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According to this tutorial over here: http://commons.wikimedia.org/wiki/User:Phys1csf4n

It is stated that $a_A + a_B + 2a_C = 0$

But I don't understand the logic used. Is there any way to prove this mathematically, instead of logically?

It holds true for the two cases presented, but how do I know that it holds true for every single case? I can not think of the solution for some reason. I do know that $T_4 = T_3 = T_1 + T_2$ and $T_1 = T_2$, but there is no way to relate the acceleration because the masses are different.

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Consider the position of the axle of the lower pulley. Call its acceleration $a_p$. You need to convince yourself that $a_p=\frac 12(a_A+a_B)$ One way to do that is to use linear algebra. First, if $a_A=a_B$, the whole system of the lower pulley and two masses is accelerating together, so $a_p=A_A=A_B=\frac 12(a_A+a_B)$. Second, if $a_A=-a_B$, the rope is just going around the pulley and the pulley is not accelerating at all, so $a_p=0=\frac 12(a_A+a_B)$ Now for general $a_A,a_B$, resolve them into $a_A+a_B, a_A-a_B$. Then the rope over the top pulley ensures $a_p=-a_C$ and you are there.

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  • $\begingroup$ What about the scenario that happens when both pulleys are moving? I.e. $m_A$ is going up relative to the lower pulley axle, $m_B$ is going down, and $m_C$ is also moving? Then doesn't $|a_C| = |a_{BP}| = |a_{AP}|$ where $AP$ and $BP$ represent the mass relative to the axle of the lower pulley. $\endgroup$ – Jason Jun 12 '14 at 20:54
  • $\begingroup$ Not necessarily. Remember the $a$'s are accelerations, while "going up" is velocity. You can have all three masses accelerating by different amounts as long as you respect the one constraint. You could also write the constraint in velocities, derived by considering the fact that the ropes are constant length. You will get $v_A+v+B+2v_C=0$, then can take a time derivative to get your acceleration equation. $\endgroup$ – Ross Millikan Jun 12 '14 at 21:02
  • $\begingroup$ Did you mean $v_a + v_b + 2v_c = 0$? $\endgroup$ – Jason Jun 12 '14 at 22:21
  • $\begingroup$ @Jason: yes, that is correct. + and _ are neighboring keys $\endgroup$ – Ross Millikan Jun 12 '14 at 22:30
  • $\begingroup$ Okay, thank you. Also, when we take the derivative of said equation we would also get $a_a + a_b + 2a_c = 0$. I can sort of see where everything fits into play now. $\endgroup$ – Jason Jun 12 '14 at 22:33
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This can be explained by thinking about slacks in the rope as the objects move -

Consider, block A is moved by $da$, upwards in time $t$, and block B by $db$. Now, block C is constrained to move according to the distances moved by A and B, as there is a rope connecting the three. Now, because of this, there is a slack $da + db$ above the bottom pulley. Now you need to push your bottom pulley upwards in order to straighten the slack in the bottom pulley's rope.

Important method Now, when you move your pulley upwards, half of the slack goes to the right side, and half to the left, so the pulley moves up (or down) only by the amount of half the slack.

Hence the bottom pulley moves up by half the total slack: $(da + db)/2$. This creates a slack on the upper pulley, and to straighten this out, block C must move down by $(da + db)/2$. Therefore, $(da + db)/2 = -dc$ (Considering the directions.)

Differentiating this twice with respect to time on both sides with appropriate limits, we get, $a_a + a_b + 2a_c = 0$.

Hence Proved!

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  • $\begingroup$ Okay, oops I didn't see the other answer. Nevertheless, hope this helps! :) $\endgroup$ – aishpr Jun 12 '14 at 21:31

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