Question

In the arrangement shown in $Fig.6.378$, all pulleys are smooth and massless. When the system is released from the rest, acceleration of blocks $2$ and $3$ relative to $1$ are $1m{s}^{-2}$ downwards and $5m{s}^{-2}$ downwards. respectively. Acceleration of block $3$ relative to $4$ is zero.
Find the absolute acceleration ofblock $4$.

• A. $2m{s}^{-2}$ upwards
• B. $1m{s}^{-2}$ downwards
• C. $3m{s}^{-2}$ downwards
• D. $1.5m{s}^{-2}$ upwards

Answer 1

The correct option is C $3m{s}^{-2}$ downwards

acceleration of block 2 & 3 to 1 are $1{ms}^{-1}$

observer is standing on 1 \& looking downwards to $a,4a_3$. So, $a_2+a_1=1m/s^2-\left(I\right)$

$a_3+a_1=5m/s^2-\left(11\right)$

now, the obserwen is standing on 4 . therefore, $a_4-a_3=0$

$a_4=a_3-\left(III\right)$

$now,consideningdinection4finding{a}_1=\frac{a5+a4}{4},$ $\frac{a_2+a_3}{2}=a_5[$ for $2^{nd}\&3^{rd}$ pully ]

$\Rightarrow 2a_1-a_4=\frac{a_2+a_3}{2}$

$\Rightarrow 4a_1=2a_4+a_3+a_2[a_4=a_3]$

$\Rightarrow 4a_1=3a_3+a_2$

$\Rightarrow 4a_1=3(5-a_1)+(1-a_1)$

$\Rightarrow 8a_1=16$

$\Rightarrow a_1=2m/s^2$

now, $a_1+a_2=1$; $a_2=-1m/s^2$

$\Rightarrow a_3+a_1=5$

Equation

(11)

$\Rightarrow a_3=5-2=3m/s^2\downarrow$

$\Rightarrow |a_3=3m|s^2$

Answer:- $3m/s^{2}downward\left(c\right).Thecorrectoptionisc$