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Two weights $W_1$ and $W_2$ are suspended from the ends to a light string passing over a smooth fixed pulley. If the pulley is pulled up at an acceleration of g, the tension in the string will be:-

(a)$\frac{4W_1W_2}{W_1+W_2}$

(b)$\frac{2W_1W_2}{W_1+W_2}$

Here' what I have tried:-

$$T-m_1g=m_1a$$ It is given net acceleration is g, therefore $$T-m_1g=m_1g$$ $$T=2m_1g$$ $$T=2W_1$$ $$W_1=\frac{T}{2}$$ Similarly, $$W_2=\frac{T}{2}$$

now $$\frac{4W_1W_2}{W_1+W_2}=\frac{(4)(\frac{T}{2})(\frac{T}{2})}{\frac{T}{2}+\frac{T}{2}}=T$$

Hence option (a) comes out to be tension T. But the answer section says the option (b)

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3 Answers 3

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Whats going on here is the following. Imagine that the pulley is fixed. With unequal weights, there should be an acceleration of the weights because they don't balance. Therefore, when the pulley is accelerated up at $g$, you can't assume that the weights are also accelerating up at $g$ as well. The weights are moving with respect to the pulley if the weights are unequal. Think about this and try the problem again.

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  • $\begingroup$ should the calculations be relative to the pulley or the relative to the ground $\endgroup$
    – Abhishek Mhatre
    Commented Oct 23, 2015 at 18:56
  • $\begingroup$ You'll want to keep it relative to the ground, since the acceleration of the pulley will add to the tension. $\endgroup$
    – tmwilson26
    Commented Oct 23, 2015 at 18:57
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Use the pulley frame as the frame of reference and it would get easy to solve the equations you'll derive. Since the pulley frame is non-inertial i.e. accelerating, you'll have to consider the pseudo forces acting on the objects to get to the answer.

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    $\begingroup$ Its generally discouraged to answer a homework problem completely for someone just so you know. $\endgroup$
    – tmwilson26
    Commented Oct 23, 2015 at 20:29
  • $\begingroup$ @tmwilson26 I did not know that. I'm new to this. $\endgroup$
    – Quark
    Commented Oct 23, 2015 at 20:30
  • $\begingroup$ No problem, I did the same thing before when I first joined, and the moderators deleted the answer that I gave. I was just giving you a heads up. $\endgroup$
    – tmwilson26
    Commented Oct 23, 2015 at 20:31
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As tmwilson26 pointed out, each weight is not necessarily accelerating at g. In fact, if the weights are unequal, they're not accelerating at g. Consider an atwood machine (this same system) where the pulley is not accelerating. What is the acceleration of each weight? Before you attack this problem, also think about the relation between the acceleration of one weight and that of the other.

Finally, remember that tension is equal along the string (by Newton's 3rd law, if one part of the string is pulling on another, the other part is pulling on the first with the same force). Your equations imply that the tension on either side is only equal when the weights are equal.

Review your calculations for a simple, non-accelerating atwood machine. Then add the acceleration of the system to your calculations.

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