Problem with two pulleys and three masses

[Erdi]

We are freeing the pulley from the other bodies. Not from the forces. What are the internal forces inside the rope that we expose when we "free" the pulley.

Im sorry, I am lost. I really don't have a clue.

 

[erobz]

Im sorry, I am lost. I really don't have a clue.

Which forces are associated with the ropes?

 

[Erdi]

Originally, the pulley B and m1

 

[erobz]

Originally, the pulley B and m1

Lets back up. What are the forces acting on $m_2$?

 

[Erdi]

That should be just the gravity and tension 1, tension 1= M*g

 

[erobz]

That should be just the gravity and tension 1, tension 1= M*g

$T_1 \neq Mg$ That would be the case if $M$ or $m_2$ were NOT accelerating, but they are.

What is the force balance on $m_2$

 

[Erdi]

its m2g

 

[erobz]

its m2g

Maybe I'm using confusing wording. That is just the force of weight. What is the sum of the forces acting on $m_2$. I think you have already written this in an earlier post.

 

[Erdi]

If that's not T - mg = ma or just T - mg , then i really don't understand at all

 

[erobz]

If that's not T - mg = ma or just T - mg , then i really don't understand at all

Ok good:

$$T_1 - mg = ma$$

Now for the other mass $M$?

 

[Erdi]

Thats also got to T1 - Mg = ma

 

[erobz]

Thats also got to T1 - Mg = ma

close. be careful about the coordinate system you chose. If $m2$ goes up is positive $a$ then what does that mean for the acceleration of $M$ here. Also, you used two different symbols for $M$

 

[Erdi]

T1 + mg = ma?

 

[erobz]

T1 + mg = ma?

You chose a coordinate system where up was positive in #46 whether you did so intentionally or not. That is upward accelerations are positive. Upward forces are positive, downward forces are negative, downward accelerations are negative. Try again. And use a capital $M$ leave the little $m$

 

[Erdi]

Well i think the formula is correct, so what you mean is:
T - Mg = M*(-a)

 

[erobz]

Well i think the formula is correct, so what you mean is:
T - mg = m*(-a)

Good. Please use capital $M$ though so you don't get confused about which hanging mass.

 

[Erdi]

Good. Please use capital $M$ though so you don't get confused about which hanging mass.

I know i edited it, guess it didnt go throug..

 

[erobz]

I know i edited it, guess it didnt go throug..

Also its $T_1$ you missed the subscript.

 

[erobz]

Now, of the forces on the LHS. Which one is supplied by the rope?

 

[Erdi]

Back to stage 1 again. That is m2g isn't it??

 

[erobz]

Back to stage 1 again. That is m2g isn't it??

The $m_2$ 's weight does not make $m_2$ go up. The other force i.e. $T_1$ (the tension) does. The tension is supplied by the rope.

I'm going to be honest here. It seems like you have some road ahead of you before you solve this problem. I got to get to bed, its midnight here. Don't lose hope, you'll figure it out. Someone will pick back up with you at some point. Goodnight.

In the meantime, it would be helpful if you learned to use Latex to format your equations. There is a guide on how to do this in the lower left corner of the new relpy box.

 

[Erdi]

Okay thanks for your help tho!

 

[Lnewqban]

I know that the tension from pulley B (T1) has to be equal the m*g of m1 for m1 to have acceleration = 0. But i can't figure how this works because the m2 is already heavier. And so the block(M) has to be negative weight?

Welcome, @Erdi !

I believe that that statement shows the misconception that is the root of your confusion.
Any mass is weightless in free fall (reason for which your guts feel funny during a roller coaster ride).

Please, see:
https://en.m.wikipedia.org/wiki/Weightlessness

The more you slowdown that free fall, the heavier the mass becomes.
If you are able to stop that mass completely (like Earth's surface and a bathroom scale do to your body), that mass "adquires" its natural or normally measurable weight when in repose.

If the same mass is suddely moved upwards (like your body in an elevator), its measurable weight increases, as weight is a force and an accelerated mass resists that acceleration showing a reaction force.

 

[Lnewqban]

Im sorry, I am lost. I really don't have a clue.

Just imagine that you are removing M and that you are grabing the rope while standing on the ground.
You keep a solid grip on that rope and m2, which is four times heavier that m1, starts moving down until m1 hits the top fixed pulley.
After that moment, your hand feels a force of m2g value pulling up.

If you slowly release your grip, the rope starts sliding up and m2 falls even lower.
Your hand starts feeling less force simultaneously as m2 is loosing weight.

If you fully release the rope, m2 becomes completely weightless and unable to exert any force on the movable pulley.
As a result, m1 is free to fall down, loosing weight at the same time.

There must be a balance point for which you keep some gripping force on the sliding rope in such a way that m2 continues falling, but m1 remains static.
At that point, your hand should feel a pulling upwards force, which value is the key to the solution of this problem.

 

[kuruman]

You might profit if you research "Atwood machine" on the web. Here is a compact treatment but there is plenty more including videos. Please read carefully and try to understand how one proceeds to solve such problems. The method is straightforward:

1. Choose a system, in this case one of the masses, say $M$.
2. Find the net force (sum of all the forces acting on it) $F_{\text{net,M}}.$
3. Set this sum equal to the mass of the system times its acceleration. This gives you one equation, $F_{\text{net,M}}=Ma$
4. Repeat for the other mass, $m_2$. Note that if $M$ accelerates up, $m_2$ must accelerate down. You should get a second equation $F_{\text{net},\text{m}_{2}}=-m_2a.$
5. Combine the two equations to find the acceleration.

See how this plan is executed in the link I provided. The link does not provide the tension, but you can find it easily with a little algebra from either $F_{\text{net,M}}=Ma$ or $F_{\text{net},\text{m}_{2}}=-m_2a$ if you have an expression for the acceleration $a$.

 

[erobz]

In addition to what the others have stated:

Might as well not waste what you have figured out so far. You are at a good point to examine the effects of $M$ on $T_1$ with a plot.

You have correctly identified the following set of equations necessary to solve for the tension in the lower rope $T_1$ as a function of $M$.

$$\begin{align} T_1 - m_2 g &= m_2 a \tag{1} \\ T_1 - Mg &= -Ma \tag{2} \end{align} $$

Solve this set of equations for $T_1$ and make a plot vs $M$. Let $M$ range from $0 \rm{kg}$ to $m_2$, study it, then let $M$ go to something arbitrarily large. Try to understand what it's telling you about the limits of the tension $T_1$.

Solving the system is a step toward completing the solution, so it's not an out of the way excursion. It should help to visually see this part of the solution.

 

[Erdi]

Thanks guys! I think i found an expression for T2 and i came to an answer for M.

 

[kuruman]

And what is your answer?

 

[erobz]

In which direction was the mass $M$ traveling?

 

[Orodruin]

You might profit if you research "Atwood machine" on the web. Here is a compact treatment but there is plenty more including videos. Please read carefully and try to understand how one proceeds to solve such problems. The method is straightforward:

1. Choose a system, in this case one of the masses, say $M$.
2. Find the net force (sum of all the forces acting on it) $F_{\text{net,M}}.$
3. Set this sum equal to the mass of the system times its acceleration. This gives you one equation, $F_{\text{net,M}}=Ma$
4. Repeat for the other mass, $m_2$. Note that if $M$ accelerates up, $m_2$ must accelerate down. You should get a second equation $F_{\text{net},\text{m}_{2}}=-m_2a.$
5. Combine the two equations to find the acceleration.

See how this plan is executed in the link I provided. The link does not provide the tension, but you can find it easily with a little algebra from either $F_{\text{net,M}}=Ma$ or $F_{\text{net},\text{m}_{2}}=-m_2a$ if you have an expression for the acceleration $a$.

It may seem, with the angst it can bring,
That an Atwood's machine's a harsh thing.
But you just need to say
That F is ma,
And use conservation of string!

https://www.physics.harvard.edu/undergrad/limericks

 

[Erdi]

I can quickly go through my math here, firstly i got:
m1: m1g -T1 = m1a
m2: m2g-T2 = m2a These are the downward "forces" (m*g) minus the upwardforces tension(T)
M: Mg - T2 = Ma
Pulley B: 2*T2 - T1, from here i can see out that T2 equals T1/2

Then i got to the acceleration relative to each other
a(m1) - a(pulleyB) = 0, a(m1) = -a(B)
a(M)-a(B)+a(m2)-a(B) = 0 => a(M)+a(m2) = 2*a(B) = -2*a(m1)

So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))

So there was a lot of math between right there to my final answer, but i ended up with:
a(m1) = (g(m1*M+m2*m1-4*M*m2)) / (M*m1+m2*m1+4*M*m2) = 0
Solved for M:
M = (m2*m1)/(4*m2-g*m1) = 0,32kg

 

[Erdi]

In which direction was the mass $M$ traveling?

So M will be traveling upwards, if I am correct.

 

[erobz]

I can quickly go through my math here, firstly i got:
m1: m1g -T1 = m1a
m2: m2g-T2 = m2a These are the downward "forces" (m*g) minus the upwardforces tension(T)
M: Mg - T2 = Ma
Pulley B: 2*T2 - T1, from here i can see out that T2 equals T1/2

Then i got to the acceleration relative to each other
a(m1) - a(pulleyB) = 0, a(m1) = -a(B)
a(M)-a(B)+a(m2)-a(B) = 0 => a(M)+a(m2) = 2*a(B) = -2*a(m1)

So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))

So there was a lot of math between right there to my final answer, but i ended up with:
a(m1) = (g(m1*M+m2*m1-4*M*m2)) / (M*m1+m2*m1+4*M*m2) = 0
Solved for M:
M = (m2*m1)/(4*m2-g*m1) = 0,32kg

$g$ shouldn't be in the final result. You also changed your variable names, and your coordinate direction from what we were working towards earlier... not making it easy to follow along.

It seems like your EoM for mass $M$ is not consistent with the coordinate direction you chose to describe the EoM of mass $m_2$?

Any chance you could try using Latex to format the math. It doesn't take much effort to learn it. It is so much easier to find\point out any errors.

LaTeX Guide

 

[erobz]

Hmm, well here's almost the rest of the math. Why shouldn't g be there?

Do the units make sense? You are subtracting a force from a mass in the denominator. Thats generally an indication something has gone haywire.

 

[kuruman]

M = (m2*m1)/(4*m2-g*m1) = 0,32kg

I got the same expression without the g for the reason already explained. The error is between the two steps below

So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))

The right-hand side of the bottom equation has dimensions of force but the left hand side has dimensions of acceleration. check your substitution from the top to the bottom equation.