Confusion in writing constraint relation for a pulley and string

Question

T1,2,3,4 are name of strings here . Pulleys are also massless as well as the string

Assumption is that B accelerates first.

When B goes down x distance T4.

A(T1) does up x distance.

I am confused how will A(T2)will go up now.

So I did by method by which I saw online.

When T 1 and T2 does up x , then B goes down x.

Then T3 goes x also.Therefore , the pulley moves down and T4 goes down again and T1 goes up again.

Not sure if this is right.because this answer does not match with my sir.

Please help

Answer is that it should be 3x(A)string lost = XB strong lost which is not the answer in online case.

I am getting confused because I think it is right

Also V IMP POINT.We say that it is xA distance by A block and XB distance by B block.Why don’t we write them 3x = x since dispatch co feed is same.Why changing variables ?

Answer 1

Your question is poorly written. I'm taking the following notation. Setting the upper pulley as origin, the vertical distances $$l_1 \ : \ \mathrm{ Vertical \ distance \ between \ the \ origin \ (upper \ pulley) \ and \ mass \ A}$$ $$l_2 \ : \ \mathrm{ Vertical \ distance \ between \ the \ origin \ (upper \ pulley) \ and \ second \ pulley}$$ $$l_3 \ : \ \mathrm{ Vertical \ distance \ between \ the \ origin \ (upper \ pulley) \ and \ mass\ B}$$ As we know the length of strings are constant thus $$l_1+l_2=\mathrm{constant}$$ $$(l_1-l_2)+(l_3-l_2)=\mathrm{constant}$$ On differentiating for acceleration $$a_A=-a_{pulley}$$ $$a_A+a_B-2a_{pulley}=0$$ $$\Rightarrow a_B=3a_{pulley}$$

Answer 2

To get the actual acceleration we can combine the observations of Young Kindaichi with force equations. Note that if the right-hand cord has a tension F, then the one on the left has a tension 2F. If mass B is going down then: ($M_B$)g – F = ($M_B$)a and 3F - ($M_A$)g = ($M_A$)(a/3). These two can be solved for, a, and, F.