Question 130

In the given arrangement, Block C begins to move down at a constant speed of $7\ \rm{cm/s} $ at time $t=0$. At the same instant, Block A is made to start moving down at constant acceleration. When it covers $20\ \rm{cm}$, it’s speed is $30\ \rm{cm/s}$. Assuming that B started from rest, find its displacement, velocity and acceleration by the time A covers that $20\ \rm{cm}$

In question 130, let acceleration of:-

  • Block A=$a$

  • Block B=$b$

  • Block C=$c$

Since the total length of the rope is constant, we have

$$x_A+2x_C+x_B=0\tag1$$ where $x_i$ is displacement of block $i$ in time $t$. Differentiating equation (1) twice with respect to time, we have


Since block C is not accelerating we have $$c=0$$

Therefore, $$a=-b$$

The value of $a$ is $-45/2\ \rm{cm/s^2}$ as obtained from equations of 1D motion with constant acceleration. Therefore value of $b=45/2\ \rm{cm/s^2}$. This means the displacement of Block B should be equal to the displacement of Block A since they both started at rest. But $20\ \rm{cm}$ length of the string is added by a motion of C. So that gives $x_B=40\ \rm{cm}\neq-x_A$, which contradicts what was derived earlier. How can I resolve this discrepancy?

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    $\begingroup$ Please write out the question instead of posting a picture of it. The picture won't show up in searches and is not accessible to everyone. $\endgroup$ – Aaron Stevens Aug 10 at 13:35
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    $\begingroup$ Also, please use MathJax to format equations. $\endgroup$ – Aaron Stevens Aug 10 at 13:56
  • $\begingroup$ But if I write out the question there will still be a diagram to portray... $\endgroup$ – Ashish Raj Shukla Aug 10 at 13:58
  • $\begingroup$ I don’t know anything about MathJax... but I tried to make my reasoning clear $\endgroup$ – Ashish Raj Shukla Aug 10 at 14:02
  • $\begingroup$ You can still put the diagram up. And I didn't say anything about your reasoning. $\endgroup$ – Aaron Stevens Aug 10 at 14:04

This isn't your fault. The question is flawed. Differentiate the eq you got, $$x_A +x_B + 2x_C =0$$ to get $$v_A +v_B + 2v_C =0$$ At $t=0$ ,$v_A =0 , v_C = 7$ then $v_B$ can not be zero. Thus Block B can not start from rest.

  • $\begingroup$ That tells me,what if B acquired a velocity equal to twice of Block C,in dt time... $\endgroup$ – Ashish Raj Shukla Aug 10 at 18:24
  • $\begingroup$ B did start from rest,but just like C,it acquired a velocity asap $\endgroup$ – Ashish Raj Shukla Aug 10 at 18:24
  • $\begingroup$ @AshishRajShukla Right. If $A$ starts from rest and $C$ starts moving down at $7\ \rm{cm/s}$, then by the relations it must be that $B$ initially moves up with velocity $14\ \rm{cm/s}$ $\endgroup$ – Aaron Stevens Aug 10 at 20:10

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