Chapter 6, Problem 6.47AP

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 20.0° with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of staticfriction exerted by the carousel on the bag. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 34.0 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 6.47AP

The force of static friction exerted by the carousel on the bag is $\text{106}\text{N}$ up the incline.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0\text{kg}$. The luggage is placed at the position $7.46\text{m}$ measured horizontally from the axis of rotation. The travel bag goes around once in $38.0\text{s}$.

Formula to calculate the centripetal force is,

$F_{\text{c}}=\frac{m{v}^2}{R}$

The horizontal force acting on the bag in horizontal direction is,

$m{a}_{\text{c}}=\frac{m{v}^2}{R}$

Substitute $\frac{2\pi R}{t}$ for $v$ in the above equation.

$\begin{array}{c}m{a}_{\text{c}}=\frac{m{\left(\frac{2\pi R}{t}\right)}^2}{R}\\ m{a}_{\text{c}}=\frac{4{\pi }^{2}m{R}^2}{R{t}^2}\\ m{a}_{\text{c}}=\frac{4{\pi }^{2}mR}{t^2}\end{array}$ . (1)

Here,

$F_{\text{c}}$ is the centripetal acceleration.

$R$ is the radius of the carousel.

$m$ is the mass of the luggage.

$t$ is the time.

Substitute $30.0\text{kg}$ for $m$, $7.46\text{m}$ for $R$, and $38.0\text{s}$ for $t$ in the above equation.

$\begin{array}{c}m{a}_{\text{c}}=\frac{4{\pi }^2\left(30.0\text{kg}\right)\left(7.46\text{m}\right)}{{\left(38.0\text{s}\right)}^2}\\ m{a}_{\text{c}}=6.118\text{N}\\ m{a}_{\text{c}}\approx 6.12\text{N}\end{array}$

Draw the free body diagram for the luggage.

From the Figure (1), the component of force in $x$ direction is,

$\begin{array}{c}{\displaystyle \sum F_{\text{x}}}=m{a}_{\text{c}}\\ F_{\text{s}}\cos \theta -N\sin \theta =m{a}_{\text{c}}\\ N=\frac{F_{\text{s}}\cos \theta -m{a}_{\text{c}}}{\sin \theta }\end{array}$ . (2)

From the Figure (1), the component of force in $y$ direction is,

$\begin{array}{c}{\displaystyle \sum F_{\text{y}}}=m{a}_{\text{y}}\\ F_{\text{s}}\sin \theta +N\cos \theta -mg=m{a}_{\text{y}}\end{array}$

Substitute $0\text{m}/{\text{s}}^{\text{2}}$ for $a_{\text{y}}$, $\frac{F_{\text{s}}\cos \theta -m{a}_{\text{c}}}{\sin \theta }$ for $N$ in the above equation.

$\begin{array}{c}{F}_{\text{s}}\sin \theta +\left(\frac{F_{\text{s}}\cos \theta -m{a}_{\text{c}}}{\sin \theta }\right)\cos \theta -mg=m\left(0\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{s}}\sin \theta +\left(F_{\text{s}}\frac{{\cos }^2\theta }{\sin \theta }\right)-m{a}_{\text{c}}\left(\frac{\cos \theta }{\sin \theta }\right)-mg=0\\ F_{\text{s}}\left[\sin \theta +\left(\frac{{\cos }^2\theta }{\sin \theta }\right)\right]=m{a}_{\text{c}}\left(\cot \theta \right)+mg\\ F_{\text{s}}\left[\left(\frac{1}{\sin \theta }\right)\right]=m{a}_{\text{c}}\left(\cot \theta \right)+mg\end{array}$

Simplify the above equation.

$F_{\text{s}}=\left[m{a}_{\text{c}}\left(\cot \theta \right)-mg\right]\sin \theta$ (3)

Substitute $30.0\text{kg}$ for $m$, $6.12\text{N}$ for $m{a}_{\text{c}}$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the equation (3).

$\begin{array}{c}{F}_{\text{s}}=\left[\left(6.12\text{N}\right)\cot 20°+\left(30.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\right]\sin 20°\\ =105.73\text{N}\\ \approx \text{106}\text{N}\end{array}$

Conclusion:

Therefore, the force of static friction exerted by the carousel on the bag is $\text{106}\text{N}$ up the incline.

To determine

The coefficient of friction between the bag and the carousel.

Answer to Problem 6.47AP

The coefficient of friction between the bag and the carousel is $0.396$.

Explanation of Solution

Given info: The metallic surface of a luggage carousel slopes downward toward outside at an angle $20.0°$ with the horizontal. The mass of luggage is $30.0\text{kg}$. The luggage is placed at the position $7.94\text{m}$ measured horizontally from the axis of rotation. The travel bag goes around once in $34.0\text{s}$.

Substitute $30.0\text{kg}$ for $m$, $7.94\text{m}$ for $R$, and $34.0\text{s}$ for $t$ in the equation (1).

$\begin{array}{c}m{a}_{\text{c}}=\frac{4{\pi }^2\left(30.0\text{kg}\right)7.94\text{m}}{{\left(34.0\text{s}\right)}^2}\\ =8.13\text{N}\end{array}$

Calculate the centripetal force for the bag.

Substitute $30.0\text{kg}$ for $m$, $8.13\text{N}$ for $m{a}_{\text{c}}$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the equation (3).

$\begin{array}{c}{F}_{\text{s}}=\left[\left(8.13\text{N}\right)\cot 20°+\left(30.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\right]\sin 20°\\ =107.352\text{N}\\ \approx \text{108}\text{N}\end{array}$

Calculate the normal force for the bag when it is on the carousel.

Substitute $108\text{N}$ for $F_{\text{s}}$, $8.13\text{N}$ for $m{a}_{\text{c}}$ and $20.0°$ in the equation (2).

$\begin{array}{c}N=\frac{\left(108\text{N}\right)\cos 20.0°-8.13\text{N}}{\sin 20.0°}\\ =272.95\text{N}\\ \approx \text{273}\text{N}\end{array}$

Formula to calculate the coefficient of friction between the bag and the carousel is,

${\mu }_{\text{s}}=\frac{F_{\text{s}}}{N}$

Substitute $108\text{N}$ for $F_{\text{s}}$, $273\text{N}$ for $N$ in the above equation.

$\begin{array}{c}{\mu }_{\text{s}}=\frac{108\text{N}}{273\text{N}}\\ =0.3956\\ \approx 0.396\end{array}$

Conclusion:

Therefore, the coefficient of friction between the bag and the carousel is $0.396$.