In Figure 5-48, three connected blocks are pulled to the right on a horizontal frictionless table by a force of magnitude${\mathit{T}}_{\mathbf{3}}\mathbf{=}\mathbf{65}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}$. If${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$,${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$, and${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{31}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$ , calculate (a) the magnitude of the system’s acceleration,(b) the tension${\mathit{T}}_{\mathbf{1}}$ , and (c) the tension${\mathit{T}}_{\mathbf{2}}$ .

$$\begin{gathered} 1)T_3=65.0N \\ 2)m_1=12.0kg \\ 3)m_2=24.0kg \\ 4)m_3=31.0kg \end{gathered}$$

Newton’s second law states that the net force acting on the object is vector sum of all the forces and equals to the product of mass and net acceleration of the object.

This problem is concerned with the application of Newton’s second law; we can apply Newton’s second law to the whole system (consisting of all three masses) as well as to the individual mass.

Formula:

F = ma (i)

Let us apply Newton’s second law to the whole system.

We have, the mass of the whole system as

$$\begin{gathered} M=m_1+m_2+m_3 \\ =12.0+24.0+31.0 \\ =67.0kg \end{gathered}$$

Now, according to the diagram shown, the tension$T_3$ is the net force. So

$$\begin{gathered} T_3=M\times a \\ 65.0N=67.0kg\times a \\ a=0.97m/s^2 \end{gathered}$$

Therefore, the acceleration of system as well as the individual mass is$0.97m/s^2$ .

By applying Newton’s second law to the mass and using equation (i),

$$\begin{gathered} T_1=m_{1}a \\ =12.0\times 0.97 \\ =11.6N \end{gathered}$$

Therefore, the tension $T_1$is 11.6 N .

Now, applying Newton’s second law to system of mass and mass and using equation (i),

$$\begin{gathered} T_2=\left(m_1+m_2\right)\times a \\ =\left(12.0kg+24.0kg\right)\times 0.97m/s^2 \\ =34.0N \end{gathered}$$

Therefore, the tension$T_2$ is 34.9 N .