Question

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate A has mass m_A, and crate B has mass m_B. The coefficient of kinetic friction between each crate and the surface is ${\mathbf{\mu }}_{\mathbf{k}}$. The crates are pulled to the right at constant velocity by a horizontal force $\overrightarrow{\mathbf{F}}$. Draw one or more free-body diagrams to calculate the following in terms of m_A, m_B and ${\mathbf{\mu }}_{\mathbf{k}}$: (a) the magnitude of $\overrightarrow{\mathbf{F}}$ and (b) the tension in the rope connecting the blocks.

Figure E5.37

Short answer

1. The magnitude of $\overrightarrow{\mathrm{F}}$is, ${\mathrm{\mu }}_{\mathrm{k}}\mathrm{g}\left({\mathrm{m}}_{\mathrm{B}}+{\mathrm{m}}_{\mathrm{A}}\right)$.
2. The tension in the connecting rope is, ${\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g}$.

Step-by-step solution

Identification of given data

The given data can be listed below as,

The mass of crate A is, m_A.

The mass of crate B is, m_B.

The horizontal force is, $\overrightarrow{\mathrm{F}}$.

The coefficient of kinetic friction is, ${\mathrm{\mu }}_{\mathrm{k}}$.

Significance of tension

Tension is a pulling force which transmitted axially with the help of rope, string or cable at the end is known as tension.

(a) Determination of magnitude of F→.

The expression for frictional force on crate A and b can be expressed as,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{A}}={\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g} \\ {\mathrm{f}}_{\mathrm{B}}={\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{B}}\mathrm{g} \end{gathered}$$

Here ${\mathrm{\mu }}_{\mathrm{k}}$ is the coefficient of kinetic friction, m_A is the mass of the crate A, m_B is the mass of the crate B.

The expression for force using Newton’s second law can be expressed as,

$\overrightarrow{\mathrm{F}}-{\mathrm{f}}_{\mathrm{B}}-{\mathrm{f}}_{\mathrm{A}}=0$

Here f_A is the force on crate A, f_B is the force on crate B.

Substitute ${\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g}$ for f_A, ${\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{B}}\mathrm{g}$for f_B in above equation.

$$\begin{gathered} \overrightarrow{\mathrm{F}}-{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g}-{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{B}}\mathrm{g}=0 \\ \overrightarrow{\mathrm{F}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{g}\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right) \end{gathered}$$

Hence, the required magnitude of $\overrightarrow{\mathrm{F}}$is, ${\mathrm{\mu }}_{\mathrm{k}}\mathrm{g}\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)$.

(b) Determination of tension in the rope connecting the blocks.

For Tension considering the block rare one block A and block B as an isolated system, the expression for tension can be expressed as,

$$\begin{gathered} \mathrm{T}-{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g}=0 \\ \mathrm{T}={\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g} \end{gathered}$$

Hence, the required tension is, ${\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g}$.