Boxes A and B are connected to each end of a light vertical rope (Fig. P4.49). A constant upward force $\mathbf{F}\mathbf{=}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{N}$is applied to box A. Starting from rest, box B descends $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$in $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$. The tension in the rope connecting the two boxes is $\mathbf{36}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. What are the masses of (a) box B, (b) box A?

The given data can be listed below as,

The value of the upward force is $\mathrm{F}=80.0\mathrm{N}$.

The distance descended by the box B is $\mathrm{s}=12.0\mathrm{m}$.

The time taken by the Box B to descend is $\mathrm{t}=4.00\mathrm{s}$.

The rope’s tension that connects the two boxes is $\mathrm{T}=36.0\mathrm{N}$.

Tension is described as a force that is transmitted in an axial direction with the help of a cable or string. Moreover, tension is also referred to as a pair of action and reaction forces.

The free-body diagram of the box B has been drawn below:

In the above diagram, the tension $\mathrm{T}$is acting in the upward direction and the weight of the box B which is ${\mathrm{W}}_{\mathrm{B}}$is acting in the downwards direction that is the product of the mass of the box B which is ${\mathrm{m}}_{\mathrm{B}}$and acceleration due to gravity $\mathrm{g}$.

The equation of the acceleration of the system is expressed as:

$$\begin{gathered} \mathrm{s}=\mathrm{ut}\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{s}-\mathrm{ut}=\frac{1}{2}{\mathrm{at}}^2 \\ 2\left(\mathrm{s}-\mathrm{ut}\right)={\mathrm{at}}^2 \\ \mathrm{a}=\frac{2\left(\mathrm{s}-\mathrm{ut}\right)}{{\mathrm{t}}^2} \end{gathered}$$

….. (1)

Here, $\mathrm{s}$is the distance descended by the box B, $\mathrm{u}$is the initial velocity of the box B, $\mathrm{t}$is the time taken by the Box B to descend and$\mathrm{a}$ is the acceleration of the system.

Since, the box B was at rest initially, then the initial velocity of the box B is zero.

The equation of the mass of the box B is calculated as:

$$\begin{gathered} {\mathrm{m}}_{\mathrm{B}}\mathrm{g}-\mathrm{T}={\mathrm{m}}_{\mathrm{B}}\mathrm{a} \\ {\mathrm{m}}_{\mathrm{B}}\left(\mathrm{g}-\mathrm{a}\right)=\mathrm{T} \end{gathered}$$

Here, ${\mathrm{m}}_{\mathrm{B}}$is the mass of the box B, $\mathrm{g}$is the acceleration due to gravity and $\mathrm{T}$is the tension exerted on the block B.

Substitute the values of the equation (1) in the above equation.

$$\begin{gathered} m_B\left(g-\frac{2\left(s-ut\right)}{t^2}\right)=T \\ {m}_B=\frac{T}{\left(g-\frac{2\left(s-ut\right)}{t^2}\right)} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{m}}_{\mathrm{B}}=\frac{36.0\mathrm{N}}{\left(9.8\mathrm{m}/{\mathrm{s}}^2-{\displaystyle \frac{2\left(12.0\mathrm{m}-\left(0\right)\mathrm{t}\right)}{{\left(4.00\mathrm{s}\right)}^2}}\right)} \\ =\frac{36.0\mathrm{N}\times \left({\displaystyle \frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}}\right)}{\left(9.8\mathrm{m}/{\mathrm{s}}^2-{\displaystyle \frac{\left(24.0\mathrm{m}\right)}{{\left(16.00\mathrm{s}\right)}^2}}\right)} \\ =\frac{36.0\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{\left(9.8\mathrm{m}/{\mathrm{s}}^2-1.5\mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{36.0\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{\left(8.3\mathrm{m}/{\mathrm{s}}^2\right)} \\ {\mathrm{m}}_{\mathrm{B}}=4.3\mathrm{kg} \end{gathered}$$

Thus, the mass of the box B is $4.3\mathrm{kg}$.

The free body diagram of the box A has been drawn below:

In the above diagram, the force $\mathrm{F}$is acting in the upward direction and the weight of the box A which is ${\mathrm{W}}_{\mathrm{A}}$is acting in the downwards direction that is the product of the mass of the box A which is ${\mathrm{m}}_{\mathrm{A}}$and acceleration due to gravity $\mathrm{g}$. The tension $\mathrm{T}$is also acting in the downwards direction.

Since, the box A was at rest initially, then the initial velocity of the box A is zero.

The equation of the mass of the box A is calculated as:

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}\mathrm{g}+\mathrm{T}-\mathrm{F}={\mathrm{m}}_{\mathrm{A}}\mathrm{a} \\ {\mathrm{m}}_{\mathrm{A}}\left(\mathrm{g}-\mathrm{a}\right)=\mathrm{F}-\mathrm{T} \\ {\mathrm{m}}_{\mathrm{A}}=\frac{\mathrm{F}-\mathrm{T}}{\mathrm{g}-\mathrm{a}} \end{gathered}$$

Here, ${\mathrm{m}}_{\mathrm{A}}$is the mass of the box A, g is the acceleration due to gravity and T is the tension exerted by the block A.

Substitute the values of the equation (1) in the above equation.

${\mathrm{m}}_{\mathrm{A}}=\frac{\mathrm{F}-\mathrm{T}}{\mathrm{g}-\left(\mathrm{g}-{\displaystyle \frac{2\left(\mathrm{s}-\mathrm{ut}\right)}{{\mathrm{t}}^2}}\right)}$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}=\frac{80.0\mathrm{N}-36.0\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2-\left(9.8\mathrm{m}/{\mathrm{s}}^2-{\displaystyle \frac{2\left(12.0\mathrm{m}-\left(0\right)\mathrm{t}\right)}{{\left(4.00\mathrm{s}\right)}^2}}\right)} \\ =\frac{44.0\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2-\left(9.8\mathrm{m}/{\mathrm{s}}^2-{\displaystyle \frac{\left(24.0\mathrm{m}\right)}{{\left(16.00\mathrm{s}\right)}^2}}\right)} \\ =\frac{44.0\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2-\left(9.8\mathrm{m}/{\mathrm{s}}^2-1.5\mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{44.0\mathrm{N}\times \left({\displaystyle \frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2-\left(8.3\mathrm{m}/{\mathrm{s}}^2\right)} \end{gathered}$$

Hence, further as:

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}=\frac{44.0\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{9.8\mathrm{m}/{\mathrm{s}}^2-\left(8.3\mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{44.0\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1.5\mathrm{m}/{\mathrm{s}}^2} \\ =29.3\mathrm{kg} \end{gathered}$$

Thus, the mass of the box B is $29.3\mathrm{kg}$.