Question

In the arrangement shown, acceleration of A is $1$ $m{s}^{-2}$ upwards acceleration of B is $7m{s}^{-2}$ upwards and acceleration of C is $2m{s}^{-2}$ upwards. Then acceleration of D has to be.

• A. $7m{s}^{-2}$ downwards
• B. $2m{s}^{-2}$ downwards
• C. $10m{s}^{-2}$ downwards
• D. $8m{s}^{-2}$ downwards

Answer 1

The correct option is D $8m{s}^{-2}$ downwards

Let,

The individual accelerations of masses A, B, C and D be $a_A,a_B,a_c\&a_D$ respectively.

The pully P is moving with acceleration $a$ upward and pully B is moving with acceleration $a$ downward.

Also,

Since, masses A and B are attached to the same pulley,

$a_a=a_b$

Similarly

$a_c=a_d$

Thus;

$a_b+a=7m/s^2$

$a_a-a=1m/s^2$

$\Rightarrow a_a+a_b=8m/s^2$

$\Rightarrow 2a_a=8m/s^2$

$\Rightarrow a_a=a_b=4m/s^2$

$\Rightarrow a=3m/s^2$

$a_c-a=2m/s^2$

$a_c=a_d=5m/s^2$

Therefore,

The acceleration of D will be given by

$a+a_d=(3+5)=8m/s^2$