Chapter 8.1, Problem 8.13P

Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C, the coefficients of friction are μ_s = 0.30 and μ_k = 0.20; between package B and the belt, the coefficients are μ_s = 0.10 and μ_k = 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.

Fig. P8.13

To determine

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

Answer to Problem 8.13P

The package C will $\underset{\_}{not\text{move}}$.

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

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

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

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

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 C:

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

Resolve the vertical component of forces.

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

Resolve the horizontal component of forces.

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

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

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

Substitute 0.30 for ${\left({\mu }_s\right)}_C$ and 37.9 N for $N_C$.

$\begin{array}{c}{\left(F_m\right)}_C=0.30\times 37.9\\ =11.37\text{N}\end{array}$

The maximum friction force is greater than the friction force.

${\left(F_m\right)}_C=11.37\text{N}>F_C=10.16\text{N}$

Therefore, the package C will $\underset{\_}{not\text{move}}$.

Therefore, the friction force in the package C is $\underset{\_}{F_C=10.16\text{N}(\nearrow )}$.

Consider Block B:

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

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 B will $\underset{\_}{\text{move}}$.

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 $N_B$.

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

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

Consider Block A and B together:

Show the free body diagram of the block A and B as in Figure 3.

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

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

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

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

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

$\begin{array}{c}{N}_A+N_B=37.9+37.9\\ =75.8\text{N}(\nwarrow )\end{array}$

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

$\begin{array}{c}{F}_A+F_B=10.16+10.16\\ =20.32\text{N}(\nearrow )\end{array}$

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

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

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

$\begin{array}{c}{\left(F_m\right)}_{A+B}={\left(F_m\right)}_A+{\left(F_m\right)}_B\\ =11.37+3.79\\ =15.16\text{N}\end{array}$

The maximum friction force is less than the friction force.

${\left(F_m\right)}_{A+B}=15.16\text{N}<F_A+F_B=20.32\text{N}$

Therefore, the packages A 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 $N_A$.

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

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