Masses and combined pulleys

[Karol]

Homework Statement

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

Homework Equations

Mass-acceleration: F=ma

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.

 

[Chestermiller]

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

 

[Karol]

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

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

 

[Chestermiller]

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 $$a_3=2a_1-a_2$$

 

[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.

 

[TSny]

View attachment 106541 The initial length of the rope equals to it's length after the change:
$$x_1+x_2+(x_1-x_2)+x_3=(x_2+\Delta_2)+(x_1+\Delta_1)+(x_1-x_2-\Delta_2+\Delta_1)+(x_3-\Delta_3)$$
$$\rightarrow~2\Delta_1=\Delta_3\rightarrow~2a_1=a_3$$

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.

 

[Chestermiller]

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.

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.

 

[Karol]

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

 

[Chestermiller]

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

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.

 

[Karol]

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.

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

 

[Chestermiller]

You didn't change anything, and it's like you've said in post #4:
View attachment 106604

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.

 

[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

 

[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:
$$a_3>0~\rightarrow~~\frac{m_3}{10m_2}>g$$

 

[Chestermiller]

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

That's not what I get.

 

[TSny]

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

 

[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

 

[Karol]

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

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:
$$a_3>0~~\rightarrow~~m_1>2m_3+\frac{m_1(m_2+m_3)}{m_1m_2+m_1m_3+4m_2m_3}$$

 

[Chestermiller]

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:
$$a_3>0~~\rightarrow~~m_1>2m_3+\frac{m_1(m_2+m_3)}{m_1m_2+m_1m_3+4m_2m_3}$$

Try again. The units don't match.

 

[Karol]

Try again. The units don't match.

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

 

[Chestermiller]

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

 

[Karol]

$$\left\{\begin{array}{l} (1)~a_3=2a_1-a_2 \\ (2)~2T+m_2g-T=m_2a_2 \\ (3)~m_1g-2T=m_1a_1 \\ (4)~ T-m_3g=m_3a_3 \end{array}\right.$$
$$(1)\& (4):~(5)~~T-m_3g=2m_3a_1-m_3a_2$$
$$(2):~~T=m_2a_2-m_2g,~~(2)\& (5):~(6):~~2a_1=\left( \frac{m_2+m_3}{m_3} \right)a_2-\left( \frac{m_2+m_3}{m_3} \right)g$$
$$(2)\& (4)~:(7)~~a_2=\frac{m_3}{m_2}a_3+\left( 1+\frac{m_3}{m_2} \right)g$$
$$(3)\& (4):~(8)~~a_1=\left( 1-\frac{2m_3}{m_1} \right)g-\frac{2m_3}{m_1}a_3$$
$$(6)\& (7)\& (8):~(9)~~a_3=\frac{m_1m_2-m_1m_3-4m_2m_3}{m_1m_2+m_1m_3+4m_2m_3}g$$
$$a_3>0~\rightarrow~~m_3<\frac{m_1m_2}{m_1+4m_2}$$
Thank you Chestermiller

 

[Vibhor]

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 $$a_3=2a_1-a_2$$

Hello ,

Let y1 be the distance of m1 from fixed pulley , y2 be distance of m2 and y3 be distance of m3 .Let h be fixed distance between m2 and middle pulley .

Since length of string is constant ,

y1 + y1 - y2 - h + y3 - y2 - h + y2 = constant

Differentiating twice I get ,

a3 = a2 - 2a1 . This has sign difference from the result you have obtained . I surely must be making a mistake somewhere .

Could you please tell my error .

Thanks

 

[TSny]

a3 = a2 - 2a1 . This has sign difference from the result you have obtained .

In the original figure, $a_3$ is defined as positive in the upward direction while $a_1$ and $a_2$ are defined as positive in the downward direction.

 

[Vibhor]

Oh !

In that case what I have done is correct assuming downward is positive for all the masses ?

 

[TSny]

In that case what I have done is correct assuming downward is positive for all the masses ?

Yes. Looks good.

 

[Vibhor]

Yes. Looks good.

Thanks .

You very well know I invariably have a hard time with moving pulleys

 

[Vibhor]

In these types of problems of moving blocks and pulleys , or any other problems ( strings massless and pulley massless , frictionless ) from laws of motion chapter which is covered before rotational mechanics , we assume that pulleys rotate such that strings do not slip .

Suppose the pulley groove is frictionless such that string slips over it , does that change our result ?

I think the results remain same irrespective of whether string slips over the pulley or if pulley rotates along with string .

Am I correct ?

Thanks

 

[TSny]

I think the results remain same irrespective of whether string slips over the pulley or if pulley rotates along with string .

Am I correct ?

Yes, as long as the pulleys are treated as massless and frictionless when they rotate.