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that is, if a denote acceleration in C.G.S. units, a =

hence, when a and m are each unity, F will be unity.

F

m

Again, by the nature of uniform acceleration, we have ย -- at, v denoting the velocity due to the acceleration a, continuing for time t.

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As a particular case, if m = 1, v = 1, t = 1, we have F = 1.

25. The force represented by the weight of a gramme varies from place to place. It is the force required to sustain a gramme in vacuo, and would be nil at the earth's centre, where gravity is nil. To compute its amount in dynes at any place where g is known, observe that a mass of 1 gramme falls in vacuo with acceleration g. The force producing this acceleration (namely, the weight of the gramme) must be equal to the product of the mass and acceleration, that is, to g.

The weight (when weight means force) of 1 gramme is therefore g dynes; and the weight of m grammes is mg dynes.

26. Force is said to be expressed in gravitation-measure when it is expressed as equal to the weight of a given mass. Such specification is inexact unless the value of g is also given. For purposes of accuracy it must always be remembered that the pound, the gramme, &c., are, strictly speaking, units of mass. Such an expression as "a force of 100 tons" must be understood as an abbrevia

tion for "a force equal to the weight [at the locality in question] of 100 tons."

27. The name poundal has recently been given to the unit force based on the pound, foot, and second; that is, the force which, acting on a pound for a second, gene

I

rates a velocity of a foot per second. It is of the

g

weight of a pound, g denoting the acceleration due to gravity expressed in foot-second units, which is about 32 2 in Great Britain.

To compare the poundal with the dyne, let x denote the number of dynes in a poundal; then we have

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28. The C.G.S. unit of work is called the erg. It is the amount of work done by a dyne working through a distance of a centimetre.

The C.G.S. unit of energy is also the erg, energy being measured by the amount of work which it represents.

29. To establish a rule for computing the kinetic energy (or energy due to the motion) of a given mass moving with a given velocity, it is sufficient to consider the case of a body falling in vacuo.

When a body of m grammes falls through a height of h centimetres, the working force is the weight of the body -that is, gm dynes, which, multiplied by the distance worked through, gives gmh ergs as the work done.

But

the velocity acquired is such that 22gh. Hence we have gmh=mv2.

The kinetic energy of a mass of m grammes moving with a velocity of v centimetres per second is therefore m2 ergs; that is to say, this is the amount of work which would be required to generate the motion of the body, or is the amount of work which the body would do against opposing forces before it would come

to rest.

measure.

30. Work, like force, is often expressed in gravitationGravitation units of work, such as the footpound and kilogramme-metre, vary with locality, being proportional to the value of g

One gramme-centimetre is equal to g ergs.

One kilogramme-metre is equal to 100,000 g ergs.

=

One foot-poundal is 453'59 × (30'4797) 421390 ergs.

One foot-pound is 13,825 g ergs, which, if g be taken as 981, is 1.356 × 107 ergs.

31. The unit rate of working is I erg per second. Watt's "horse-power" is defined as 550 foot-pounds per second. This is 7.46 × 109 ergs per second. The "force de cheval" is defined as 75 kilogrammetres per second. This is 7.36 × 109 ergs per second. We here assume g=981.

Examples.

1. If a spring balance is graduated so as to show the masses of bodies in pounds or grammes when used at the equator, what will be its error when used at the poles, neglecting effects of temperature?

Ans. Its indications will be too high by about the total weight.

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2. A cannon-ball, of 10,000 grammes, is discharged with a velocity of 45,000 centims. per sec. Find its kinetic energy.

Ans. × 10000 × (45000)2 = 1.0125 × 1013 ergs.

3. In last question find the mean force exerted upon the ball by the powder, the length of the barrel being 200 centims.

Ans. 5'0625 x 101o dynes.

4. Given that 42 million ergs are equivalent to I gramme-degree of heat, and that a gramme of lead at 10° C. requires 15.6 gramme-degrees of heat to melt it; find the velocity with which a leaden bullet must strike a target that it may just be melted by the collision, supposing all the mechanical energy of the motion to be converted into heat and to be taken up by the bullet.

We have 1215'6 × J, where J = 42 × 106. Hence 721310 millions; 362 thousand centims. per v = second.

=

5. With what velocity must a stone be thrown vertically upwards at a place where g is 981 that it may rise to a height of 3000 centims. ? and to what height would it ascend if projected vertically with this velocity at the surface of the moon, where g is 150?

Ans. 2426 centims. per second; 19620 centims.

Centrifugal Force.

32. A body moving in a curve must be regarded as continually falling away from a tangent. The accelera

tion with which it falls away is

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r

v denoting its velocity

and the radius of curvature. The acceleration of a body in any direction is always due to force urging it in that direction, this force being equal to the product of mass and acceleration. Hence the normal force on a body of m grammes moving in a curve of radius r centimetres, with velocity v centimetres per second, is mv2

r

dynes. This force is directed towards the centre of curvature. The equal and opposite force with which the body reacts is called centrifugal force.

If the body moves uniformly in a circle, the time of revolution being T seconds, we have v =

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2π1

Ti

r, and the force acting on the body is

If n revolutions are made per minute, the value of T is

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1. A body of m grammes moves uniformly in a circle of radius 80 centims., the time of revolution being second. Find the centrifugal force, and compare it with

the weight of the body.

Ans. The centrifugal force is m x

× 80=50532 m dynes.

of a

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The weight of the body (at a place where g is 981) is 981 m dynes. Hence the centrifugal force is about 52 times the weight of the body.

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