Q.
In the figure acceleration of $A$ is $1\, m / s ^{2}$ upwards, acceleration of $B$ is $7\, m / s ^{2}$ upwards and acceleration of $C$ is $2\, m / s ^{2}$ upwards. The acceleration of $D$ will be
Laws of Motion
Solution:
Acceleration of pulley $P$
$a_{P}=\frac{a_{A}+a_{B}}{2}=\frac{1+7}{2}=4\, m / s ^{2}$ (upward)
Acceleration pulley $Q$ will be $4\, m / s ^{2}$ downwards.
$a_{Q}=\frac{a_{D}-a_{C}}{2}$
$4=\frac{a_{D}-2}{2}$
$\Rightarrow a_{D}=10\, m / s ^{2}(\downarrow)$
