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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 $\endgroup$ May 28, 2019 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. $\endgroup$ May 29, 2019 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 $a_{3}=\frac{(m_{2}-m_{1})}{m_{3}}a-\frac{(m_{1}+m_{2}-m{3}}{m_{3}}g$. $\endgroup$ May 29, 2019 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. $\endgroup$ Jun 21, 2019 at 21:24
  • $\begingroup$ The problem evidently assumes massless pulleys and no slip of the ropes rotating the pulleys. $\endgroup$
    – John Darby
    Jul 23, 2022 at 19:39

1 Answer 1

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The pulley system constrains the positions of the masses to be related because the masses are connected to ropes which can not change length. For example, $m_1$ and $m_2$ are connected by a rope of length $L_A$ so their motion is constrained such that if one mass moves up then the other must move down, and vice versa. As a direct consequence there is a constraint on their acceleration, which you derived in your work.

The tension in the ropes had no part in the derivation of the constraint, but this constraint doesn't totally determine the motion of the system either. In general to determine the specific values for the accelerations you need as many constraint equations as you have unknowns. In this case you have 3 unknown accelerations so you need 3 constraints. You have derived one constraint equation so far, and the other 2 equations come from Newton's 2nd law applied to each of the 2 ropes, and this is where the tension in the ropes as well as the values of the masses enter into the problem.

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