Figure shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are${\mathit{m}}_{\mathbf{A}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathit{k}\mathit{g}$,${\mathit{m}}_{\mathbf{B}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$, and${\mathit{m}}_{\mathbf{c}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$. When the blocks are released, what is the tension in the cord at the right?

It is given that, the masses of the blocks are,

$$\begin{gathered} m_A=6kg \\ {m}_B=8kg \\ {m}_c=10kg \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagram the tension in the cord can be found.

Formula

Newton’s second law is,

$F_{net}=\sum _{}ma$ (i)

where, $F_{net}$is the net force, $m$is mass and $a$is an acceleration.

The total mass of the system is,

$$\begin{gathered} m=m_A+m_B+m_C \\ =24\mathrm{kg} \end{gathered}$$

When the blocks are released, due to the difference in the masses of A and C, the system get accelerated with an acceleration

Using equation (i),

$$\begin{gathered} m_{C}g-m_{A}g=ma \\ a=\frac{\left(m_{c}g-m_{a}g\right)}{m}=\frac{\left(m_C-m_A\right)}{m} \\ a=\frac{\left(9.8\right)\left(10-6\right)}{24} \\ a=1.63\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Free body diagram for mass c is,

$$\begin{gathered} m_{c}g-T=m_{c}a \\ T=m_{c}g-m_{c}a \end{gathered}$$

From free body diagram,

$$\begin{gathered} T=m_c\left(g-a\right) \\ T=10\left(9.8-1.63\right) \\ T=81.7\mathrm{N} \end{gathered}$$

Hence, tension in the chord at right when the blocks are released is 81.7 N .