Chapter 7, Problem 9P

To determine

The force of wind on person.

To compare: The maximum force of friction between person and ground to the force exerted by wind on person.

Answer to Problem 9P

The force of wind acting on person is $F_{wind}=833.4\text{ N }$

The force acting on person due to wind greater than the force of friction.

Explanation of Solution

Given:

Mass of person, $M=70\text{ kg}$

Speed of wind, $v_{wind}=100\text{ km/h}$

The rate of striking of air on person, $R=40\text{ kg/s}$

Formula used:

  $F=\frac{\Delta p}{t}$

Where, $\Delta p$ = change in momentum

  $t$ = time interval

  $F$ = force

Calculation:

The wind strikes a person at a rate,

  $\begin{array}{l}\frac{m_{wind}}{t}=\left(40\right)\times \left(1.5\right)\times \left(0.5\right)\\ \frac{m_{wind}}{t}=30\text{ kg/s}\end{array}$

The speed of wind in m/s is,

  $v_{wind}=\left(100\right)\left(\frac{5}{18}\right)\text{ m/s}$

The force experienced by the person due to motion of wind is,

  $\begin{array}{l}{F}_{wind}=\left(\frac{m_{wind}}{t}\right)v_{wind}\\ F_{wind}=\left(30\text{ kg}\text{.s}\right)\left(100\times \frac{5}{18}\text{ m}\text{.s}\right)\\ F_{wind}=\left(30\right)\left(27.78\right)\\ F_{wind}=833.4\text{ N (1)}\end{array}$

The maximum friction force acting on the person is,

  $\begin{array}{l}f={\mu }_{s}Mg\\ f=\left(1\right)\left(70\right)9.8\\ f=686\text{ N (2)}\end{array}$

From, equation (1) and (2),

  $F_{wind}>f$