Question

In Figure 5.72 ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{12}\mathbf{\hspace{0.33em}}\mathbf{kg}$and ${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{\hspace{0.33em}}\mathbf{kg}$. The kinetic coefficient of friction between ${\mathbf{m}}_{\mathbf{1}}$and the floor is $\mathbf{0}\mathbf{.}\mathbf{3}$and that between ${\mathbf{m}}_{\mathbf{2}}$and the floor is $\mathbf{0}\mathbf{.}\mathbf{5}$. You push with a force of magnitude $\mathbf{F}\mathbf{=}\mathbf{110}\mathbf{\hspace{0.33em}}\mathbf{N}$. (a) What is the acceleration of the center of mass? (b) What is the magnitude of the force that ${\mathbf{m}}_{\mathbf{1}}$exerts on ${\mathbf{m}}_{\mathbf{2}}$?

Short answer

(a) The acceleration of the center of mass is$2.95\mathrm{m}/{\mathrm{s}}^2$.

(b) The magnitude of the force that $m_1$exerts on$m_2$ is $39.27\hspace{0.33em}\mathrm{N}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The value of the first mass is,${\mathrm{m}}_1=12\hspace{0.33em}\mathrm{kg}$ .
• The value of the second mass is,${\mathrm{m}}_2=5\hspace{0.33em}\mathrm{kg}$ .
• The coefficient of friction between the floor and the first mass is ${\mu }_1=0.3$.
• The coefficient of friction between the floor and the second mass is${\mu }_2=0.5$ .
• The force required to push the masses are$\mathrm{F}=110\hspace{0.33em}\mathrm{N}$ .

Significance of the coefficient of friction

The coefficient of friction is described as the ratio between the friction and the normal force. However, the coefficient of friction mainly depends on surface roughness and nature of the material used.

(a) Determination of the acceleration of the center of mass

The equation of the net force is expressed as:

${\mathrm{F}}_{\mathrm{net}}=\mathrm{F}-({\mathrm{\mu }}_1{\mathrm{m}}_1\mathrm{g}+{\mathrm{\mu }}_2{\mathrm{m}}_2\mathrm{g})$

Here,$F_{net}$is the net force, F is the force required to push the masses, ${\mathrm{\mu }}_1$is the coefficient of friction between the floor and the first mass,$m_1$is the mass of the first mass, g is the acceleration due to gravity,${\mu }_2$is the coefficient of friction between the floor and the second mass and$m_2$is the mass of the second mass.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=110\mathrm{N}-\left(0.3\times 12\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2+0.5\times 5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =110\mathrm{N}-\left(35.28\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2+24.5\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right) \\ =110\mathrm{N}-\left(59.78\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right)\right) \\ =50.22\mathrm{N} \end{gathered}$$

The equation of the acceleration of the center of mass is expressed as:

$\mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{net}}}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

Here, a is the acceleration of the center of mass,$F_{net}$ is the net force,$m_1$ is the mass of the first mass and $m_2$is the mass of the second mass.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{a}=\frac{50.22\mathrm{N}}{12\mathrm{kg}+5\mathrm{kg}} \\ =\frac{50.22\mathrm{N}}{17\mathrm{kg}} \\ =2.95\mathrm{N}/\mathrm{kg} \\ =2.95\mathrm{N}/\mathrm{kg}\times \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2/1\mathrm{N} \\ =2.95\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration of the center of mass is $2.95m/s^2$.

(b) Determination of the magnitude of the force

The equation of the magnitude of the force that the first mass exerts on the second mass is expressed as:

${\mathrm{F}}_1={\mathrm{m}}_2\mathrm{a}+{\mathrm{\mu }}_2{\mathrm{m}}_2\mathrm{g}$

Here, $F_1$is the magnitude of the force that the first mass exerts on the second mass, a is the acceleration of the center of mass, g is the acceleration due to gravity,${\mu }_2$ is the coefficient of friction between the floor and the second mass and$m_2$ is the mass of the second mass.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_1=\left(5\mathrm{kg}\right)\left(2.95\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.5\right)\left(5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =14.77\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2+24.5\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =39.27\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =39.27\mathrm{N} \end{gathered}$$

Thus, the magnitude of the force that$m_1$ exerts on$m_2$ is$39.27\hspace{0.33em}\mathrm{N}$ .