Question

In the given situation, two blocks A and B are arranged as shown in the figure. M is a fixed pulley and N is a movable pulley. The system is released from rest. What is the relationship between accelerations for the blocks A and B?

• A. a_B = 2a_A
• B. a_A = 2a_B
• C. a_B = -2a_A
• D. a_A = -2a_B

Answer 1

The correct option is C

a_B = -2a_A

Let us consider the system shown in the figure. It is clear from the figure that the position of A is governed by the position of centre of movable pulley. Let at any instant the block B is at $y_B$ and center of movable pulley is $y_A$ from the reference line (dotted line). The total length of the cord:

$Y_B+2Y_A=I_0$ ............(i)

$I_0$ is the length of the string (remain constant). Assume acceleration of blocks and pulleys. Differentiating equation (i) with respect to time, we get

$\frac{d{y}_B}{dt}+\frac{2d{y}_A}{dt}=\frac{d{l}_0}{dt}$ ............(ii)

As $i_0$ is constant, $\frac{d{l}_0}{dt}=0$

and $\frac{d{y}_B}{dt}=v_B$ and $\frac{d{y}_A}{dt}=-v_A$

as $Y_A$ is decreasing since block A is assumed tobe with acceleration $a_A$

the equation (ii) become $v_B-2v_A=0$ ...........(iii)

Differentiating once more we get $a_B-2a_A=0$ or $a_B=+2a_A$

This means block B goes down with twice the acceleration with which A goes upwards. $\Rightarrow a_B=-2a_A$

Alternate solution II If mass A goes up by a distance x, we can observe that the string lengths ab and cd are slack. Due to the weight of block B, this length (ab + cd = 2x) will go on this side and block B, will descend by a distance 2x. As in equal time duration,

B has travelled a distance twice that of A. So $v_B=-2v_A$ and $a_B=-2a_A$

Alternate solution III

Displacement of the pulley is the average displacement of both sides of the pulley.

$\Rightarrow x_A=\frac{x_p+0}{2}$

$\Rightarrow x_P=2x_A$

Now for pulley 2

$0=\frac{x_P-x_B}{2}$

$X_P=X_B$

$\Rightarrow 2X_A=X_B$ Lets take care of direction

A goes up so $x_A$ is +ve

B goes down so $X_B$ is -ve

$X_B=-2X_A$

$V_B=-2V_A$

$a_B=-2a_A$