Chapter 5.3, Problem 40PP

(a)

To determine

To Find: The average value of $\varphi$ .

Answer to Problem 40PP

  $\varphi ={19.2}^{\text{o}}$

Explanation of Solution

Given:

Value of $\varphi$ measured in 5 trials: ${\text{21}}^{\text{o}},{17}^{\text{o}},{21}^{\text{o}},{18}^{\text{o}},{19}^{\text{o}}$

Formula used:

  $\text{average = }\frac{\text{sum of observations}}{\text{number of observations}}$

Calculation:

  $\begin{array}{l}\varphi \text{ = }\frac{{\text{21}}^{\text{o}}+{17}^{\text{o}}+{21}^{\text{o}}+{18}^{\text{o}}+{19}^{\text{o}}}{5}\\ \varphi \text{= 19}{\text{.2}}^{\text{o}}\end{array}$

Conclusion:

Thus, the average value is: $\varphi ={19.2}^{\text{o}}$

(b)

To determine

To Find: The value of coefficient of static friction.

Answer to Problem 40PP

  ${\mu }_{\text{s}}=0.35$

Explanation of Solution

Given:

Value of $\varphi$ measured in 5 trials: ${\text{21}}^{\text{o}},{17}^{\text{o}},{21}^{\text{o}},{18}^{\text{o}},{19}^{\text{o}}$

Formula used:

The coefficient of static friction is:

  ${\mu }_{\text{s}}=\tan \varphi$

Calculation:

Use the average value of $\varphi ={19.2}^{\text{o}}$ from part (a).

  $\begin{array}{l}{\mu }_{\text{s}}=\tan ({19.2}^{\text{o}})\\ {\mu }_{\text{s}}=0.35\end{array}$

Conclusion:

Thus, the coefficient of static friction is 0.35.

(c)

To determine

The value of coefficient of kinetic friction.

Answer to Problem 40PP

  ${\mu }_{\text{k}}=0.14$

Explanation of Solution

Given:

Acceleration of the top book, $a=\text{1}{\text{.3 m/s}}^2$

Formula used:

The force of kinetic friction:

  $F_k={\mu }_{k}mg\cos \varphi$

Here, ${\mu }_k$ is the coefficient of kinetic friction, m is the mass and g is acceleration due to gravity.

Calculation:

  $\begin{array}{l}{F}_k={\mu }_{k}mg\cos \varphi \\ ma={\mu }_{k}mg\cos \varphi \\ {\mu }_k=\frac{a}{g\cos \varphi }\\ {\mu }_{\text{k}}=\frac{1.3}{9.8\times \cos (19.2)}\\ {\mu }_{\text{k}}=0.14\end{array}$

Conclusion:

The coefficient of kinetic friction is 0.14.