# Two Blocks Sliding, Newtons 3rd

1. Mar 18, 2009

### myxomatosii

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

The coefficient of static friction is 0.55 between the two blocks in Figure P8.32. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force FT causes both blocks to cross a distance of 5.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

http://img27.imageshack.us/img27/5121/p832.gif [Broken]

2. Relevant equations

I think a few of these may be unnecessary, if I knew exactly how the problem worked I'd probably know why.

a=accel
v=velocity
Δx=displacement
Δt=change in time

fkkn and fkkm7g and fssm4g
F=ma
vf=vi+aΔt
Δxf=Δxi+viΔt+.5a(Δt)2
vf2=vi2+2aΔx

3. The attempt at a solution

(Verbal Explanation)
Realized that the greatest force would have to be equal to static friction created by 4kg block.
Set greatest force static friction could produce to the tension force.
Used tension force to find acceleration produced across floor with a 7kg block.
Calculated final velocity using acceleration and displacement.
Calculated time using final velocity and acceleration.

m4=4kg mass
m7=7kg mass
Using the top box's maximum static friction I found the force...

fssm4g=FT

..to be 21.56N, I then plugged that force into my acceleration equation...

ax=((F/m7)-μkg)

..which gave me 1.12m/s2 as the acceleration across the 5m displacement

I then used that acceleration in the kinematic equation...

vf2=vi2+2aΔx

..giving me a vf of 3.35m/s

..finally I used the equation

vf=vi+aΔt

to solve for time, giving me 2.99s, which turned out to be wrong.

Is my methodology incorrect or just my math somewhere along the way?

Last edited by a moderator: May 4, 2017
2. Mar 19, 2009

### alphysicist

Hi myxomatosii,

I don't believe that is correct. If you think about the force diagram for the 4kg block by itself, the only horizontal forces it has are the static frictional force and the tension force; if those are equal then its horizontal acceleration would be zero. But here you want it to accelerate.

I think here you'll need to use two force diagrams to determine the acceleration; you can make three different force diagrams (3kg by itself, 4kg by itself, and the total 7kg mass) and so you can choose the simpler two of the three.

Last edited by a moderator: May 4, 2017
3. Mar 19, 2009

### myxomatosii

Well yes I do want acceleration but I want the whole system to accelerate, and the 4kg block needs to remain immobile.

I know I am approaching this problem wrong, I really do not understand Newton's Third Law as far as interacting bodies are concerned..

I would appreciate any further explanation you can provide, I am going to be checking other areas of this site until then..

I seem to be posting repeated questions to figure out the same concept in multiple problems, I would really like (and NEED to) to understand interacting bodies.

4. Mar 19, 2009

### alphysicist

The idea is that you can draw three different foce diagrams here, and they will all have the same acceleration. So, for example, if you just focus on the 4kg block by itself, you can get an equation that has two unknowns (tension and acceleration). You can then do the same for the entire 7kg mass, getting another equation with the same two unknowns.

Again, the important point here is that the acceleration of the 4kg mass is the same acceleration as that of the entire object (and the same as the 7kg mass by itself); if it wasn't they would not be moving together and the 4kg object would slide off. Once you have two equations with two unknowns here, you can solve for both of them.

5. Mar 19, 2009

### myxomatosii

I will try this and post what I get.. if anything. I've been scouring the web for videos with a good explanation of Newton's Third Law.

My book does me no good, I've read the chapter three times thoroughly and it only makes general statements about somewhat related problems.. I really don't know what to do about learning this material.

I'm afraid if I just do not understand then there is not much I can do, my average in the class is very good. It baffles me that I am missing something which is apparently so obvious.

6. Mar 19, 2009

### sArGe99

If you've drawn the free body diagrams and marked all forces correctly, then the problem is half-done. Determining the direction in which the frictional force between blocks acts on either blocks might be slightly tricky.
Note : Frictional force between blocks acts on both of them.

7. Mar 19, 2009

### alphysicist

Newton's third law for these types of problems is essentially that if I push an object with some force, it pushes back with that same force. This is especially helpful in drawing force diagrams for the types of problems that are in your post.

So for example, you found that the normal force on the 4kg block is 39.2 N upwards. So when you draw the force diagram for the top block, you include a 39.2 N upward force.

Newton's third law then shows that when you go to draw the force diagram for the bottom 3kg block by itself, then the 4kg block is pushing downwards on the bottom block with a force of 39.2 N. So the force diagram for the 3kg block would include a 39.2 N downwards force.

Also, you found the static frictional force on the 4kg block was 21.56N, and to the left. Then the static frictional force on the 3kg block would be 21.56N to the right. (Because the static frictional force on each block comes from the other.) And this would allow you to draw the separate force diagrams with the correct force directions.

8. Mar 19, 2009

### myxomatosii

block A sits on top of block B

Is the normal force pushing down on block B from A the reason why one can simply use 2 diagrams and not worry about not taking into account the weight of the entire system? Since that weight is taken care of by those "normal forces". (Assuming one's free-body diagrams were drawn correctly)

9. Mar 19, 2009

### sArGe99

No the normal force would obviously counteract the weight, right?

10. Mar 19, 2009

### alphysicist

No, you would still include the weight, because that is the force from the earth on the object. It's just that you only include forces for the object that you are considering. So if you are drawing the 4kg object force diagram, you only include forces that act on the 4kg object, and you would include the weight of the 4kg object. Since the weight of the 3kg object is the force from the earth on the 3kg, it would not be included.

The reason we only need two diagrams here is just that the third diagram is redundant. If you add or subtract two of the equations from the force diagrams from each other you should get the third, so there is no new information in the third equation.

11. Mar 19, 2009

### sArGe99

Wel, all this has made me a bit confused.
I did try this one out, is the answer something like 1.93 s?
I needed only two equations to solve this one, marking the forces correctly is the most important part of it. I found out the acceleration first and then used a kinematic relationship to find the time taken.

12. Mar 19, 2009

### myxomatosii

Sorry I was unclear when I said this I believe. I mean you DO take into account the weights (of course) you just take them separately, anyway. I'm about to try it out... I'll see what I get with my new approach.

(I only have one submission left)

13. Mar 19, 2009

### myxomatosii

Thank you for your effort! :!!)

14. Mar 19, 2009

### myxomatosii

Thats what I meant to say.

Lol, you say potato, I say po.. LJSLDFJLDSJF:LSF:SDFJSF.

(I ramble in unknown tongues but I swear I know what I'm doing most of the time!)

Ok, going to approach the problem from this direction. Its 5am and I need to go to bed soon, going to set my alarm for six hours so I can get up early and work on this some more.

15. Mar 19, 2009

### myxomatosii

Ok here is my force diagram, my two free-body diagrams.

top mass = m2
bottom mass = m1

w=weight(mg)
f=friction
F=Tension Force

Free Body Diagram: m2 (top mass)

Up: n2 , n1 on 2

Down: w2

Left: fs1 on 2

Right: FTension

Free Body Diagram: m1 (bottom mass)

Up: n1

Down: n2 on 1 , w1

Left: fk1

Right: fs2 on 1

Could someone perhaps look at these and see if I've got the idea right?

I'm doubtful, so if you see a force I missed please point it out..

16. Mar 19, 2009

### alphysicist

Oops, I involuntarily went away from lack of sleep.

Normal forces are forces of contact. The top block is in contact with only one other surface (that of m1), so it has only one normal force on it. (The bottom block is in contact with two surfaces, the top block and the table, so it has two normal force, like you correctly have below.)

I think the rest looks good. You already have the diagram for the total object (m1+m2), right? You can then choose which two force diagrams look simpler and find two equations from them.

(By the way, I'm being sloppy in my language. When I say get "the equation" from the force diagram, I mean each diagram gets two equations--one vertical and on horizontal--and you can then eliminate a normal force that occurs in both equations and so end up with one equation for each diagram. But I think you were already doing that.)

17. Mar 19, 2009

### myxomatosii

I think that this has put me in the right direction, going to try to solve the system now.. its my last attempt :yuck:

18. Mar 20, 2009

### myxomatosii

Got it! 1.956 was right.

Learned a few things solving this problem, thanks all~

19. Mar 20, 2009

### alphysicist

I'm glad it worked out! I know how aggravating it is to be down to your very last attempt on a problem.

20. Mar 20, 2009

### myxomatosii

Yea.. lol, I am trying to apply all the things I learned on this problem to another at the moment, I can't seem to figure out what I'm doing wrong. I won't give up until time runs out tho :grumpy: