There was a question which I came across in newtons law of motion chapter in the physics book by H.C.Verma which was based upon simple atwood machine (atwood machine consists of two unequal masses connected by string going over a clamped smooth , light pulley).

The question asks us to consider a case where the heavier mass of the atwood machine were stopped for a moment 2 seconds after system is set into motion, and to find time elapsed before string is tight again.

I am willing to do the question by myself but I was unable to understand the condition of tight string. When do we say that the string is tight? Is it at a particular tension or is it when minimum value of tension is just greater than 0?

  • 2
    $\begingroup$ HCV is not so good for concepts. Try the book by Resnick and Halliday. $\endgroup$
    – Shubham
    Oct 20, 2015 at 18:32

2 Answers 2


Starting with tightness requirement:

"...Find time elapsed before string is tight again."

Clearly, at some low enough tension a string would still be stretchable to overcome curls in its fibers. The string will be tight/taut provided each mass has a weight in excess of that required to stretch the string to normal length, i.e. no further stretching and below breaking force.

The following sequence of events will happen:

a. both masses begin motion $\rightarrow$ b. 2 seconds elapse $\rightarrow$ c. heavier mass stopped $\rightarrow$ d. string slackens $\rightarrow$ e. lighter mass stops upward motion $\rightarrow$ f. lighter mass drops back down $\rightarrow$ g. lighter mass stops (rope tight/taut)

Assuming the two fore-mentioned are met, and as stated in the question, the pulley is smooth (no friction), then at start of motion, the two masses accelerate at a rate:

$a = F/(m_1+m_2) \tag1$

where the mass weight imbalance is

$F = (m_2-m_1)g\tag2$ At time 2 seconds after start of motion, the velocity of the masses, the heavier one $m_2$ falling, the lighter one $m_1$ rising, is:

$v = u+at \Rightarrow v = 2a \tag3$

substituting Equation $2$ into $1$ and $1$ into $3$:

$v = 2g{{m_2-m_1}\over{m_1+m_2}} \tag4$

As one stops the falling of the heavier mass, then the lighter mass $m_1$ will continue to rise as it decelerates due to gravity. The string loses its tautness. In this process, the final velocity from Equation $4$ above becomes $m_1$'s initial velocity, and its final velocity this time is $0$.

$v = u + at \Rightarrow t=(v-u)/a \Rightarrow t = (0 - (2g{{m_2-m_1}\over {m_1+m_2}}))/g \tag5$ therefore time until lighter mass $m_1$ stops upward motion after the heavier mass $m_2$ is stopped is:

$t = {{2(m_1-m_2)} \over{m_1+m_2}}\tag6$

Finally, the lighter mass will drop to the 'rope tight' position. Now note that since there is no friction in the pulley, and gravity acts symmetrically on the lighter mass as it rises vs as it falls:

(downward velocity of the lighter mass after falling back down to 'rope tight' position) $=$ (upward velocity of lighter mass just after heavier mass was stopped):

$v = u + at$ where v and u are switched for the drop journey of $m_1$ vs its rise journey. The time is therefore the same, and total time is twice the rise time (from Equation 6):

$t_{m_1 rise-and-fall} = 2(t_{m_1 rise}) = {{4(m_1-m_2)} \over{m_1+m_2}}\tag7$


Before the block was stopped, both the heavier and light blocks were moving with the same acceleration, a= (Net Force)/(Total Mass). When you stop the heavier block, the net force gets a different magnitude and direction.

In this condition the direction is opposite to the earlier one. Thus, the lighter block will accelerate downward with g at the instant the heavier block is stopped. But the velocity of the lighter block will remain in the upward direction for a while, when the string connecting the heavier block slackens for a while. But after sometime, since acceleration is g downwards, the lighter body moves downwards, and a time comes when the string connecting the lighter block again becomes taut.

Also, regarding understanding of tightness in strings:

If you are having a string with zero tension it is neither taut nor slackened. If tension is more than zero its taut. If less than zero(or you may say, when tension is undefined) the string is said to be slack.**enter image description here**

  • $\begingroup$ I think in the 2nd line, you meant acceleration=(force)/(net mass)... $\endgroup$ Oct 21, 2015 at 6:39
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    $\begingroup$ I think this answer is pretty good except the wording needs clarification. I don't understand the last sentence of the first paragraph. Perhaps replace with: "Now stop the heavy block." I don't understand the first sentence of the second paragraph. Perhaps remove it completely, and start the next sentence "Now the lighter block ...". With those changes, I understand the answer. Without those changes (or something like them) I find the answer confusing. $\endgroup$
    – garyp
    Aug 19, 2016 at 14:24
  • $\begingroup$ @SchrodingersCat What is the meaning of "net mass"? I don't think I've ever heard the phrase "net mass". It looks ok to me the way it is. $\endgroup$
    – garyp
    Aug 19, 2016 at 14:25

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