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

 

This question can be found in Vector Mechanics for Engineers: Statics and Dynamics, 11th edition, chapter 12, question 12.2.

Leave a comment

Your email address will not be published.