The value of g at any latitude C may be obtained

The value of g at any latitude θ may be obtained from the formula $g=32.09(1+0.0053\sin^2\theta )$ ft/$s^2$ which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly spherical. Knowing that the weight of a silver bar has been officially designated as 5 lb, determine to four significant figures (a) the mass in slugs, (b) the weight in pounds at the latitudes of 0°, 45°, and 60°.

Solution:

a) A slug = $1\,\text{lb}\cdot \text{s}^2/\text{ft}$

Use the following equation (Newton’s second law):

$W=mg$

(Where $W$ is weight, $m$ is mass, and $g$ is acceleration due to gravity)

$m=\dfrac{W}{g}$

$m=\dfrac{5}{32.2}$

$m=0.1553\,\text{lb}\cdot \text{s}^2/\text{ft}$

b) Use the equation given to us to figure out the acceleration of gravity at the specific angles.

At $\theta=0^0$;

$g=32.09(1+0.0053\sin^2(0^0))$

$g=32.09$ ft/$s^2$

Using our previous equation, we can figure out the weight.

$W=mg$

$W=(0.1553)(32.09)=4.9836$ lb

At $\theta=45^0$;

$g=32.09(1+0.0053\sin^2(45^0))$

$g=32.175$ ft/$s^2$

$W=mg$

$W=(0.1553)(32.175)=4.9968$ lb

At $\theta=60^0$;

$g=32.09(1+0.0053\sin^2(60^0))$

$g=32.22$ ft/$s^2$

$W=mg$

$W=(0.1553)(32.22)=5.004$ lb