Chapter 8.1, Problem 8.14P

To determine

Find whether any of the package moves and the friction force acting on each package.

Answer to Problem 8.14P

The packages A, C, and B will $\underset{\_}{\text{move}}$.

The friction force in the package C is $\underset{\_}{F_C=7.58\text{N}(\nearrow )}$.

The friction force in the package A is $\underset{\_}{F_A=7.58\text{N}(\nearrow )}$.

The friction force in the package B is $\underset{\_}{F_B=3.03\text{N}(\nearrow )}$.

Therefore,

Explanation of Solution

Given information:

The mass of the package A, B, and C is $m_A=m_B=m_C=4\text{kg}$.

The static coefficient of friction between packages A and C and the belt is

${\left({\mu }_s\right)}_A={\left({\mu }_s\right)}_C=0.30$.

The static coefficient of friction between package B and belt is ${\left({\mu }_s\right)}_B=0.10$.

The kinetic coefficient of friction between packages A and C and belt is ${\left({\mu }_k\right)}_A={\left({\mu }_k\right)}_C=0.20$.

The kinetic coefficient of friction between package B and belt is ${\left({\mu }_k\right)}_B=0.08$.

Calculation:

Consider the acceleration due to gravity as $g=9.81{\text{m/s}}^2$.

Consider Block B:

Show the free body diagram of the block B as in Figure 1.

Resolve the vertical component of forces.

$\begin{array}{l}+\nwarrow {\displaystyle \sum F_y=0}\\ N_B-m_{B}g\cos 15°=0\\ N_B-4\times 9.81\cos 15°=0\\ N_B=37.9\text{N}(\nwarrow )\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}\nearrow {\displaystyle \sum F_x=0}\\ F_B-m_{B}g\sin 15°=0\\ F_B-4\times 9.81\sin 15°=0\\ F_B=10.16\text{N}(\nearrow )\end{array}$

Find the maximum friction force ${\left(F_m\right)}_B$ using the relation.

${\left(F_m\right)}_B={\left({\mu }_s\right)}_B{N}_B$

Substitute 0.10 for ${\left({\mu }_s\right)}_B$ and 37.9 N for $N_B$.

$\begin{array}{c}{\left(F_m\right)}_B=0.10\times 37.9\\ =3.79\text{N}\end{array}$

The maximum friction force is less than the friction force.

${\left(F_m\right)}_B=3.79\text{N}<F_B=10.16\text{N}$

Therefore, the package C will move.

Consider Block A, B, and C together:

Show the free body diagram of the block A, B, and B as in Figure 2.

The normal force in package A is $N_A=37.9\text{N}(\nwarrow )$.

The normal force in package C is $N_C=37.9\text{N}(\nwarrow )$.

The normal force in package B is $N_B=37.9\text{N}(\nwarrow )$.

The friction force in package A is $F_A=10.16\text{N}(\nearrow )$.

The friction force in package C is $F_C=10.16\text{N}(\nearrow )$

The friction force in package B is $F_B=10.16\text{N}(\nearrow )$.

Find the total normal force in package A, B, and C as follows;

$\begin{array}{c}{N}_A+N_B+N_C=37.9+37.9+37.9\\ =113.7\text{N}(\nwarrow )\end{array}$

Find the total friction force in package A, B, and C as follows;

$\begin{array}{c}{F}_A+F_B+F_C=10.16+10.16+10.16\\ =30.48\text{N}(\nearrow )\end{array}$

The maximum friction force in package A is ${\left(F_m\right)}_A=11.37\text{N}$.

The maximum friction force in package C is ${\left(F_m\right)}_C=11.37\text{N}$.

Find the maximum friction force ${\left(F_m\right)}_B$ using the relation.

${\left(F_m\right)}_B={\left({\mu }_s\right)}_B{N}_B$

Substitute 0.10 for ${\left({\mu }_s\right)}_B$ and 37.9 N for $N_B$.

$\begin{array}{c}{\left(F_m\right)}_B=0.10\times 37.9\\ =3.79\text{N}\end{array}$

The maximum friction force in package B is ${\left(F_m\right)}_B=3.79\text{N}$.

Find the maximum friction force ${\left(F_m\right)}_{A+B+C}$ using the relation.

$\begin{array}{c}{\left(F_m\right)}_{A+B+C}={\left(F_m\right)}_A+{\left(F_m\right)}_B+{\left(F_m\right)}_C\\ =11.37+3.79+11.37\\ =26.53\text{N}\end{array}$

The maximum friction force is less than the friction force.

${\left(F_m\right)}_{A+B+C}=26.53\text{N}<F_A+F_B+F_C=30.48\text{N}$

Therefore, the packages A, C, and B will $\underset{\_}{\text{move}}$.

Find the friction force in the package A using the kinetic relation.

$F_A={\left({\mu }_k\right)}_A{N}_A$

Substitute 0.20 for ${\left({\mu }_k\right)}_A$ and 37.9 N for $N_A$.

$\begin{array}{c}{F}_A=0.20\times 37.9\\ =7.58\text{N}(\nearrow )\end{array}$

Find the friction force in the package B using the kinetic relation.

$F_B={\left({\mu }_k\right)}_B{N}_B$

Substitute 0.08 for ${\left({\mu }_k\right)}_B$ and 37.9 N for $N_B$.

$\begin{array}{c}{F}_B=0.08\times 37.9\\ =3.03\text{N}\end{array}$

Find the friction force in the package C using the kinetic relation.

$F_C={\left({\mu }_k\right)}_C{N}_C$

Substitute 0.20 for ${\left({\mu }_k\right)}_C$ and 37.9 N for $N_C$.

$\begin{array}{c}{F}_C=0.20\times 37.9\\ =7.58\text{N}(\nearrow )\end{array}$

Therefore, the friction force in the package A is $\underset{\_}{F_A=7.58\text{N}(\nearrow )}$.

Therefore, the friction force in the package B is $\underset{\_}{F_B=3.03\text{N}(\nearrow )}$.

Therefore, the friction force in the package C is $\underset{\_}{F_C=7.58\text{N}(\nearrow )}$.