# Calculating time/force applied on two blocks using coefficient of friction

1. Nov 5, 2012

### mesa

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

The coefficient of static friction between two blocks is .60. The coefficient of kinetic friction between the lower block and the floor is .20. If a horizonatal force is applied to the upper block (m=4kg) what is the minimum amount of time the two blocks (m=7kg) can cover 5 meters without the top block slipping?

3. The attempt at a solution

First I calculated the maximum force that could be applied to the top block:
.60*g*4kg = 23.5N
Then I figured the force of friction between the floor and both blocks:
.20*g*7kg = 13.7N

Since the force of friction between the floor and lower block opposes the maximum force applied to the top block I subtracted the two giving 9.8N, using this I divided by the total mass of 7kg giving an acceleration of appx 1.4m/s^2. From here solved for t = √(2d/a) = 2.67 s which is incorrect :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 5, 2012

### Sourabh N

I believe you were calculating the resultant force on the lower block. As such, you should divide the force by the mass of the lower block to get the acceleration.

3. Nov 5, 2012

### mesa

Yup, that works but I am having a hard time seeing it, what am I missing here?

4. Nov 5, 2012

### Sourabh N

A free body diagram.

EDIT: Another way to look at it: For lower block, the static friction from upper block and kinetic friction from ground, are the agents of motion.

Last edited: Nov 5, 2012
5. Nov 5, 2012

### mesa

LOL, okay I am going to draw this sucker out again, I think I see what you are saying :)

6. Nov 5, 2012

### mesa

Shoot I am still not getting this one, what if the two blocks were glued together and we pulled on the top block with the same 23.5N force would the answer still be the same?

7. Nov 5, 2012

### Sourabh N

Yes!

You would be wondering what difference does friction make, then?
If you apply a force larger than 23.5N, the boxes with static friction would start moving separately, since static friction can't keep them together anymore. Those bound by a glue, on the other had, stay together forever.

8. Nov 5, 2012

### mesa

I am sure I am missing something simple here, okay if the problem were exactly the same and the two blocks were glued together and we put a 23.5N force on the set and they had a kinetic coefficient of friction of .20 with a total mass of 7kg then wouldn't that break down like this:

23.5N = 7kg*a so a=3.35m/s^2, however there is still kinetic friction to account for between the floor and the blocks opposing this motion so:

fk= 7kg*g*.20 = 13.73N
F-fk = 9.8N so acceleration would be 1.4m/s^2

I'm screwing something up but am having a hard time seeing it :P
How does the static friction cancel out vs glue? That's basically the difference right?
feeling a little more stoopid than usual today :/

9. Nov 5, 2012

### Sourabh N

Okay I see one of the things going wrong there (infact you had the same thing in your original question, I couldn't catch it). As I said: FREE BODY DIAGRAM!

Look at my FBD for the two scenarios: http://i.imgur.com/yav8u.png

The external force (F ext in the figure) is NOT equal to the static frictional force (F st fr). If that were the case, the system wouldn't move at all!