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Pulley System

Hello there, I'm trying to understand how Pulleys work. The question asked me to calculate the acceleration constraint that relates the accelerations for each of the masses $m_1, m_2, m_3$. The two ropes are rope $A$ and $B$. The math I did is as follows:

Derivate twice: $$L_B = y_A + y_3$$ $$a_A = -a_3$$ Derivate twice: $$L_A = 2y_A - y_1 - y_2$$ $$2a_A = a_1 + a_2$$ Substitute: $$2(-a_3) = a_1 + a_2$$

This is the correct answer, I'm simply following the method that my professor told us to follow, however this does not make much intuitive sense for me. I just used the lengths provided by the graph. I understand that the tension of Rope A is twice of Rope B, but that didn't help much at all.

How is the acceleration constraint affected by the pulley system? If someone could point me in the right direction or give me resources I'd appreciate it.

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  • $\begingroup$ Using f=ma will give you the tensions, but you need to know the masses; this problem as stated doesn't give you enough information to make that statement about the tension. hth $\endgroup$ – GdbF137 May 28 at 17:07
  • $\begingroup$ I would argue there's enough information to solve the problem symbolically. I would recommend drawing a free body diagram for each mass. Hint: $m_{1}$ and $m_{2}$ do not appear in your answer. Also, how do you know your answer is correct? If it is correct, then you haven't stated the problem correctly. $\endgroup$ – Cinaed Simson May 29 at 3:56
  • $\begingroup$ I took the positive direction to be down in the $3$ force diagrams: $(1)-T+m_{1}g=-m_{1}a\space(2)-T+m_{2}g=m_{2}a\space (3)-2T+m_{3}g=m_{3}a_{3}$. Note, the tension in rope B was twice the tension in rope A, and on pulley A, when $m_{1}$ is accelerating up, $m_{2}$ is accelerating down (or vica versa) so I arbitrarily set $a_{1}=-a$ and $a_{2}=a$. Solving for $a_{3}$ the $T$ disappears and my result is $a_{3}=\frac{(m_{2}-m_{1})}{m_{3}}a-\frac{(m_{1}+m_{2}-m{3}}{m_{3}}g$. $\endgroup$ – Cinaed Simson May 29 at 7:50
  • $\begingroup$ A proper question will give you some constraint on the masses, which gives you symbolic solutions in a single mass scaling variable. $\endgroup$ – GdbF137 Jun 21 at 21:24

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