Question

In $Fig.6.48$ blocks A and B are connected together by a string and placed on a smooth inclined plane. B is connected to C (which is suspended vertically) by another string which passes over a smooth pulley fixed to the plane. The mass of A is $m_A=1kg$ and mass of B is $m_B=2kg$.
$a$. If the system is at rest, find the mass of C.
$b$. If the mass of C is twice the mass calculated in (a), then find the acceleration of the system.

Answer 1

From the free-body diagram of A :
$T_1$ : Tension in string between A and B
$N_A$ : Normal reaction between A and incline
As A is at rest, net force parallel and perpendicular to inclined should be balanced.
$N_A=m_{A}g\cos \theta =10\cos {30}^0=5\sqrt{3}N$
$T_1=m_{A}g\sin \theta =10\sin {30}^0=5N$
From the free-body diagram of B :
As B is connected to both strings, two tensions $T_1$ and $T_2$ will act on it.
$T_2$ is the tension (force) of string between B and C acting upwards
$T_1$ is the tension of string between A and B acting downwards
Balancing Forces:
$N_B=m_{B}g\cos {30}^0=20\cos {30}^0=10\sqrt{3}N$
$T_2=T_1+m_{B}g\sin {30}^0=5+20\sin {30}^0$
$=5+20\times \frac{1}{2}=15N$
Force diagram of C :
$T_2$ is the pulling force of string on block C, therefore,
$m_{C}g=T_2$ hence $m_C=1.5kg$
$b$. In this case, mass of $C=3kg$
Let the acceleration of the system be a. We can assume an arbitrary direction of motion. Let the blocks move up the arbitrary and block C move downward.
From FBD of A : $T_1-10\sin {30}^0=1a$......(i)
From FBD of B : $T_2-T_1-20\sin {30}^0=2a$.....(ii)
From FBD of C : $30-T_2=3a$.......(iii)
Solving (i), (ii), and (iii), $a=2.5m{s}^{-2}$