Question

Mass A is connected with a string which passes through a fixed pulley. The other end of the string is connected with a movable pulley N. The block B is connected with another string which passes through the pulley N as shown in figure. Find the relation between acceleration of A & B.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The positions of blocks and movable pulley are taken from fixed pulley are shown in figure. The length of string is constant.

For string '1':$y_1+y_p=L_1$ ............(i)

$L_1$= length of string 1

Assume accelerati­­­ons of blocks and pulleys. Here block A has $+a_1$ and B has $-a_2$ as going down also $a_p$ is in downward direction

Differentiating equation (i) with respect to time

$\frac{d{y}_1}{dt}+\frac{d{y}_p}{dt}=\frac{d{l}_2}{dt}$ or$-v_A+v_P=0$

$rightarrow{v}_A=v_P$ which implies $a_A=a_P$

$\frac{d{y}_1}{dt}=-v_A$ as $y_1$ is decresing (A is going up)

$\frac{d{l}_2}{dt}=0$ As length of string 2 is constant

For string '2': $(y_0-y_p)+(y_2-y_p)=I_2$ .........(ii)

Differentiating equation (ii) with respect to time

$0-\left(v_p\right)+\left(v_2\right)-\left(v_p\right)=0\Rightarrow v_2=2v_{p}or{v}_B=+2v_A\Rightarrow a_B=+2a_A$

This means block will go down with twice the acceleration with which A goes up. So $a_B=-2a_A$

Alternate solution I

Consider the situation shown in figure. If we consider that mass A is goes up by distance x, pulley N, which is attached to the same string, will go down by the same distance x. Due to this, the string which is connected to mass B will now have free lengths ab and cd (ab = cd = x) which will go on the side of mass B due to its weight as the other end is fixed at point P. Thus mass B will go down by 2x, Hence, its speed and acceleration will be twice that of block B.

$v_B=-2v_A$ and $a_B=-2a_A$

Alternate solution II

Displacement of the pulley is the average displacement of both sides of the pulley. Take displacement in +y axis as +ve and in - y axis as - ve.

Displacement of pulley M = 0

As its fixed

Let's assume block A moves up to $x_A$ and the other side moves down by $x_p$ so $-x_p\Rightarrow \frac{x_A-x_P}{2}=0,x_A=x_P$

Now for pulley P.

$x_A=x_P=$

with sign $2x_A=x_B$ only magnitude

$-2x_A=x_B$ as $X_A$ is up +ve

$\Rightarrow -2v_A=v_B$ $X_B$ is down -ve

$\Rightarrow -2a_A=a_B$