# Masses and combined pulleys

1. Sep 23, 2016

### Karol

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

1) Which mass accelerates faster than g
2) What's m1's acceleration
3) In which condition m3 will move upwards, if they start from rest

2. Relevant equations
Mass-acceleration: F=ma

3. The attempt at a solution
$$\left\{\begin{array}{l} (1)~a_3=2a_2+2a_1 \\ (2)~2T_3+m_2g-T_3=m_2a_2 \\ (3)~m_1g-2T_3=m_1a_1 \\ (4)~ T_3-m_3g=m_3a_3 \end{array}\right.$$
$$\rightarrow~a_2=\frac{m_1m_2+4m_2m_3+3m_1m_3}{m_1m_2+4m_2m_3-2m_1m_3}g$$
The fraction is always greater than 1, so a2 accelerates faster than g.
$$a_1=\frac{m_1m_2-6m_2m_3-2m_1m_3}{m_1m_2+4m_2m_3-2m_1m_3}g$$
3) eq' (1):
$$a_3=2a_2+2a_1~~\rightarrow~~a_3=\left( \frac{2m_1m_2-2m_2m_3+m_1m_3}{m_1m_2+4m_2m_3-2m_1m_3} \right)2g$$
I made eq' (4) as m3 ascending, so if a3 is positive-it ascends. from (5) two conditions are necessary, either both, the nominator and denominator are positive or are they negative. the first scenario:
$$\left\{ \begin{array}{l} (6)~2m_1m_2-2m_2m_3+m_1m_3>0 \\ (7)~m_1m_2+4m_2m_3-2m_1m_3>0 \end{array}\right.$$
From (6) i get (i multiply by 2): $4m_1m_2+2m_1m_3>4m_2m_3$
From (7): $m_1m_2+4m_2m_3>2m_1m_3$
I combine those 2 to get: $5m_2>0$, but this is always true.

2. Sep 23, 2016

### Staff: Mentor

I don't match your kinematic equation (1).

3. Sep 26, 2016

### Karol

The initial length of the rope equals to it's length after the change:
$$x_1+x_2+(x_1-x_2)+x_3=(x_2+\Delta_2)+(x_1+\Delta_1)+(x_1-x_2-\Delta_2+\Delta_1)+(x_3-\Delta_3)$$
$$\rightarrow~2\Delta_1=\Delta_3\rightarrow~2a_1=a_3$$

4. Sep 26, 2016

### Staff: Mentor

Call $y_1$ the distance between the center of the upper pulley and the center of the middle pulley, $y_2$ the distance between the center of the middle pulley and the center of the bottom pulley, and $y_3$ the distance between the center of the middle pulley and mass 3. Then the total length of rope (excluding the parts wrapped on the pulleys, which are constant) is $$L=2y_1+2y_2 +y_3$$ Since the length L is constant, $$2\frac{d^2y_1}{dt^2}+2\frac{d^2y_2}{dt^2}+\frac{d^2y_3}{dt^2}=0$$In terms of these parameters, $$a_1=\frac{d^2y_1}{dt^2}+\frac{d^2y_2}{dt^2}$$
$$a_2=\frac{d^2y_1}{dt^2}$$
$$a_3=-\left(\frac{d^2y_1}{dt^2}+\frac{d^2y_3}{dt^2}\right)$$
From these equations, if follows that $$a_3=2a_1-a_2$$

Last edited: Sep 28, 2016
5. Sep 26, 2016

### mpresic

I worked the problem and I agree with Karol, concerning the constraint equation. I have a hard time believing the the delta l2 cancels out but I used graph paper and tried to draw the system and the system at a later time and I got Karol's constraint equation. Mass 2 still has greater than acceleration g.

6. Sep 26, 2016

### TSny

You've defined $x_3$ as the displacement of $m_3$ from one of the moving pulleys. So, the acceleration determined by $x_3$ will be the acceleration of $m_3$ relative to that pulley rather than relative to the earth frame of reference.

7. Sep 27, 2016

### Staff: Mentor

Show us what you did. TSny already demonstrated an error in Karol's result, so this must be a common feature of yours as well.

If there is something wrong with the simple and straightforward analysis I presented in post #4, please point it out.

8. Sep 28, 2016

### Karol

Why don't you take into account $\ddot y_2$ in:
$$a_3=-\left(\frac{d^2y_1}{dt^2}+\frac{d^2y_3}{dt^2}\right)$$

9. Sep 28, 2016

### Staff: Mentor

In post.#4, I meant to say that y3 is the distance between the middle pulley and m3. I've gone back and changed it.

Last edited: Sep 28, 2016
10. Sep 28, 2016

### Karol

You didn't change anything, and it's like you've said in post #4:

11. Sep 28, 2016

### Staff: Mentor

Yes, I did. I changed the word "lower" to "middle" in referring to y3. y3 is the distance between the center of the middle pulley and m3.

12. Sep 28, 2016

### Karol

Yes, that fixes it, thank you mpresic, TSny and Chestermiller.
But i forgot i didn't finish. m1's acceleration is $a_1=\frac{a_3-a_2}{2}$
For m to accelerate upwards, from the relation: $a_3=2a_1-a_2~~\rightarrow~~2a_1>a_2$
Now i will translate it to masses

Last edited: Sep 28, 2016
13. Sep 28, 2016

### Karol

$$\left\{\begin{array}{l} (1)~a_3=2a_1-a_2 \\ (2)~2T_3+m_2g-T_3=m_2a_2 \\ (3)~m_1g-2T_3=m_1a_1 \\ (4)~ T_3-m_3g=m_3a_3 \end{array}\right.$$
$$\rightarrow~a_1=-\frac{m_3}{5m_2},~~a_3=\frac{m_3}{10m_2}-g$$
For m to move upwards, according to the direction in the drawing, a3 must be positive:
$$a_3>0~\rightarrow~~\frac{m_3}{10m_2}>g$$

14. Sep 28, 2016

### Staff: Mentor

That's not what I get.

15. Sep 28, 2016

### TSny

I agree with the 4 equations in post #14, but not with the results for a1 and a3.

16. Sep 28, 2016

### mpresic

I now agree with the four equations of Chestermiller, but I did have the wrong constraint equation earlier, I had a3 = 2 a1 (without a2). I looked at the length relative to the moving (pully), and not relative to the fixed (top) pulley. I now agree with the newest 4 equations

17. Sep 29, 2016

### Karol

Thank you TSny,:
$$\left\{\begin{array}{l} (1)~a_3=2a_1-a_2 \\ (2)~2T_3+m_2g-T_3=m_2a_2 \\ (3)~m_1g-2T_3=m_1a_1 \\ (4)~ T_3-m_3g=m_3a_3 \end{array}\right.$$
$$(2)\& (4):~(5)~a_2=\frac{m_3}{m_2}a_3+\frac{m_3}{m_2}g+g$$
$$(3)\& (4):~(6)~a_3=\left( \frac{m_1-2m_3}{2m_3} \right)g-\frac{m_1}{2m_3}a_1$$
$$(1)\& (5)\& (6):~(7)~\left( \frac{m_1m_2+2m_3^2}{2m_2m_3} \right)g+\frac{m_3}{m_2}a_3=\left( 2+\frac{m_1}{2m_3} \right)a_1$$
$$(6)\& (7):~(8)~a_1=\frac{(m_2+m_3)g}{m_1m_2+m_1m_3+4m_2m_3}$$
$$(6)\& (8):~a_3=\frac{1}{2m_3}\left( m_1-2m_3-\frac{m_1(m2+m_3)}{m_1m_2+m_1m_3+4m_2m_3} \right)g$$
For m3 to move up it must be positive, in accordance with my drawing:
$$a_3>0~~\rightarrow~~m_1>2m_3+\frac{m_1(m_2+m_3)}{m_1m_2+m_1m_3+4m_2m_3}$$

18. Sep 30, 2016

### Staff: Mentor

Try again. The units don't match.

19. Sep 30, 2016

### Karol

$$\left\{\begin{array}{l} (1)~a_3=2a_1-a_2 \\ (2)~2T_3+m_2g-T_3=m_2a_2 \\ (3)~m_1g-2T_3=m_1a_1 \\ (4)~ T_3-m_3g=m_3a_3 \end{array}\right.$$
$$(2)\& (4):~(5)~a_2=\frac{m_3}{m_2}a_3+\left( 1+\frac{m_3}{m_2} \right)g$$
$$(3)\& (4):~(6)~a_3=\left( \frac{m_1}{2m_3}-1 \right)g-\frac{m_1}{2m_3}a_1$$
$$(1)\& (5)\& (6):~(7)~a_1=\left( \frac{m_1m_2+2m_3^2}{m_2(m_1+4m_3)} \right)g+\left( \frac{2m_3}{m_1+4m_3} \right)a_3$$
$$(6)\& (7):~(8)~a_3=\frac{m_1^2-2m_1m_3-10m_3^2-m_1m_2}{4m_3(m_1+2m_3)}$$
For m3 to move up it must be positive, in accordance with my drawing:
$$a_3>0~\rightarrow~m_1^2>2m_1m_3+m_1m_2+10m_3^2$$

20. Sep 30, 2016

### Staff: Mentor

I get $$m_3<\frac{m_1m_2}{(m_1+4m_2)}$$