A Treatise on Coal Mining by International Correspondence Schools

Author:International Correspondence Schools
Language: eng
Format: epub
Publisher: The Colliery engineer co .
Published: 1900-03-25T05:00:00+00:00

48 MECHANICS. § 16

Suppose that the body is revolved horizontally, so that the action of gravity upon it will always be the same. According to the first law of motion, a body put in motion tends to move in a straight line unless acted upon by some ^^ other force, causing a change in the direc-

j^""^^"^^^ ^ tion. When the body moves in a circle, / \ the force that causes it to move in a cir-

cle instead of a straight line is exactly equal to the tension of the string. If the string were cut, the pulling force that Fig. ftsr. draws it away from the straight line would

be removed, and the body would then ** fly off at a tangent ;" that is, it would move in a straight line tangent to the circle, as shown in Fig. G27.

1898. Since, according to the third law of motion, every action has an equal and opposite reaction, we call that force which acts as an equal and opposite force to the pull of the string the centrifugal force, and it acts away from the center of motion.

The other force, or the pull of the string towards the center, is called the centripetal force, and it acts towards the center of motion. It is evident that these two forces, acting in opposite directions, tend to pull the string apart, and, if the velocity be increased sufficiently, the string will break. It is also evident that no body can revolve without generating centrifugal force.

The value of the centrifugal force, expressed in pounds, of any revolving body is calculated by the following rule:

Rule. — TJic centrifugal force is equal to the continued

product of .OOOriJf.^ the 7ceight of the body in pounds^ the radius in feet (taken as the distance between the center of gravity of the body and the center about which it revolves)^ and the square of the number of revolutions per minute.

Let F = centrifugal force in pounds;

^F= weight of revolving body in pounds;

R = radius in feet of circle described by center of

gravity of revolving body; N = revolutions per minute of revolving body.

Then, F= .00034 W R N\ • (112.)

In calculating the centrifugal force of fly-wheels, it is the usual practice to consider the rim of the wheel only, and not take the arms and hub of the wheel into account. In this case, R would be taken as the distance between the center of the rim and the center of the shaft.

Example. —A crank-pin weighing 65 pounds revolves in a circle whose radius is 21 inches. The number of revolutions is 180. What is the centrifugal force set up by the pin ?

Solution. — 21 in. = 1^ ft. Using formula 112,

F= .00084 X 65 X If X 180« = 1.253.07 lb. Ans.

SPECIFIC GRAVITY.

1899. The specific gravity of a t>ody is the ratio between its weight and the weight of a like volume of water.

Since gases are so much lighter than water, it is usual to take the

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