2 Blocks Stacked, Pullied, Bottom Accelerating, Newton's 3rd

[myxomatosii]

Unfortunately this is due today, I will go work on another problem and watch these boards.
And its not a last minute thing, I have been working on this since it was assigned...

Homework Statement

The lower block in Figure P8.28 is pulled on by a rope with a tension force of F = 28 N. The coefficient of kinetic friction between the lower block and the surface is 0.34. The coefficient of kinetic friction between the lower block and the upper block is also 0.34. What is the acceleration of the 2.0 kg block?

http://img27.imageshack.us/img27/6212/abaabab.gif

Homework Equations

F=ma
f_k=μ_kn

The Attempt at a Solution

I drew out my free-body diagrams. Here are the formulas I gained from them (if they were right).

m₁ (Top Mass)

F_x1=f_{k2 on 1}-T_{2 on 1}

F_y1=n₁-m₁g

m₂ (Bottom Mass)

F_x2=F-T_{1 on 2}-f_{k1 on 2}-f_k2

F_y2=n₂-mg-n_{1 on 2}

etc etc etc...

Question is, after I have formulas

T_{2 on 1}=μ_k(m₂g+m₁g)

T_{1 on 2}=F-μ_km₁g-μ_k(m₂g+m₁g)

T_{2 on 1} is coming out to be 9.996N

while

T_{1 on 2} is coming out 14.672N

Am I missing something? They are connected so would the tensions not be the same? I am thinking I have a concept mixed up or are they not supposed to be the same?

Anyway I ask because I tried to use formula F_x2=m₂a_x2 and it gave me the wrong acceleration, so I am curious if anyone can see my problem?

 

[Doc Al]

There's only one tension in the rope, of course. You need that assumption to solve for the acceleration and the tension, so I don't know how you managed to solve for the tensions.

I suspect a sign error. Show your Newton's 2nd law equations for both masses. What is the relationship between the accelerations of the two masses?

 

[myxomatosii]

There's only one tension in the rope, of course. You need that assumption to solve for the acceleration and the tension, so I don't know how you managed to solve for the tensions.

You have confirmed what I thought, I knew my tensions were wrong.. I'm going to look over things.

I suspect a sign error. Show your Newton's 2nd law equations for both masses.

Sorry, I am unsure of which part of the equation you are asking for, I don't typically hear parts of my problem described that way.

I thought I included almost all of my work other than the free-bodies above?

What is the relationship between the accelerations of the two masses?

The two masses accelerations of course are bound to each other, if the bottom block moves at 1m/s^2, the top block will as well.

 

[myxomatosii]

AHHHHH!I was setting the F=ma=0

As if my acceleration was zero to solve for the tensions..stupid stupid stupid stupid stupid!

ok, I'm going to lay out my a=F/m and see what it looks like, i have a feeling its going to just confuse me even more since i see that the tensions are going to be in the middle of the acceleration equation and I'm still not going to know them

 

[Doc Al]

AHHHHH!


I was setting the F=ma=0

As if my acceleration was zero to solve for the tensions..

D'oh!

ok, I'm going to lay out my a=F/m and see what it looks like, i have a feeling its going to just confuse me even more since i see that the tensions are going to be in the middle of the acceleration equation and I'm still not going to know them

Write your equations in the form:
∑F = ma

You'll have two unknowns and two equations--which works out fine. (Hint: Combine them so as to eliminate the tension.)

Be careful with signs. Hint: Assume the bottom mass has an acceleration of a to the right.

 

[myxomatosii]

D'oh!



Write your equations in the form:
∑F = ma

You'll have two unknowns and two equations--which works out fine. (Hint: Combine them so as to eliminate the tension.)

Be careful with signs. Hint: Assume the bottom mass has an acceleration of a to the right.

Ok my equations for acceleration are.


a_x1=(f_k21-T₂₁)/m₁

a_x2=(F-T₁₂-f_k12-f_k2)/m₂

Yet I am still unsure of the values of T₁₂ and ₂₁.

I know that the accelerations both have to be the same.. I know that both tensions have to be the same.. could I create some sort of equation with (m₁+m₂)

Sec, I'll try it and see if I come up with anything.. assuming it can be done? Idk. Just thinking..

 

[Doc Al]

Ok my equations for acceleration are.


a_x1=(f_k21-T₂₁)/m₁

a_x2=(F-T₁₂-f_k12-f_k2)/m₂

Please rewrite your equations in the form I suggested. Call the acceleration of the bottom block a. (Don't clutter your equations with uneeded subscripts.)

Yet I am still unsure of the values of T₁₂ and ₂₁.

For one thing, call the single tension by a single symbol: T.

 

[myxomatosii]

a=(f_k21-T)/m₁

a=(F-T-f_k12-f_k2)/m₂

 

[Doc Al]

Write your equations in the form:
∑F = ma

(not a = ∑F/m)

And express those friction forces in terms of μ, m, and g.

 

[myxomatosii]

Write your equations in the form:
∑F = ma

(not a = ∑F/m)

And express those friction forces in terms of μ, m, and g.

F_x1=μ_km₁g-T

F_x2=F-T-μ_k(m₂g+m₁g)-μ_k(m₂g+m₁g)

 

[Doc Al]

F_x1=μ_km₁g-T

F_x2=F-T-μ_k(m₂g+m₁g)-μ_k(m₂g+m₁g)

What happened to the "ma" part?

 

[myxomatosii]

What happened to the "ma" part?

F_x1=μ_km₁g-T=m₁a

F_x2=F-T-μ_k(m₂g+m₁g)-μ_k(m₂g+m₁g)=m₂a

Like that? I just left it as F before since they mean the same thing.

 

[Doc Al]

F_x1=μ_km₁g-T=m₁a

F_x2=F-T-μ_k(m₂g+m₁g)-μ_k(m₂g+m₁g)=m₂a

Two comments:

(1) Sign problem in your first equation. You are using a convention of right = positive, left = negative. Good! But if the acceleration of the bottom block is +a, what's the acceleration of the top block?

(2) One of your friction forces in your second is incorrect. (Too many masses.)

 

[myxomatosii]

Two comments:

(1) Sign problem in your first equation. You are using a convention of right = positive, left = negative. Good! But if the acceleration of the bottom block is +a, what's the acceleration of the top block?

(2) One of your friction forces in your second is incorrect. (Too many masses.)

F_x1=μ_km₁g-T=m₁(-a)

F_x2=F-T-μ_k(m₁g)-μ_k(m₂g+m₁g)=m₂a

I see what you are saying, how is this?

 

[Doc Al]

μ_km₁g-T=m₁(-a)

F-T-μ_k(m₁g)-μ_k(m₂g+m₁g)=m₂a

I see what you are saying, how is this?

Perfect. So how can you combine those equations to eliminate T?

 

[myxomatosii]

Perfect. So how can you combine those equations to eliminate T?

I'll work this out really fast, thanks for your quick responses! I'll edit this post and add my formula once I get it.

edit: Okay so T=am₁+μ_km₁g

I plug this into the larger equation and solve for a?


another edit: I tried what I mentioned above and came up with. Oh wait I think I may have messed up a sign..

a=(F-μ_k(m₂g+m₁g))/(m₂+m₁m₂)

maybe it should be

a=(F-2(μ_km₁g)-μ_k(m₂g+m₁g))/(m₂+m₁m₂)

2.268m/s², that doesn't seem too outrageous.

 

[Doc Al]

edit: Okay so T=am₁+μ_km₁g

I plug this into the larger equation and solve for a?

That will work. Another way to eliminate T is to subtract one equation from the other.

another edit: I tried what I mentioned above and came up with. Oh wait I think I may have messed up a sign..

a=(F-μ_k(m₂g+m₁g))/(m₂+m₁m₂)

maybe it should be

a=(F-2(μ_km₁g)-μ_k(m₂g+m₁g))/(m₂+m₁m₂)

Both are incorrect, but the second one is almost there. Your mistake is in the denominator.

Also, simplify that expression a bit. Combine like terms.

 

[myxomatosii]

I had kept refreshing that other page and didn't see that it had scrolled over to the second page.. give me a minute to get it back out.

edit: new equation, starting over i have

a=((F-μ_km₁g-am₁)/m₂) + μ_kg

edit 2:then solving for "a" i get..

a=((F-μ_km₁g)/(m₂(1+m₁)) - μ_kg

This equation gives 2.835 m/s²

 

[Doc Al]

edit: new equation, starting over i have

a=((F-μ_km₁g-am₁)/m₂) + μ_kg

edit 2:then solving for "a" i get..

a=((F-μ_km₁g)/(m₂(1+m₁)) - μ_kg

Still not quite right. When you get terms like 1 + m or m + m², you know something must be wrong since the units don't match.

Try my idea: Subtract your initial two equations to eliminate T, then solve the resulting equation for a.

 

[myxomatosii]

This was a big mess, deleted and continued below.

 

[Doc Al]

Your final formula is incorrect. You've made several errors along the way, but since you have edit on top of edit, it's difficult to know what to comment on.

Why don't you edit the entire thing so that all your errors are corrected. Then indicate when you are done.

 

[myxomatosii]

-2a = (2(μ_km₁g)-F+μ_km₂g) / (m₁-m₂)

This is wrong?

 

[Doc Al]

-2a = (2(μ_km₁g)-F+μ_km₂g) / (m₁-m₂)

This is wrong?

Yes, it's wrong. The closest you've come is the version that I said was almost there in post 17. That only has a single mistake in the denominator.

Comments on the current version:
Where did the factor of 2 come in on the left side?

How did you get m1 - m2?

 

[myxomatosii]

Yes, it's wrong. The closest you've come is the version that I said was almost there in post 17. That only has a single mistake in the denominator.

Oh, I can't see it because I don't know what I did wrong, that's why I took a different approach with the subtraction method you said, I'm still not quite sure I had the idea right.

Comments on the current version:
Where did the factor of 2 come in on the left side?

How did you get m1 - m2?

It made sense at the time but it does not belong, I was thinking I had 2 "a"s when in fact I was just "undistributing".a = (2(μ_km₁g)-F+μ_k(m₂g+m₁g) / (m₁-m₂)

That one feels solid to me, I know it isn't but I can't seem to find the hole I made..

How did you get m1 - m2?

Well on the left when I subtracted ..

am₁-am₂=[the equation for m1 you said was good]-[the equation for m2 you said was good]

I tried to solve for a and I did this ..

a(m₁-m₂)=[the equation for m1 you said was good]-[the equation for m2 you said was good]

Then I had to get it on the other side.

 

[myxomatosii]

Anyway, thanks for trying to help me, I just didn't get it.

I did try though, I was supposed to be at work 3 hours and 10 minutes ago and I'm only leaving now.. lol.

So yea, thanks and what can I say? Oh well.

 

[Doc Al]

You just need to take it slow and easy. I'd be able to point out your error in two seconds if you just wrote, step by step, what you did and left it alone without editing.

Well on the left when I subtracted ..

am₁-am₂=[the equation for m1 you said was good]-[the equation for m2 you said was good]

But you are really subtracting [-m₁a] - [m₂a], not [m₁a] - [m₂a]. (That's probably your main error.)

If you just do the subtraction carefully, making sure your signs are correct, you'll be fine.

 

[myxomatosii]

You just need to take it slow and easy. I'd be able to point out your error in two seconds if you just wrote, step by step, what you did and left it alone without editing.
But you are really subtracting [-m₁a] - [m₂a], not [m₁a] - [m₂a]. (That's probably your main error.)

If you just do the subtraction carefully, making sure your signs are correct, you'll be fine.

I'm going to look back on this when I have time. I do realize how urgent it is for me to realize and memorize what you just said in order not to make that same error again. Physics is very cumulative.. we all know, plus I want to be able to help others out on the forums.

Anyway, I will keep in mind what you said about showing all of my work next time Doc~

Thanks =)

 

[myxomatosii]

You just need to take it slow and easy. I'd be able to point out your error in two seconds if you just wrote, step by step, what you did and left it alone without editing.
But you are really subtracting [-m₁a] - [m₂a], not [m₁a] - [m₂a]. (That's probably your main error.)

If you just do the subtraction carefully, making sure your signs are correct, you'll be fine.

I believe you pointed that out earlier in the problem that I had messed up the signs with my acceleration. Then it appears I did it again in the same problem, doh!