WHAT IS THE WEIGHT OF A BODY? 63

sprinkle some iron filings over the paper. What happens to thefilings? Where are the iron filings concentrated? Draw the pat-tern made by the filings.Explanation:The iron filings will arrange themselves along the lines of force

of the electromagnet. The pattern will be similar to that of apermanent magnet.

12. Materials needed:apparatus shown in Fig. 3

Connect the apparatus to two or more dry cells. What happensto the suspended bar magnet when the pushbutton is presseddown? Why does it stay in a certain position? Can you make thebar magnet turn?Explanation:This experiment could be called a synthesis of the preceding

experiments. It is a simple demonstration of the electric motorprinciple. WTien the pushbutton is pressed down, the electric cir-cuit through the electromagnet is completed. The magnetic fieldof the electromagnet will repel and attract that of the bar magnet,causing the bar magnet to align itself according to the polarityof the electromagnet. If the pushbutton is pressed and releasedat the proper interval, the bar magnet can be made to turn.

WHAT IS THE WEIGHT OF A BODY?

JULIUS SUMNER MILLER2116 Benecia Ave., West Los Angeles 25, Calif.

It seems to be acceptable practice for textbook writers and teachersto define the weight of a body as the force of gravitational attractionbetween the body and the earth, that is, a force mg, which is theearth’s attraction upon a mass m, directed vertically downward.This is not exactly true. Let us show it.Let m be a mass on the earth’s surface at a latitude 0. The total

gravitational pull of the earth on m is q, say, and this is a force directedapproximately to the earth’s center. If, now, the earth did not rotatethis mass would be in equilibrium under the action of two forces:

(1) the force q (=mgo) directed toward 0.(2) The push of the earth, which is equal and opposite to this, say T7o.

But the earth does rotate, and m revolves daily on a circle of radiusr at latitude 6. It (the mass) has, therefore, an acceleration toward P;it is pulled toward the center of that circle by a suitable force, ofmagnitude mi^/r. Where does this force come from? The answer to this

64 SCHOOL SCIENCE AND MATHEMATICS

is the crux of our inquiry. The answer is: it has to be supplied byq. q is thereby diminished, and what is left of q, after mv^/r is pro-vided for, is the force W. This W is the weight of m. It is the forcethat gives rise to the acceleration g, and we come to W=mg. Thusthe weight of a body is not the total gravitational pull of the earthon it; it is that pull diminished by the force required to keep it in itscircular path. At the poles, naturally, W and q are identical.

Let’s look at it another way: The force q{=mg)o is unaltered by theearth’s rotation, but Wo (the negative of it) is changed. How is itchanged? It must now be such in magnitude and direction that, whenadded vectorially to mgo it will provide mv^/r, toward P. It wouldthen be a force F, say. And the opposite of F is W, the true mg of m.And its direction is not toward 0.

It is left as an exercise to show that analytically we may write

g^gQ�^R cos2 6, very nearly.

The diminution in g is obviously greatest at the equator (6=0).In this equation, co is the angular velocity of the earth, and R the

radius. Approximately, R= 6.4 X 106 meters,

co = 1 rev/day= 7.3 X 10-5 rad/sec.

At the equator, therefore, W and q differ by about ^%.

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