# Friction between two blocks

1. Nov 12, 2009

### EV33

1. The problem statement, all variables and given/known data
A 2kg block sits on a 4kg block that is on a frictioness table. The coefficients of friction between the blocks are U(s)= .3, U(k)=.2

a. What is the max horizontal force that can be applied to the 4 kg block if the 2 g block is not to slip?

2. Relevant equations

Fr = μN

3. The attempt at a solution

N= (2kg)(g)= 19.62 N

Fr = μN= (19.62)(.3)=5.886

I chose to use the coefficient for static friction because kinetic frition is for sliding objects, and we do not want the block of 2 kg to be sliding.

With this information I came to the conclusion that the max force that could be applied to the 4kg block without the 2 kg block slipping would be 5.886N because that is the frictional force between the two objects.

The answer in the back of my book is 18N, and I can't figure out how to arrive at this answer.

2. Nov 12, 2009

### ideasrule

The problem isn't quite so simple because the bottom block isn't stationary; it's going to be accelerating under the force of friction between the first and second blocks. Try drawing a free-body diagram for each of the blocks and writing out Newton's second law.

3. Nov 12, 2009

### EV33

How do you convert the horizontal force on the bottom block to the horizontal force between the blocks? Is there another equation using parallel forces rather than normal forces?

4. Nov 18, 2009

### EV33

The forces I see...

2 kg block
1. Normal force: 2g by the 4kg block
2. gravity force of -2g by the the earth
3. Friction force: -5.886 would be the maximum static friction between the two without the2kg block slipping. This force allows the top block to move with the bottom block
4. Friction force by bottom block to top block... I would assume it has to be equal to 5.886 or greater.

4kg block
1. normal force: 6g by the surface
2. Gravity of both blocks added together: -6g
3. friction: 5.886 would be the max before sliding would start.
4.Normal force by hand...

So for my second attempt at this I would say I need the force on the 4kg block times the static friction coefficient to be equal to 5.866, and then when I solve for the force I get 19.62.

F(.3)=5.886
F=19.62

This is still incorrect, what am I doing wrong here?

5. Nov 19, 2009

### EV33

6. Nov 24, 2009

### EV33

ok I have a new solution...

I decided that block of 2kg can only accelerate at an acceleration that is equal to the friction force...

5.886=2a
a=2.943

and from here I figured the push can only cause an acceleration equal to this...

f/6=2.943
f=17.658

so my my final answer is the bottom block can be pushed with a force of 17.658 N.

The correct answer is supposed to be 18. so I was wondering if I got an answer that is just luckily close or if that is the right answer but my book rounded? If not could someone point me in the right direcion please. Thank you.

7. Nov 25, 2009