Question

In the arrangement shown in figure, pulleys are small, light and frictionless. Threads are inextensible and mass of blocks A, B and C is $m_1=5$kg, $m_2=4$ kg and $m_3=2.5$ kg respectively. Coefficient of friction for both the planes is $=0.50$ . Acceleration of block A (in $m/s^2$) is:

• A. $3$
• B. $2$
• C. $4$
• D. None of these

Answer 1

Answer: (C)

$4$

Let's say,

acceleration of A is $a_A$

acceleration of B is $a_B$

acceleration of C is $a_C$

tension is $T$

acceleration due to gravity is g

Dividing gravity into 2 components as $mgsin\left(x\right)$ along the incline and mgcos(x) perpendicular to the incline. Using $f=\mu N.$

Writing equations for these accelerations-

$m_A{a}_A=T+m_{A}gsin\theta -{\mu m}_{A}gcos\theta$

$m_B{a}_B=T-{\mu m}_{B}gcos\theta$

$a_C=\frac{a_A+a_B}{2}$

$m_C{a}_C=m_{C}g-2T$

putting values of $sin\left(37\right)=$ $\frac{3}{5}$ and solving the equations, we find that,

$T=14.78N,$ which is less than the maximum frictional force provided by the floor on $B,$ hence $B$ will be stationary.
$a_B$$=0$

writing new equations with $B$ as stationary

$m_A{a}_A=T+m_{A}gsin\theta -{\mu m}_{A}gcos\theta$

$a_C=\frac{a_A}{2}$

$m_C{a}_C=m_{C}g-2T$

solving these,

we get

$T$$=0N$

$a_A=4m{s}^{-2}$ and

$a_C$ $=$ 2 $m{s}^{-2}$