Chapter 6, Problem 25PQ

An ice cube with a mass of 0.0507 kg is placed at the midpoint of a 1.00-m-long wooden board that is propped up at a 50° angle. The coefficient of kinetic friction between the ice and the wood is 0.133. a. How much time does it take for the ice cube to slide to the lower end of the board? b. If the ice cube is replaced with a 0.0507-kg wooden block, where the coefficient of kinetic friction between the block and the board is 0.275, at what angle should the board be placed so that the block takes the same amount of time to slide to the lower end as the ice cube does? You may find a spreadsheet program helpful in answering this question.

To determine

The time does the ice cube takes to slide to the lower end of the board.

Answer to Problem 25PQ

The time does the ice cube takes to slide to the lower end of the board is $0.387\text{s}$.

Explanation of Solution

Consider the axis is along the incline.

Free body diagram

Write the expression for the net force acting on the ice cube.

    $F_{\text{net}}=mg\sin \theta -{\mu }_{k}N$                                                                                (I)

Here, $F_{\text{net}}$ is the net force acting on the ice cube, $m$ is the mass of the ice cube, $g$ is the acceleration due to gravity, $\theta$ is the incline angle, ${\mu }_k$ is the coefficient of the kinetic friction, and $N$ is the normal force.

Write the expression for the normal force.

    $N=mg\cos \theta$                                                                                           (II)

Write the expression for the net force in terms of Newton’s law.

    $F_{\text{net}}=ma$                                                                                                (III)

Here, $a$ is the acceleration.

Use (II) and (III) to rewrite (I).

    $\begin{array}{c}ma=mg\sin \theta -{\mu }_{k}mg\cos \theta \\ a=g\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$                                                                       (IV)

Write the expression for the time taken to slide down the lower end.

    $t={\left(\frac{2S}{a}\right)}^{1/2}$ (V)

Here, $t$ is the time taken to slide down the lower end and $S$ is the distance travelled.

Use equation (IV) to rewrite (V).

    $t={\left(\frac{2S}{g\left(\sin \theta -{\mu }_k\cos \theta \right)}\right)}^{1/2}$ (VI)

Conclusion:

Ice slides down from the mid-point of the board of length $1.00\text{m}$ ,

    $\begin{array}{c}S=\frac{1.00\text{m}}{2}\\ =0.500\text{m}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ , $0.500\text{m}$ for $S$ , $50°$ for $\theta$, and $0.133$ for ${\mu }_k$ in (VI) to calculate $t$.

    $\begin{array}{c}t={\left(\frac{2\left(0.500\text{m}\right)}{\left(9.81\text{m}/{\text{s}}^2\right)\left(\sin 50°-\left(0.133\right)\cos 50°\right)}\right)}^{1/2}\\ =0.387\text{s}\end{array}$

Therefore, the time does the ice cube takes to slide to the lower end of the board is $0.387\text{s}$.

To determine

The angle the board be placed so that the block takes the same amount of time to slide to the lower end as the ice cube does.

Answer to Problem 25PQ

The angle the board be placed so that the block takes the same amount of time to slide to the lower end as the ice cube does is $56.4°$ .

Explanation of Solution

Rewrite the equation (VI) to solve for $\theta$.

    $\left(\sin \theta -{{\mu }^{\prime }}_k\cos \theta \right)=\frac{2S}{g{t}^2}$ (VII)

Here, ${{\mu }^{\prime }}_k$ is the kinetic friction between wooden block and wooden board.

Conclusion:

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ , $0.500\text{m}$ for $S$ , $50°$ for $\theta$, and $0.275$ for ${{\mu }^{\prime }}_k$ in (VII) to calculate $t$.

    $\begin{array}{c}\left(\sin \theta -\left(0.275\right)\cos \theta \right)=\frac{2\left(0.500\text{m}\right)}{\left(9.81\text{m}/{\text{s}}^2\right){\left(0.387\text{s}\right)}^2}\\ \left(\sin \theta -\left(0.275\right)\cos \theta \right)=0.68063\end{array}$ (VIII)

Substitute different values for $\theta$ in (VIII) to make the left hand side of the equation (VIII) closer to its right hand side value.

Tabular column shows the result for various values of $\theta$

$\theta \left(\text{in degree}\right)$	Result
$56.1$	$0.676632$
$56.2$	$0.678003$
$56.3$	$0.679372$
$56.4$	$0.680739$
$56.5$	$0.682103$
$56.6$	$0.683466$
$56.7$	$0.684826$

From the above table for the angle $56.4°$ the result for Left hand side $0.680739$ is closer to $0.68063$ that is to the tenth of the degree.

Therefore, the angle the board be placed so that the block takes the same amount of time to slide to the lower end as the ice cube does is $56.4°$ .