Figure 6-32 shows three crates being pushed over a concrete floor by a horizontal force of magnitude $\mathbf{440}\mathbf{}\mathbf{N}$. The masses of the crates are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{kg}$, ${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{kg}$, and ${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$.The coefficient of kinetic friction between the floor and each of the crates is $\mathbf{0}\mathbf{.}\mathbf{700}$. (a) What is the magnitude ${\mathbf{F}}_{\mathbf{32}}$of the force on crate 3 from crate 2? (b) If the crates then slide onto a polished floor, where the coefficient of kinetic friction is less than $\mathbf{0}\mathbf{.}\mathbf{700}$, is magnitude ${\mathbf{F}}_{\mathbf{32}}$more than, less than, or the same as it was when the coefficient was $\mathbf{0}\mathbf{.}\mathbf{700}$?

$$\begin{gathered} {\mathrm{m}}_1=30.0\mathrm{kg} \\ {\mathrm{m}}_2=10.0\mathrm{kg} \\ {\mathrm{m}}_3=20.0\mathrm{kg} \end{gathered}$$

Force, $\mathrm{F}=440\mathrm{N}$

Coefficient of kinetic friction between the Crates and floor,${\mathrm{\mu }}_{\mathrm{k}}=0.700$

To find the force on box 3 due to box 2, find the acceleration of the system and then by using the Newton’s 2nd law of motion, the force ${}^{{\mathbf{F}}_{\mathbf{32}}}$can be found. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\sum ma$

where, F is the net force, m is mass and a is an acceleration.

Free Body Diagram of the system $\left({\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3\right)$and 3rd crate:

By using Newton’s 2nd law of motion along the vertical direction to the system,

$$\begin{gathered} N-(m_1+m_2+m_3)g=0 \\ N=(m_1+m_2+m_3)g \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} f_k={\mu }_{k}N \\ ={\mu }_k\left(m_1+m_2+m_3\right)g \end{gathered}$$

Now, applying Newton’s 2nd along horizontal direction to the system,

$$\begin{gathered} {f_k}^{\text{'}}={\mu }_k{N}^{\text{'}} \\ ={\mu }_k{m}_{3}g \\ {F}_{32}-{f_k}^{\text{'}}=m_{3}a \\ {F}_{32}=m_{3}a+{f_k}^{\text{'}} \\ =(20.0)(0.47)+(0.700)(20.0)(9.81) \\ =147N \end{gathered}$$

Using the Newton’s 2nd law along vertical and horizontal direction to the 3rd crate,

$$\begin{gathered} N\text{'}-m_{3}g=0 \\ N\text{'}=m_{3}g \end{gathered}$$

Then frictional force,

$$\begin{gathered} {f_k}^{\text{'}}={\mu }_k{N}^{\text{'}} \\ ={\mu }_k{m}_{3}g \\ {F}_{32}-{f_k}^{\text{'}}=m_{3}a \\ {F}_{32}=m_{3}a+{f_k}^{\text{'}} \\ =(20.0)(0.47)+(0.700)(20.0)(9.81) \\ =147N \end{gathered}$$

Hence, the magnitude ${}^{F_{32}}$of the force on crate 3 from crate 2 is ${}^{147\mathrm{N}}$

As from the above,

$$\begin{gathered} F_{32}=m_{3}a+{f_k}^{\text{'}} \\ =m_3\left(\left(\frac{F-{\mu }_k(m_1+m_2+m_3)g}{(m_1+m_2+m_3)}\right)+{\mu }_{k}g\right) \\ =m_3\left(\frac{F-{\mu }_k(m_1+m_2+m_3)g+{\mu }_k(m_1+m_2+m_3)g}{(m_1+m_2+m_3)}\right) \\ =m_3\left(\frac{F}{(m_1+m_2+m_3}\right) \end{gathered}$$

From the above expression of $F_{32}$it can be said that it does not depend on the kinetic friction between the crate and the table.

Thus, the magnitude of $F_{32}$ remains the same.