Finding Tensile Force in Pulley Problem

[Ryino]

Homework Statement

See Picture, The problem gives three boxes(M₁, M₂, and M₃) and a few pulleys, no value for the coefficient of friction is given. None of the masses are known, and we are told to assume M1 and M3 are sliding. The objective is to find F_t

(if you can't see that well enough i made this)

Homework Equations

F=ma
F_f=µma

The Attempt at a Solution

First I started of by drawing free body diagrams for all of the boxes, Using these(and assuming M₁ and M₂ are sliding towards the middle) I came up with

M₁:
F_NET=F_t-F_f
M₁a=F_t-µM₁a
F_t=M₁a+µM₁a

M₂ Would then have the same equations except M₁ would be M₂ because it also has
F_NET=F_t-F_f

(I feel like that might be wrong but I'm not sure)
I'm not sure about the signs on all of those either.

Then I have for M₃
F_NET=F_g-F_t

At this point I'm not sure what to do, (the lack of values is killing me) I would think putting F_t from the M₁ equation into the F_t equation would be a good start, but then everything except mass would cancel wouldn't it? So that doesn't seem right. I feel like I'm going in the right direction and I need to use these equations simultaneously some how but I'm not sure what to do for the next step as I don't see how this is going to get me F_t.

Also I'm starting to think M₂ Net force is different
F_NET=F_f-F_t (Since the ropes would be moving in opposite directions) and that would also change the equation for M₂ to

M₂a=µM₂a-F_t
F_t=µM₂a+M₂a

But even if that's right, what would i do with it...?

Any help is appreciated, I'm currently stumped.

 

[Doc Al]

First I started of by drawing free body diagrams for all of the boxes, Using these(and assuming M₁ and M₂ are sliding towards the middle) I came up with

M₁:
F_NET=F_t-F_f
M₁a=F_t-µM₁a
F_t=M₁a+µM₁a

OK, but be careful to distinguish the accelerations of each mass--they aren't the same. Call this acceleration a₁.

M₂ Would then have the same equations except M₁ would be M₂ because it also has
F_NET=F_t-F_f

(I feel like that might be wrong but I'm not sure)
I'm not sure about the signs on all of those either.

OK. Use a₂.

Then I have for M₃
F_NET=F_g-F_t

Consider M3 and the pulley as a single object and apply Newton's 2nd law as before. (Careful with the tension force here.)

Key point: How does the acceleration of M3 relate to a₁ and a₂? Figure that out.

Hint: Set up these equations correctly and you'll have 3 equations and 3 unknowns. (Treat the masses and μ as knowns.)

(You're on the right track.)

 

[Ryino]

Consider M3 and the pulley as a single object and apply Newton's 2nd law as before. (Careful with the tension force here.)

If they are both treated as the same object then it would have two tensile forces on it correct?

Giving you F_NET=M₃a₃-F_t1-F_t2
and a₃=g=9.81m/s^2

Both of the tensile forces will be equal I assume so,
M₃a=M₃g-(2F_t)

Right?

 

[Doc Al]

If they are both treated as the same object then it would have two tensile forces on it correct?

Right.

Giving you F_NET=M₃a₃-F_t1-F_t2
and a₃=g=9.81m/s^2

Well, I wouldn't use a₃ to represent g; a₃ is the acceleration of M₃.

Both of the tensile forces will be equal I assume so,
M₃a=M₃g-(2F_t)

Right?

Right.

Also I'm not really sure how to go about this part of it, wouldn't it all be equal?

Play around with it, or figure it out by considering that the length of the rope doesn't change. Example: If M₁ moves 1 m to the right and M₂ moves 1 m to the left, how far does M₃ move down?

 

[Ryino]

I Solved for a₃ in the net force equation for M₃ and got
a₃=9.81-(F_t1-F_t2)/M₃
then I substituted the values i got for tension for the first and second box earlier and came up with

a₃=9.81m/s^2 - ((M₁a₁+µM₁a₁)/M₃)-((M₂a₂+µM₂a₂)/M₃)

assuming i did my algebra right...

Is that what you meant by relating all three accelerations?Im looking at what i said and it makes me think, If a₃ = 9.81-2F_t/M₃
M₁=M₂ and a₁=a₂.
Although what you said earlier to label the accelerations different makes it seem like they shouldn't be equal.Edit:::

I Think this whole post is going in the wrong direction, read below post

 

[Ryino]

Okay, so... If the Left box moves 1 meter right, and the Right box moves 1 meter left, The middle box moves down 1 meter So, (a₁+a₂)/2=a₃

 

[Doc Al]

So, (a₁+a₂)/2=a₃

Exactly! This is the constraint equation relating the motion of the three masses. Now you should have all you need to solve for the tension.

(It's worth your time to learn how to derive this a bit more rigorously from the fact that the string length is fixed.)

 

[Ryino]

If you solve for a₁ in the first equation and a₂ in the second equation

Plug the results into (a₁+a₂)=a₃ and then substitute that into the the equation for M₃

Using that solve for F_t

And you end up getting

F_t=[2m₁m₂m₃(1+u)] / [m₃(m₁+m₂) + 4m₁m₂]

I think i did that right but my teacher said that i will have M₁ M₂ M₃ g and u. when I solve for F_t. My equation is missing the g.
What could i be doing wrong

 

[Doc Al]

If you solve for a₁ in the first equation and a₂ in the second equation

Plug the results into (a₁+a₂)=a₃ and then substitute that into the the equation for M₃

Using that solve for F_t

OK, except that a₃ = (a₁ + a₂)/2

And you end up getting

F_t=[2m₁m₂m₃(1+u)] / [m₃(m₁+m₂) + 4m₁m₂]

That cannot be correct since the units don't match. The right hand side has units of mass, not force. (And yes, you should have a g in your answer.)

To find out where you went wrong, show the equations that you used.

 

[Ryino]

Okay, for some reason i accidently substituted g for a₁ and a₂ on the right of the first two equations, I am reworking it now with the correct variables there, and i also see one other mistake...

This time through I am using the equations

M₁a₁=Ft-µM₁a₁ (1st equation)
M₂a₂=Ft-µM₂a₂ (2nd equation)
M₃a₃=F_g-F_t or M₃a₃=M₃g-F_t (3rd equation)
a₃ = (a₁ + a₂)/2 (4th equation) that's what I meant last time I just forgot to write the divided by 2 as I was typing it fast sorry

Ill edit this post when i get an answer with these equations

 

[Doc Al]

M₃a₃=F_g-F_t or M₃a₃=M₃g-F_t (3rd equation)

Careful: The tension pulls up on M₃ twice.

 

[Ryino]

Careful: The tension pulls up on M₃ twice.

Right. that's what i have written down, forgot that two when i typed it in. Other than that do all the equations look good? I'm still working on the algebra to solve it for F_t.

 

[Doc Al]

Other than that do all the equations look good?

Yep; looks good.

 

[Ryino]

I feel like I am really bad at algebra right now, so far i have

1/2m₁(1+u)+1/2m₂(1+u)=g/F_t-2/m₃

Everything i do to this makes it feel like I am going backwards, if i multiply by ones on the right side to get a common denominator it takes that straight back to m₃g-2F_t

Any suggestions on the next step? I'm still trying

 

[Doc Al]

Not sure how you ended up with F_t in the denominator.

Why don't you follow the same approach you described in post #8?

 

[Ryino]

i had F_t/2m₁(1+u) + F_t/2m₂(1+u)=g-2F_t/m₃ and i tried dividing by F_t

thats how it got on the bottom

With that i continued to 1/M₁+1/M₂=2g(1+u)/F_t - 4(1+u)/M₃

I can multiply by Ft here to get it back on top on both the fractions on the left and the last fraction

And originally i solved for a1, a2, and substituting that into (a1+a2)/2=a3 and then substituted a3 into m3a3=m3g-2F_t

Thats where i started trying to get F_t

 

[Doc Al]

i had F_t/2m₁(1+u) + F_t/2m₂(1+u)=g-2F_t/m₃ and i tried dividing by F_t

Since you want to solve for F_t, don't divide by it. Instead, start by collecting all terms with F_t on the left side of the equation. Then just factor out the F_t and simplify the result.

 

[Ryino]

I think i got it F_t=2gm₁m₂m₃(1+u)/(m₂m₃-m₁m₃+4m₁m₂(1+u))Is that correct? god i hope so

Edit:: I think i can pull out an M3 from the denominator and make it m3(m2-m1)+4m1m2(1+u)

 

[Doc Al]

You are very close but I think you made an error somewhere. (As a sanity check, you might want to consider the special case where all masses are equal and there's no friction. In that case the tension will be mg/3.)

 

[Ryino]

Im trying to set all the masses equal and set my u to 0, and i can't find out where i messed up.

I substituted like that at the point where i got my last equation and it came to 2gm/5.

...

Can you see anything I'm doing wrong?from this point i would get the Ft on the left and factor/solve

Edit:: Wow I feel stupid it does work out to gm/3... For some reason i thought 4+1+1 was equal to 5 >.<. I'm working it out again

 

[Ryino]

I think i got it


F_t=2gm₁m₂m₃(1+u)/(m₂m₃-m₁m₃+4m₁m₂(1+u))


Is that correct? god i hope so

Edit:: I think i can pull out an M3 from the denominator and make it m3(m2-m1)+4m1m2(1+u)

Okay, i went back through it and, somehow i turned a positive into a negative in the middle of all my algebra for no reason. So now i have the same thing except

F_t=2gm₁m₂m₃(1+u)/(m₂m₃+m₁m₃+4m₁m₂(1+u))

 

[Doc Al]

Okay, i went back through it and, somehow i turned a positive into a negative in the middle of all my algebra for no reason. So now i have the same thing except

F_t=2gm₁m₂m₃(1+u)/(m₂m₃+m₁m₃+4m₁m₂(1+u))

That's almost what I get. How did you get the (1+μ) in the denominator?

Unfortunately, I missed a few errors (typos, I presume) in your earlier equations:

M₁a₁=Ft-µM₁a₁ (1st equation)
M₂a₂=Ft-µM₂a₂ (2nd equation)

The friction force terms should be μMg, not μMa.

(Algebra's a pain in the butt, ain't it? )

 

[kevtimc]

Unfortunately, I missed a few errors (typos, I presume) in your earlier equations:

The friction force terms should be μMg, not μMa.

(Algebra's a pain in the butt, ain't it? )

Darn, I thought I was going to be the first to catch that one

 

[Ryino]

Unfortunately, I missed a few errors (typos, I presume) in your earlier equations:

The friction force terms should be μMg, not μMa.

Unfortunately I used, μMa meaning I probably got it wrong. I turned it in today so i don't have any of the work with me. When I get it back (Wednesday maybe?) I'll go back over and finish out the problem the right way.

Hopefully I'll get partial extra credit for what I had.