Chapter 1, Problem 55P

The weight of bodies may change somewhat from one location to another as a result of the variation of the gravitational acceleration g with elevation. Accounting for this variation using the relation in Prob. 1-31, determine the weight of an 65-kg person at sea level (z = 0), in Denver (z = 1610 m), and on the top of Mount Everest (z = 8848 m).

To determine

The weight of an $65\text{kg}$ person at sea level.

The weight of an $65\text{kg}$ person in Denver.

The weight of an $65\text{kg}$ person on the top of Mount Everest.

Answer to Problem 55P

The weight of an $65\text{kg}$ person at sea level is $637.455\text{N}$.

The weight of an $65\text{kg}$ person in Denver is $637.065\text{N}$.

The weight of an $65\text{kg}$ person on the top of Mount Everest is $635.505\text{N}$.

Explanation of Solution

Given information:

The mass of a person is $65\text{kg}$.

The gravitation constant function is given as.

  $g=a-bz$  .......(I)

Here, elevation above sea level is $z$, constant $a$ is $9.807\text{m}/{\text{s}}^{\text{2}}$, and constant $b$ is $3.32\times {10}^{-6}{\text{s}}^{\text{-2}}$.

For sea level $z$ is $0\text{m}$, in Denver $z$ is $1610\text{m}$, and on the top of Mount Everest $z$ is $8848\text{m}$.

Write the expression for the weight of the person.

  $W=mg$  .......(II)

Here, weight of the person is $W$, mass of the person is $m$, and gravitation constant is $g$.

Calculation:

Substitute $9.807\text{m}/{\text{s}}^{\text{2}}$ for $a$, $3.32\times {10}^{-6}{\text{s}}^{\text{-2}}$ for $b$, and $0\text{m}$ for $z$ in Equation (I).

  $\begin{array}{c}{g}_{atsealevel}=\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)-\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)\left(0\text{m}\right)\\ =9.807\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $9.807\text{m}/{\text{s}}^{\text{2}}$ for $a$, $3.32\times {10}^{-6}{\text{s}}^{\text{-2}}$ for $b$, and $1610\text{m}$ for $z$ in Equation (I).

  $\begin{array}{c}{g}_{inDenver}=\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)-\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)\left(1610\text{m}\right)\\ =9.801\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $9.807\text{m}/{\text{s}}^{\text{2}}$ for $a$, $3.32\times {10}^{-6}{\text{s}}^{\text{-2}}$ for $b$, and $8848\text{m}$ for $z$ in Equation (I).

  $\begin{array}{c}{g}_{ontopofMountEverest}=\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)-\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)\left(8848\text{m}\right)\\ =9.777\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $65\text{kg}$ for $m$, and $9.807\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (II).

  $\begin{array}{c}{W}_{atsealevel}=\left(65\text{kg}\right)\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(637.455\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =637.455\text{N}\end{array}$

Substitute $65\text{kg}$ for $m$, and $9.801\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (II).

  $\begin{array}{c}{W}_{atsealevel}=\left(65\text{kg}\right)\left(9.801\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(637.065\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =637.065\text{N}\end{array}$

Substitute $65\text{kg}$ for $m$, and $9.777\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (II).

  $\begin{array}{c}{W}_{atsealevel}=\left(65\text{kg}\right)\left(9.777\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(635.505\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =635.505\text{N}\end{array}$

Conclusion:

The weights are calculated according to expression W= mg and the elevation at sea level and the constant function at gravitation.