Chapter 10, Problem 21P

(a)

To determine

The force that needs to be applied in the horizontal direction.

Explanation of Solution

Given:

Weight of the block is 150 N.

Volume of block $\left(V\right)$ is $50\text{ cm}\times 30\text{ cm}\times 20\text{ cm}$.

Velocity of the block $\left(V\right)$ is $0.8\text{m}/\text{s}$.

Friction coefficient $\left(f\right)$ is 0.27.

Calculation:

Draw the free body diagram as in Figure (1).

Apply horizontal force equilibrium condition.

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_1-F_f\cos 20°-F_{N1}\sin 20°=0\end{array}$

Apply vertical force equilibrium condition.

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{N1}\cos 20°-F_f\sin 20°-W=0\end{array}$

Determine the friction force.

  $F_f=f{F}_{N1}$

From the above expressions;

  $\begin{array}{c}{F}_{N1}=\frac{W}{\cos 20°-f\sin 20°}\\ =\frac{150\text{ N}}{\cos 20°-0.27\sin 20°}\\ =177\text{ N}\end{array}$

Determine the force that needs to be applied in the horizontal direction.

  $\begin{array}{c}{F}_1=F_f\cos 20°+F_{N1}\sin 20°\\ =\left(0.27\times 177\text{ N}\right)\cos 20°+\left(177\text{ N}\right)\sin 20°\\ =105.5\text{ N}\end{array}$

Thus, the force that needs to be applied in the horizontal direction is $\underset{\_}{177\text{ N}}$.

(b)

To determine

The percentage reduction in the required force.

Explanation of Solution

Given:

Thickness of the oil film $\left(h\right)$ is 0.4 mm.

Dynamic viscosity $\left(\mu \right)$ is $0.012\text{N}\cdot \text{s}/{\text{m}}^2$.

Calculation:

Draw the free body diagram as in Figure (2).

Determine the shear force.

  $\begin{array}{c}{F}_{\text{shear}}={\tau }_w{A}_s\\ =\mu A_s\frac{V}{h}\\ =\left(0.012\text{N}\cdot \text{s}/{\text{m}}^2\right)\left(0.5\text{ m}\times 0.2\text{ m}\right)\frac{0.8\text{m}/\text{s}}{4\times {10}^{-4}\text{m}}\\ =2.4\text{ N}\end{array}$

Apply horizontal force equilibrium condition.

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_2-F_{\text{shear}}\cos 20°-F_{N2}\sin 20°=0\end{array}$

Apply vertical force equilibrium condition.

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{N2}\cos 20°-F_{\text{shear }}\sin 20°-W=0\end{array}$

  $\begin{array}{c}{F}_{N2}=\frac{\left(F_{\text{shear}}\sin 20°+W\right)}{\cos 20°}\\ =\frac{2.4\text{ N}\times \sin 20°+150\text{ N}}{\cos 20°}\\ =160.5\text{ N}\end{array}$

Determine the force that needs to be applied in the horizontal direction.

  $\begin{array}{c}{F}_2=F_{\text{shear}}\cos 20°+F_{N2}\sin 20°\\ =\left(2.4\text{ N}\right)\cos 20°+\left(160.5\text{ N}\right)\sin 20°\\ =57.2\text{ N}\end{array}$

Determine the percentage reduction in the required force.

  $\begin{array}{c}\text{Percentage reduction in required force}=\frac{F_1-F_2}{F_1}\times 100\%\\ =\frac{105.5-57.2}{105.5}\times 100\%\\ =45.8\%\end{array}$

Thus, the percentage reduction in the required force is $\underset{\_}{45.8\%}$.