# In this pulley-block system, acceleration of blocks-A and B is same then how is their displacement different?

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

$$a+2c+b=0$$

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?

• 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. – Aaron Stevens Aug 10 at 13:35
• Also, please use MathJax to format equations. – Aaron Stevens Aug 10 at 13:56
• But if I write out the question there will still be a diagram to portray... – Ashish Raj Shukla Aug 10 at 13:58
• I don’t know anything about MathJax... but I tried to make my reasoning clear – Ashish Raj Shukla Aug 10 at 14:02
• You can still put the diagram up. And I didn't say anything about your reasoning. – 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.
• @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 $14\ \rm{cm/s}$ – Aaron Stevens Aug 10 at 20:10