Chapter 6, Problem 80A

(a)

To determine

The speed of a 97 kg person at the equator.

Answer to Problem 80A

The speed of a $97\text{ kg}$ person at the equator is $464\text{ m/s}$ .

Explanation of Solution

Given:

When the earth rotates an object (man) on the equator undergoes distance equal to the circumference of earth equator 24 hours.

Formula used:

  $v=\frac{d}{t}$

v is the velocity

dis the Distance

t is the time

Calculation:

  $v=\frac{d}{t}$

Substituting the values.

  $\begin{array}{c}v =\frac{2\pi \left(6.38\times {10}^6\right)}{24\times 60\times 60}\\ v=464\text{ m/s}\end{array}$

This is speed of any object on the earth’s equator due to the earth's rotation.

Conclusion:

The speed of a $97\text{ kg}$ person at the equator is $464\text{ m/s}$ .

(b)

To determine

The force needed to accelerate the person in the circle.

Answer to Problem 80A

  $\text{3}\text{.3 N}$

Explanation of Solution

Given:

The speed of a $97\text{ kg}$ person at the equator is $465.4\text{ m/s}$ .

Formula Used:

  $F=\frac{m{v}^2}{r}$

F is the centripetal force

mis the mass

vis the velocity

r is the radius.

Calculation:

The centripetal force needed to accelerate the body is given by

  $F=\frac{m{v}^2}{r}$

Substituting the values

  $F=\frac{\left(\text{97 kg}\right){\left(\text{464 m/s}\right)}^2}{\text{6}{\text{.38×10}}^{\text{6}}\text{ m}}$

  $F=3.3\text{}\text{ N}$

Conclusion:

The force needed to accelerate the person in the circle $F=3.3\text{}\text{ N}$ .

(c)

To determine

The weight of the person.

Answer to Problem 80A

The weight of the person is $F=9.5\times {10}^2\text{ N}$

Explanation of Solution

Given:

Mass $m=\text{97 kg}$

Acceleration due to gravity $g=9.8{\text{ m/s}}^{-2}$ .

Formula used:

Weight of the person

  $F=mg$

Where F is the force

m is the mass

g is the Acceleration due to gravity

Calculation:

  $F=mg$

Substituting the values

  $F=\left(\text{97 kg}\right)\left(9.8{\text{ m/s}}^{-2}\right)$

  $F=9.5\times {10}^2\text{ N}$

Conclusion:

The weight of the person is $F=9.5\times {10}^2\text{ N}$

(d)

To determine

The normal force of the earth on the person

Answer to Problem 80A

  $F_N=947.3\text{ N}$

Explanation of Solution

Given:

The weight of the person is $F=9.5\times {10}^2\text{ N}$

The centrifugal force is $F=3.3\text{}\text{ N}$

Formula used:

Normal weight:

  $F_N=w-F_C$

Calculation:

Substituting the values

  $F_N=950.6\text{}\text{ N-3}\text{.3 N}$

  $F_N=947.3\text{ N}$

Conclusion:

The normal force of the earth on the person that is the person’s apparent weight is $F_N=947.3\text{ N}$ .