Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 181: 44

$T1 = m1a$ $T2 - T1 = m2a$ $m3g - T2 = m3a$ a = $4.9 m/s^2$

By analyzing the forces acting on each block and utilizing Newton's Second Law, we can solve the problem and find the acceleration of the masses. On block ($M1$), there are three forces acting on the block. These are the Normal Force of the table on the block($N$), the force of gravity downward($mg$) and the tension force exerted by the rope connected to the block($T1$). On block m2, there are four forces exerted on the block. These are the Normal Force of the table on the block($N$), the force of gravity downward($mg$), the Tension Force($T2$) in the positive direction and the Tension Force($T1$) in the negative direction. Finally, on block($T3$), there are only two forces acting on the block. These are the force of gravity downward($mg$) and the tension force upward($T2$) Using Newton's Second Law, which states that the net component forces acting on an object equals an object's mass($m$) times acceleration($a$), we can write relations for all three blocks. For block $m1$, $T1$ is the only force acting horizontally on the block, $T1 = m1a$. For block $m2$, $T2$ acts in the positive horizontal direction and $T1$ acts in the negative horizontal direction. As a result, $T2 - T1 = m2a$. For block $m3$, $T2$ acts in the negative direction because the block accelerates downward. $m3g$ then acts upward. As a result, $m3g - T2 = m3a$. By using algebraic manipulation, $T2 = a(m1 + m2)$ and $m3g = (m1 + m2 + m3)a$. Since $m1 = 1 kg$, $m2 = 2kg$, $m3 = 3kg$ and g is $9.8 m/s^2$, $a$ is $4.9m/s^2$.