enter image description here

enter image description here Using any method find the constraint relation between mass 1 and mass 2.

Assumptions: No friction , massless string and pulley.

Q: Find constraint relation between 1kg and 2kg block.

what I tried so far

So, I used a method in which I consider the length of string. Length of red string is $2x_1$. Now , we can notice that one string is long joining 1kg and other joining the pulley is short. But since the assumptions are there. We can consider it like this. If you need a proof , then let me know. I shall give an example too in edit afterwards.

So , then length of string for 2nd pulley is $2x_p$.

I assume that the Length of string for $2kg $block is $2x_2$. However, I feel this is wrong since the same string is connected to 1kg block as well.

So, I get is $2x_1+2x_2+2x_p $but my answer is coming wrong.

So, it would be helpful if you can help me just correct where my mistake is or share another method if you’d like that.

another simpler example

EDIT: an example enter image description here

Here , $a_1 , a_2 , a_3 $ are accelerations of $1kg , 2kg , 3kg $block respectively.

  • $\begingroup$ A similar question was posted a few days ago - check that. $\endgroup$ – Solar Mike May 3 at 11:39
  • $\begingroup$ I don't think you have posted the question as asked (or it was asked badly). Which constraint relationship are you looking for? The one between the relative heights of the m1 (it is a '1'?) and m2? Why have you labelled all the strings with 'T'? Please edit and post the whole of the original question. In the posted example answer what is 'a'? $\endgroup$ – Transistor May 3 at 12:48
  • $\begingroup$ Constraint relation to be found is between 1kg and 2kg. $\endgroup$ – Srijan M.T May 3 at 12:51
  • $\begingroup$ Is 5a1+a2 the answer? $\endgroup$ – r13 May 3 at 17:36
  • $\begingroup$ No.@r13 it is a2 = -7a1 $\endgroup$ – Srijan M.T May 3 at 17:49

Using the following coordinates:

enter image description here

i.e.: using pulley one (top left as the reference point:

  • $x_1$ the distance of mass 1 from pulley 1
  • $x_2$ the distance of mass 2 from pulley 1
  • $P_2$ the distance of pulley 2 from pulley 1
  • $P_3$ the distance of pulley 3 from pulley 1

then the equations are the following:

  • for rope wrapped around pulley 1 $$L_1 = x_1 + P_2$$
  • for rope wrapped around pulley 2 $$L_2 = (x_1 - P_2) + (P_3-P_2)$$
  • for rope wrapped around pulley 3 $$ L_3 = (x_1 - P_3) + (X_2 - P_3)$$

From those relationships you get:

$$\begin{cases} P_2 = L_1 - x_1\\ P_3 = L_2 - x_1 +2 P_2 \\ L_3 = x_1 + x_2 - 2 P_3 \end{cases} \rightarrow \begin{cases} P_2 = L_1 - x_1\\ P_3 = L_2 - x_1 +2 (L_1 - x_1) \\ L_3 = x_1 + x_2 - 2 P_3 \end{cases} \Rightarrow \begin{cases} P_2 = L_1 - x_1\\ P_3 = L_2 +2 L_1 - 3 x_1 \\ L_3 = x_1 + x_2 - 2 (L_2 +2 L_1 - 3 x_1) \end{cases} $$ The final equation is what you want:

$$L_3 = x_1 + x_2 - 2 (L_2 +2 L_1 - 3 x_1)$$ $$L_3 = x_1 + x_2 - 2 L_2 -4 L_1 + 6 x_1$$ $$L_3 + 2 L_2 + 4 L_1 = 7x_1 + x_2 $$

By differentiating with respect to time twice you get $$0 = 7\ddot{x}_1 + \ddot{x}_2 $$ $$0 = 7a_1 + a_2 $$

Additionally there should be a way to obtain this by the energies. Probably by obtaining the equations of motion through langrange's method.

  • $\begingroup$ Why didn’t you write x1+p2=2x1 $\endgroup$ – Srijan M.T May 4 at 14:50
  • $\begingroup$ Because they are not equal. $\endgroup$ – NMech May 4 at 15:33
  • $\begingroup$ What about the example in my edit. I took extra length as C right. Why didn’t you write the same then ? $\endgroup$ – Srijan M.T May 4 at 15:33
  • $\begingroup$ I did not check the simpler example. $\endgroup$ – NMech May 4 at 15:37
  • $\begingroup$ Well , then check and do let me know about it. 👍🏻 $\endgroup$ – Srijan M.T May 4 at 15:38

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