Archived Atwood Machine, I determining position of the masses.

[chickenman444]

Homework Statement

I know how to do this type of problem, I'm just confused as to how to determine the distance the (4m) mass goes up .

Homework Equations

N/A

The Attempt at a Solution

So I was thinking that if mass (5m) and mass (3m) go in opposite directions then mass (4m) will remain where it is. If mass (5m) and mass (3m) both go down the same distance (x=y), then mass (4m) will go up x or y. But what if x≠y? I was tried to have both mass (5m) and mass (3m) go down so that mass (4m) can go up. I figured if mass (5m) goes down by -2x, then mass (3m) goes up by 2x = y which would still have mass (4m) where it is. So then I thought about moving mass (5m) down another -4x so in total -6x but also have mass (3m) go down -4x = -y so then mass (4m) goes up 4x. I don't know if that makes sense, but this whole thing is confusing me.

 

[haruspex]

If the 5m goes down h relative to the pulley, then the 3m goes up h relative to it.
If the pulley goes down by d then x=d+h, y=d-h, and 4m goes up by d.

 

[mastermechanic]

I solved this question like;

Let's say 5m mass has traveled "X" distance(downward).Firstly, we need to calculate the acceleration of 5m and 3m masses.

$$ 5mg- 3mg = 8m . a => 2mg= 8m . a => a = \frac {g}{4} $$

and for the rope tension passes over 3m mass $$ F_{T} - 3mg = 3m.\frac {g}{4} = > F_{T}=\frac {15mg}{4}$$

and for the rope tension passes over the right pulley $$ 2. F_{T} = \frac {15mg}{2}$$ which exerts to 4m mass (upward)

Now if we consider acceleration for the 4m mass; $$ \frac {15mg}{2} - 4mg = 4m . a => a = \frac {7g}{8}$$

Okay now we know the acceleration of 4m mass and if we know how much time did the 5m spend for "x" distance. We can know the distance 4m traveled in this time interval.

For the time; $$ X= \frac {1}{2}at^2$$ => $$a_{5m} = \frac {g}{4}(m/s^2)$$ $$t = \sqrt{\frac {8X}{g}}$$

In this time interval 4m mass should have traveled ;

$$ X_{4m} = \frac {1}{2}at^2$$ => $$X=\frac {1}{2}\frac {7g}{8}.[\sqrt{\frac{8X}{g}}]^2$$ and we get $$ X_{4m} = \frac {7X}{2} $$

$$\frac{7X}{2}$$ is the distance that 4m has traveled in terms of x.

I don't know if I'm correct.

 

[haruspex]

Firstly, we need to calculate the acceleration of 5m and 3m masses.

You cannot do that in isolation. The 4m mass is relevant. E.g., if we set that to zero then the 5 and the 3 will both have acceleration g.

 

[mastermechanic]

You cannot do that in isolation. The 4m mass is relevant. E.g., if we set that to zero then the 5 and the 3 will both have acceleration g.

Then, can we do $$ 8mg-4mg = 12m. a $$ $$ a = \frac{g}{3}$$ which is whole system's acceleration and also as I've found $$ a_{5m} = \frac {g}{4}$$.
So sum of accelerations $$a_{5m}= \frac {7g}{12}$$ and we do this $$ X= \frac {1}{2}at^2$$ again. Is it possible?

 

[haruspex]

whole system's acceleration

Different parts have different accelerations, so you need to define what you mean by that.

$a_{5m} = \frac {g}{4}$​

Why? And is that the acceleration in the lab frame or relative to the accelerating pulley?

 

[mastermechanic]

Different parts have different accelerations, so you need to define what you mean by that.

Why? And is that the acceleration in the lab frame or relative to the accelerating pulley?

First, I've counted the 5m and 3m as a one mass. So 4m mass should go upward because of the force inequality.This movement will give the 5m&3m (which I counted as one mass) $$\frac {g}{3}$$ and both 5m and 3m mass has $$\frac {g}{3}$$ and now if we consider the 5m&3m as separate they also have addional acceleration $$\frac {g}{4}$$ which you've asked. And all I've done is adding up these two accelaration.

 

[haruspex]

I've counted the 5m and 3m as a one mass. So 4m mass should go upward because of the force inequality.

if we consider the 5m&3m as separate they also have addional acceleration

You are still making the same mistake. You cannot partition the problem like that.
Consider e.g. making the 3m zero. According to your method, the 4m should accelerate upwards at (5-4)/(5+4)g = g/9; but clearly it would accelerate downwards at g.

There is one sure way to approach rigid body problems: consider the forces and acceleration of each body separately. Relate them by forces that must correspond (action = -reaction) and by accelerations that must interrelate to preserve certain distances (lengths of strings, in this case).

Let the tension in the lower string be T.

• What is the tension in the other string?
• What is the relationship between the three accelerations, just based on strings staying the same length?
• Write out the ΣF=ma equation for each mass.