Chapter 5, Problem 19P

To determine

ToDetermine:

1. The normal force exerted by the road on the car.
2. The normal force exerted by the car on the $\text{62}\text{.0}\text{kg}$ driven.
3. The car speed at which the normal force on the driver equals zero.

Answer to Problem 19P

Solution:

1. The normal force exerted by road on the car $=\text{5965}\text{.23}\text{N}$ .
2. the normal force exerted by a car on the driver $=\text{379}\text{.33}\text{N}$ .
3. the speed of the car for which normal force on the driveris $0=29.37\text{m/s}$ .

Explanation of Solution

Given:

Mass of sports car (including the driver), $m=975\text{kg}$

Radius rounded top of a hill, $r=88.0\text{m}$

Speed of car, $v=18.0\text{m/s}$

Formula used:

Centripetal acceleration $=\frac{v^2}{r}$

(Actingtowards the center)

Newton’s second law of motion.

  ${\displaystyle \sum f}=ma$

Calculation:

  

  $\begin{array}{l}{\displaystyle \sum f}y=m{a}_c\left(\text{Newton's second law}\right)\\ 975g-N=975\left(\frac{v^2}{r}\right)\\ 975g-N=975\left(\frac{{18}^2}{88.0}\right)\\ N=975\times 9.8-975\left(\frac{{18}^2}{88.0}\right)\\ N=5965.23\text{N}\end{array}$

Hence normal force exerted by road on the spot car $=596.23\text{N}$ .

(b)

  

Apply Newton’s second law.

  $\begin{array}{l}62.0g-N=62\left(\frac{v^2}{r}\right)\\ 62\times 9.8-N=62\left(\frac{{18}^2}{88.0}\right)\\ N=62\times 9.8-\frac{62\times {18}^2}{88.0}\\ N=379.33\text{N}\end{array}$

Normal force exerted on man by sports car $=379.33\text{N}$ .

(c)

  

The speed of car and driver are same because both move together.

Given N=0

So, apply Newton's second law.

  $\begin{array}{l}62.0g=62\frac{v^2}{r}\\ v=\sqrt{rg}\\ r=88.0\text{m}\\ g=9.80\\ \text{ =}\sqrt{88\times 9.80}\\ \text{ =29}\text{.37m/s}\end{array}$

The speed of the car for which normal reaction on man or car is zero $=29.37\text{m/s}$ .

When a car crosses the round top of a hill, then the normal reaction is always less than the weight of car because resultant of the weight of the car and normal reaction provides centripetal acceleration. If speed is further increased at a particular speed, normal reaction becomes zero.