# Kinetic friction & static friction

1. Sep 11, 2009

### MechaMZ

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

http://img182.imageshack.us/img182/8174/p560.gif [Broken]

An aluminum block, m1 = 5.00 kg, and a copper block, m2 = 15.36 kg, are connected by a light string over a frictionless pulley. They sit on a steel surface as shown in the figure below, where θ = 30.0°. (See the table below for the appropriate coefficients of friction).

http://img182.imageshack.us/img182/2475/5table60.jpg [Broken]

3. The attempt at a solution
When they are released from rest, will they start to move?
The answer is NO, but how do I know it is not going to move? Should I taking the friction force against the opposite force individually for each block, or total friction force versus the total force against the frictional forces?

Determine the sum of the magnitudes of the forces of friction acting on the blocks.
The answer is 75.3 N, but how to get this number?
I've tried
fs = NUs and fs = NUk
but the answer is still wrong =(

Last edited by a moderator: May 4, 2017
2. Sep 11, 2009

### Hootenanny

Staff Emeritus
Let's take the first question first:
You need to consider the sum of the forces acting on each block, individually. That is you need to sum the forces acting on m1 and then do the same for m2.

3. Sep 11, 2009

### MechaMZ

for m1,
the sum of horizontal forces
= T-fs
= T - (0.61 x 49.05)
T - 29.9205

for m2,
the sum of horizontal forces
= 150.68sin30 - T - fs
= 150.68sin30 - T - (0.53)(150.68cos30)
= 75.34 - T - 69.16

so how to analysis from the result above?

4. Sep 11, 2009

### Hootenanny

Staff Emeritus
The next step is to apply Newton's second law to both equations, as you would do for a standard pulley question. You will find that you have two equations, with two unknowns (T & a). You need to solve these two equations to find a.

5. Sep 11, 2009

### MechaMZ

I think should get a zero for acceleration since the object is not moving, but my answer for acceleration is -1.16m/s2

my working steps,
F=m1a
T - 29.9205 = ma
T - 29.9205 = 5a
T = 5a + 29.9205 -------(1)

F-=m2a
75.34 - T - 69.16 = 15.36a
a = (6.18 - T) / 15.36 --------(2)

(1) >> (2)
a = (6.18 - 5a - 29.9205)/ 15.36
a = -1.16m/s2

where did I wrong?

6. Sep 11, 2009

### Hootenanny

Staff Emeritus
Your method looks fine. All I can suggest is that you do not round at intermediate steps, i.e. store the results of any intermediate calculations in your calculator and then use them directly for later steps.

7. Sep 11, 2009

### MechaMZ

Okay, for the second question, how do I solve it?
Determine the sum of the magnitudes of the forces of friction acting on the blocks.
The answer is 75.3 N, but how to get this number?

8. Sep 11, 2009

### Hootenanny

Staff Emeritus
I'm not sure. Using either the coefficients of static or kinetic friction yield numbers different from 75.3N. Let me see if I can get another pair of eyes on this.

Where did you get these "correct" answers from?

9. Sep 11, 2009

### MechaMZ

but i need to understand the concept, it is more important

10. Sep 11, 2009

### Hootenanny

Staff Emeritus
I have just run through the problem myself, starting from scratch and obtained an acceleration of -1.17 m/s2 (3sf). I would suggest that the answer sheet is wrong. However, I've requested that some of the other Homework Helpers come and take a look at the question, in case I've missed something.

11. Sep 11, 2009

### Count Iblis

Quick comment (I'll do it myself shortly). In this problem you do not know what the friction forces are as a ratio of the normal forces.

If you have some object on a surface and you pull it with some force, then if the force is not too large, the object will stay at rest. So, the static friction force adjust itself, until some limit is overcome. That limit is given by the static friction coefficient times the normal force.

So, what you cannot do is to a priori take the friction force to be the friction coefficient times the normal force. Instead, what you need to do is perform a stability analysis. Assume that the whole thing is at rest. Then compute the friction force that would be necessary for that. Then compute the ratio with the normal force to see if this ratio exceeds the static friction coefficient.

In this case you have two objects connected to each other, but the method is essentially the same (the ratio needs to be less than the static friction coefficient for one of the two blocks for the set up to be stable).

12. Sep 11, 2009

### Hootenanny

Staff Emeritus
Very good point, Count. I hadn't even thought of that! It is unlikely that the frictional forces for both masses will be at their maximum value. It's a very interesting problem that I've never come across before.

13. Sep 11, 2009

### D H

Staff Emeritus
The answer sheet is correct, MechaMZ. Hootenanny made a rare mistake here.

A quick sanity check here. (Rhetorical question) What does this acceleration mean? You have set things up so positive is to the right. This negative acceleration means that the aluminum block is pulling the copper block uphill solely due to friction acting on the two blocks. That doesn't make a bit of sense, which in turn means you did something wrong.

14. Sep 11, 2009

### MechaMZ

so how i still can't figure out the reason why the acceleration is abnormal, perhaps my normal reaction force was wrong.

15. Sep 11, 2009

### D H

Staff Emeritus
Where you went wrong was assuming that the friction force was equal to the normal force times the static coefficient of friction. That is not how static friction works. If the object is not moving, the net force on the object is zero. This means that the net normal force is zero, and so is the net horizontal (better: parallel) force. This latter comprises the force due to friction plus the sum of the horizontal components of all other forces. The frictional force is simply equal but opposite to the horizontal component of the total of these non-frictional force. The coefficient of friction does not say what the frictional force is. It instead puts an upper limit on how big that force can be.

Think of it this way: Suppose the aluminum block is not connected by a string to the copper block. It is just sitting all by its lonesome on a horizontal surface. The only forces acting on it are vertical. There is no horizontal force acting on it, period. The frictional force is zero in this simple example.

Now connect the string and gradually increase tension. Up to a point, the force due to friction will exactly counterbalance the tension. The block won't move. The point at which the block does start moving is when the tension is equal to the normal force times the coefficient of friction.

16. Sep 12, 2009

### MechaMZ

Hi DH,

could you please show me in a mathematical way to prove when they are released from, they will not moving?

Isn't not moving if m1's tension equals or lesser than fs=(0.61x49.05), and m2's tension + 150.68sin30 equals or lesser than fs = (0.53 x 150.68cos30)?

thank you

17. Sep 12, 2009

### D H

Staff Emeritus
That's correct. The tension on m1 is of course the same as the tension on m2.

What is the source of this tension?