

### nperk7288

I really need help quick; I still have a big lab to do for tomorrow as well so I will have to leave to do that in a couple minutes.

8. Jul 4, 2005

### Pyrrhus

Well you got 2 cases, the first case when the 47 N force is applied, you should be able to get the coeffiecient of friction, by using Newton's 1st Law. For the 2nd case, you should calculate the horizontal force on the lower block.

On your first case you don't got the applied force on the lower block, and the friction force on the top of the bottom block is pointing right.

9. Jul 4, 2005

### nperk7288

How does Newton's frist law help to get the coefficient of Friction?

10. Jul 4, 2005

### Pyrrhus

Are you familiar with Newton's Laws? Pick up the textbook and read the chapter,

11. Jul 4, 2005

### OlderDan

The statement of the first problem does not make sense. Nothing is said about applying a force to the lower block in the first part. If the upper block slides relative to the lower block without applying any additional force (it will with equal coefficients), it can also be slid back without any additional force on the lower block.

For number 2, the normal force is constant, so the friction force is constant. The amount of cloth hanging over the edge determines the force opposing the frictional force acting on the part of the cloth on the table. You can treat the cloth like a string with tension in the cloth acting between the two sections of cloth.

12. Jul 4, 2005

### nperk7288

I know the laws but that has to do with an object staying at rest or in motion unless acted on by a net force

13. Jul 4, 2005

### nperk7288

There is no force applied to the lower block in the first part. Only to top block is moving. The problem doesn't want to slide the top block back, the person just resets it manually and then wants to slide the lower one...

---------
-----------
---------> | Box 1 |
---------
-----------
---------
| Box 2 |
---------------------------floor

part two .... boxes are returned to normal....
-----------
| Box 1 |
-----------
------> | Box 2 |
---------------------------floor
Now the person wants to ONLY move the bottom box

14. Jul 4, 2005

### Pyrrhus

Yes, now analyze both cases, with newton's 1st Law, the first case should give you the coefficient of static friction, and with it, you can find the force applied on the 2nd case.

15. Jul 4, 2005

### nperk7288

I am still completely lost... I really don't understand this.... I'm about to break down!

16. Jul 4, 2005

### OlderDan

OK Sorry. Now I get it.

17. Jul 4, 2005

### Pyrrhus

Here, start off by the top block

$$\sum F_{x} = ... = 0$$

18. Jul 4, 2005

### nperk7288

That equals fs + FP

19. Jul 4, 2005

### nperk7288

How does that give me the coefficient though?

20. Jul 4, 2005

### Pyrrhus

Remember the signs.

and also

$$F_{f} = \mu N$$