Question

The pulley and block arrangement is as shown in the figure.

If string is massless and inextensible, pulley and surfaces are frictionless. Find the horizontal component of the acceleration of COM for the system of blocks. Take $g=10{\text{m/s}}^2$.

• A. $1.875{\text{m/s}}^2$
• B. $5.625{\text{m/s}}^2$
• C. $7.5{\text{m/s}}^2$
• D. $3.75{\text{m/s}}^2$

Answer 1

The correct option is A $1.875{\text{m/s}}^2$

FBD of (block $2\text{kg}$) and ($6\text{kg}+$pulley system):


For $2\text{kg}$ block:
$\Rightarrow$n Forces in vertical direction is balanced i.e
$N_A=m_{A}g$
Newton's $2^{nd}$ of motion in direction of acceleration:

$T=m_{A}a=2\times a=2a...\left(i\right)$

For $6\text{kg}$ block and pulley as system, the system will remain in equilibrium in horizontal direction, because horizontal tension and normal reaction $N_B$ from fixed block balances each other:

$T=N_B............\left(ii\right)$

equation of dynamics for $6\text{kg}$ block in direction of acceleration;

$m_{B}g-T=m_{B}a$
$6g-T=6a...\left(iii\right)$

From equation $\left(ii\right)$ and $\left(iii\right)$:

$6g-2a=6a$
$\Rightarrow 6g=8a$

$\therefore a=\frac{3}{4}g=7.5{\text{m/s}}^2$
$[a_A{]}_{\text{horizontal}}=7.5{\text{m/s}}^2,[a_B{]}_{\text{horizontal}}=0$

($\because$ block $B$ moves in vertical direction)

For system of blocks the $a_{\text{COM}}$ is given by:
$[a_{\text{COM}}{]}_{\text{horizontal}}=\frac{m_A(a_A{)}_{\text{horizontal}}+m_B(a_B{)}_{\text{horizontal}}}{m_A+m_B}$

$\Rightarrow [a_{\text{COM}}{]}_{\text{horizontal}}=\frac{2\times 7.5+6\times 0}{2+6}=\frac{15}{8}$

$\therefore [a_{\text{COM}}{]}_{\text{horizontal}}=\frac{15}{8}=1.875{\text{m/s}}^2$