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at rest, and if in motion, the mass continues to move with uniform velocity.

III. When the force applied to a certain mass is doubled or trebled, etc., then the acceleration produced is doubled or trebled etc.

IV. When the mass to which a certain force is applied is doubled, or trebled etc., then the acceleration is halved, or divided by three, etc.

V. Every force applied to a mass produces in that mass its proper acceleration in its own direction, independently of any other motion which the mass may have.

Thus, by III, Forces are measured by the number of celos which they respectively produce in 1 lb.

By IV, masses are measured by the inverse of the number of celos produced in them respectively by 1 poundal.

And by V, when two equal and opposite external forces act on a particle, they produce equal and opposite accelerations; in which case the particle continues at rest, or moving with constant velocity.

34. A force which is applied to a mass A by another mass B is said to be external to A.

35. Art. 31 gives a definition of force in terms of mass; in other words, Art. 31 states the lawt which connects the ideas of

force and mass. It will be found to combine in one statement Newton's first and second Laws of motion (See Chapter XI].

+ Considered as a definition of force Art. 31 assumes that we know what mass is. We must in fact accept the definition of the mass of a body as the quantity of

Suppose however we take our definition of force as the push which when applied to a steel spring compresses it.

We should in this case choose as our unit force, that force which applied to a certain spring compresses it in a certain manner.

The statement of Art. 31 would be arranged as a definition of mass as follows.

Mass is that which, when free from the application of any force, perseveres in its state of rest or of uniform velocity; and in which the application of force produces acceleration in the direction of the force, so that the mass of a body varies inversely as the acceleration produced in it by a given force, and varies as the force required to produce a given acceleration.

We should choose as our unit mass the mass in which the unit force produces unit acceleration.

matter in it.

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36. The meaning of these definitions of Force and Mass may be stated thus : The motion of a given mass is always that of uniform velocity in a straight line,except during any interval in which it is acted on by external force. When external force acts on a mass its motion is that of acceleration in the direction of the force, so long as the force is in action, but no longer. The acceleration is proportional to the force which produces it; so that when the force changes the acceleration changes; when the force ceases the acceleration ceases, that is, the velocity becomes uniform.

By definition, 1 poundal acting on 1 lb. produces in it i celo.

Therefore i poundal acting continuously for i second on 1 lb. adds I velo to its velocity. By definition 2 poundals acting on i lb. produce in it 2 celos.

and 2 poundals acting on 2 lbs. produce in it i celo.

Example i. A force of 3 poundals is applied continuously to the mass 1 lb. which at the beginning of a certain interval has 10 velos, the force having the same direction as the velos: find (i) when the mass will have 25 velos ; (ii) how far it will go in 12 seconds. By Art. 32, 3 poundals acting on i lb. produce in it 3 celos, 3 celos produce 15 velos in 5 seconds.

(i) The 1 lb. has 3 celos; the velocity of the 1 lb. at the middle of the 12 seconds is therefore (10+3x6) velos; that is, 28 velos; therefore in 12 secs. it passes over 28 x 12 ft., or, 336 ft. (ii)

Example ii. A particle of 4 lbs. having 256 velos is acted on continuously by a force of 32 poundals in the direction opposite to that of the velos; find (i) when and where it will have 224 velos, (ii) when and where it will have 224 velos in the opposite direction.

32 poundals acting on 4 lbs. produce in it 32, that is 8, celos.
Hence the particle has – 8 celos.
So that, if after x seconds the particle has 224 velos, we have

224=256 - 8x; whence, x=4. In these 4 seconds the particle passes over

4 times (256 – 2 x8) ft. or 960 feet.

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(i)

If after x seconds the particle has – 224 velos,

we have – 224 = 256 - 8x; whence, x=60. In these 60 seconds the particle passes over

(ii) 60 times (256 – 30 x 8) ft. or 960 ft. Thus, the particle after 4 seconds has 224 velos; it continues in the same direction until the action of the force, producing in it a negative acceleration, brings it to rest ; then the action of the force gradually produces in it velocity in the opposite direction, and it will be found that, as it returns, it repasses each point with the same velocity in the opposite direction. [Sce Lock's Elementary Trigonometry cap. yIII.]

EXAMPLES. X. NOTE. In these and in all future examples, in which a force is applied to a mass, it is supposed that the force is so applied, that the shape of the mass need not be considered ; in some cases, the body is supposed to be so small that its size need not be considered; such a body is called a particle.

Find the acceleration in each of the following 6 cases.
1. 3 poundals acting on 6 lbs.
2. 32 poundals acting on 1 lb.
3. 32 poundals acting on 64 lbs.
4. 64 poundals acting on lbs.
5. 48 poundals acting on i cwt. [=112 lbs.]
6. 7 poundals acting on 4

cwt. Find the velocity acquired and the distance passed over in each of the 6 following cases where the mass is initially at rest.

7. 3 poundals acting on 6 lbs. for 10 seconds.
8. 12 poundals acting on 1 lb. for
9. 18 poundals acting on 27 lbs. for 3 secs.
10. 32 poundals acting on 56 lbs. for 9 secs.
11. 56 poundals acting on 2 cwt. for 12 secs.
12. 64 poundals acting on 2 lbs. for 4 sec.

13. Find the velocity acquired and the distance passed over in each of the 3 following cases, the force having the same direction as the initial velocity.

(i) 4 poundals acting for 3 secs, on 3 lbs. having initially 4 velos. (ii) 5 poundals acting for 4 secs. on jo lbs. having initially 20 velos.

22 poundals acting for 33 secs. on 1 lb. having initially 4 velos.

4

secs.

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14. 20 poundals act on a mass of 5 lbs. (initially at rest) for 3 seconds and then cease; how far will the mass go in 10 seconds more?

15. 64 poundals act for 4 seconds on a mass of 2 lbs. initially at rest, after which the force is suddenly reversed; find how far the mass. goes in 8 seconds from rest.

16. A particle of 1 lb. has 96 velos and is acted on by 32 poundals in the direction opposite to the velos; when will it have moved over 128 ft.? and find its velocities at those times,

17. A particle of 1 lb. has 64 velos, and is acted on by 32 poundals in the direction opposite to the velos; when will it be at rest? when will it be .moving in the opposite direction with

32 velos?

18. A particle of 1 lb. acted on by a constant force moves in a certain second over 20 ft., and in the next second but one it moves over 128 ft.; find the force.

19. A particle of 1 lb. is at rest, when being acted on by a constant force, it moves over 16 ft. in the first second of its motion ; find the force.

20. A mass of 20 lbs. is acted on by a constant force, and in the fifth and seventh seconds of its motion it passes over 108 ft. and 140 ft. respectively: find its initial velocity and the magnitude of the force.

21. A particle acted on by a constant force of 20 poundals passes over 72 ft. while its velocity increases from 16 to 20 velos; find its

mass.

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22. A particle acted on by 20 poundals has a velocity of miles per hour after 12 seconds from rest; find its mass.

23. A train of 90 tons on a horizontal plane at rest is acted on by a force of 1000 poundals; the force acts for one minute and then ceases ; find how far the train travels before coming again to rest, supposing the friction of the rails etc. to be equivalent to a retarding force of 300 poundals. [N.B. 1 ton=2240 lbs.]

24. A train of 100 tons attains from rest a velocity of 40 miles per hour in going i mile; the breaks are then applied and the train comes to rest in 440 yds. Compare the retarding force exerted by the breaks with the force exerted by the engine, (neglecting the rotary motion, and the ordinary friction of the wheels etc.).

25. A force of p poundals acts for n minutes on m lbs.; find the velocity generated.

26. Find how many poundals will produce in 10 lbs. a velocity of 30 velos after going 25 yards from rest.

27. A train of 100 tons is impelled along a horizontal railroad by a horizontal force of 15040 poundals; neglecting friction etc., how long will it take to go 4 miles from rest? and how many miles per hour will it be going at the end of 4 miles ?

37. The force which produces a celos in m lbs. is

ma poundals.
For, to produce i celo in mass i lb. 1 poundal is required ;

to produce i celo in mass m lbs. m poundals are required;

to produce a celos in mass m lbs. ma poundals are required. Example. A force f poundals acts continuously on a particle m lbs. which at a certain instant has u velos in the direction of the force; find (i) the acceleration, (ii) the velocity after t seconds, (iii) the distance it passes over in t seconds, (iv) how many seconds it takes to go s feet.

i poundal acting on 1 lb. produces i celo.
f poundals acting on 1 lb. produces f celos.

f
f poundals acting on m lbs. produce celos.

(i)

m

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f

ft In t seconds celos produce - velos, therefore after t seconds the

ft velocity is

velos.

(ii)

Il +

m

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the velocity at the middle of the interval of t secs. is and therefore the distance passed over in t seconds is

(n+3 ) velos,

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Let the particle take x seconds to pass over s feet, then by (iii)

fra S=ux + 3

m

from which quadratic equation x may be found.

(iv)

EXAMPLES. XI.

1. How many poundals are required to produce g celos in m lbs.?

2. How far will a ton [2240 lbs.] acted on by f poundals go from rest in t seconds ?

3. How many poundals are required to produce g celos in k tons ?

4. 1000 poundals acts upon a train of ķ tons on a horizontal plane; the friction of the rails etc. is equivalent to a retarding for of 250 poundals, how long will the train take to get up from rest a speed of m miles

per hour?

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