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celos;

Example iii. How long does a stone take in falling from rest under the action of gravity through a vertical distance of 128 ft.?

By Arts. 43—46 the stone has an acceleration of 32 celos downwards.

Let t seconds be the time of passing over 128 ft.; then,

128= x 32° 5. or,

= 8; therefore,

t=2/2= 2 X 14142=2.8284.... The interval occupied is therefore 2.8... seconds.

Example iv. 16 lbs. is acted on continuously by a force equal to the weight of 4 lbs.; find its acceleration; and find how far it moves from rest in 10 seconds. The weight of 4 lbs. = 4 32 poundals.

4 X 32 This force 4 x 32 poundals acting on 16 lbs. produces in it

16 That is, this force produces in 16 lbs. the acceleration 8 celos. In 10 seconds the 16 lbs. will pass over a distance s feet such that

s=at=x8 x 100=400. The distance required is 400

feet. NOTE. A force is said to be expressed in pounds weight, when we state the number of lbs. whose weight is equal to the force.

Example v. A mass 4.cwt. is observed to be moving with 16 celos; express in pounds weight the force which is acting upon it.

When 4 cwt. has 16 celos, it has 4x 112 x 16 pound-celos.

To produce this, 4 x 112 x 16 poundals must be acting upon it; 4 x 112 x 16 poundals are equal to the weight of 4 x 112 x 16

32 that is, the weight of 2 cwt. must be acting on the 4 cwt.

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lbs. ;

XIV.

EXAMPLES. 1. A stone is let fall from the top of a tower 256 ft. high; in how many seconds will it reach the ground?

2. A stone is let fall; how far will it fall (i) in the 2nd second, (ii) in the 4th second from rest?

3. A stone is thrown vertically upwards with a velocity of 320 velos; when will it come to rest? and how high will it go ?

4. A stone is thrown vertically upwards and returns after 4 seconds to the point from which it was thrown; with what velocity was it thrown? and how high did it go?

† Example iv. p. 16 and Example i. p. 20 should be noticed.

5. A stone is thrown vertically upwards and passes a point 48 ft. high with a velocity of 32 ft. per second; in how many seconds will it return to the point of projection?

6. A stone is let fall from a tower 256 ft. high and at the same instant another stone is projected from the foot of the tower with just sufficient velocity to carry it to the top of the tower; when and where will the stones meet?

7. A body of 1 cwt. is acted on by a force equal to the weight of 7 lbs.; find its acceleration.

8. A train of 100 tons is drawn with an acceleration of 4 celos; express in pounds weight the force exerted by the engine in addition to that necessary to overcome the friction.

9. 28 lbs. placed on perfectly smooth ice is acted on by a horizontal force equivalent to the weight of 1 lb.; find its velocity after 7 minutes.

10. Two masses consisting of a cubic foot of lead and a cubic foot of iron placed at rest on perfectly smooth ice are each acted on in the same direction by a horizontal force equal to the weight of 1 lb.; how far will they be apart after 30 seconds?

[NOTE. A cubic ft. of lead weighs about 11340 oz.; a cubic ft. of iron weighs about 7680 oz.]

11. 1 ton on a smooth horizontal plane is observed to move from rest a distance of 48 ft. in 60 seconds; if the motion is caused by a uniform horizontal force, how many lbs. would the force lift?

12. A train of 100 tons is drawn by an engine which can exert a force, which would lift a weight of 800 lbs. ; supposing the force of the friction etc. to be equivalent to a force of 100 lbs. weight, find how long the train would be in attaining from rest a velocity of 60 miles per hour.

13. A mass of 1 cwt. is placed on ice and is acted on by a horizontal force equal to the weight of 8 lbs.; supposing that a force equal to the weight of 1 lb. is sufficient to overcome the friction of the ice, find when the mass will have 100 velos.

14. Supposing the force of 8 lbs. weight of Question 13 to be withdrawn when the mass is moving at the rate of 60 miles an hour, in how many seconds will the mass be brought to rest by the retardation due to the friction?

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a

47. In Art. 28 it is assumed that we can apply a constant force to a mass.

Such a force might be obtained by taking care that the force is just sufficient to keep a certain steel spring compressed in a certain manner.

By applying the same constant force to each of two bodies in turn and observing the number of celos produced in them, we obtain a measure of the mass of the bodies.

It happens however that the experiment of Art. 43 provides a more convenient method of measuring the mass of a body. For, since the weight of a mass is a force which produces a fixed number (8) of celos in the mass itself, the weight of a mass is proportional to the mass itself; therefore, equal masses have equal weights.

By a ton of coal is meant a certain amount of mass.

That a certain quantity of coal contains a ton mass is ascertained by weighing it; that is, we ascertain that the mass of coal has the same weight as a certain other standard mass of iron, which mass we call a ton.

NOTE. In the language in common use, the ideas of mass and weight are not clearly distinguished. For example, a piece of metal is said to be

'a 7 lbs. weight'; here, the word weight means, a piece of metal whose mass is 7 lbs., whose weight is used for the purpose of comparing the weights of other

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masses.

Or again,

'a train weighing 100 tons’; here, the word weighing means, 'whose weight is equal to that of 100 tons.' Again, a string is often said to be light, meaning weightless, or

of too small weight to be observed; such a string must also be of too small mass to be observed. The student however must carefully remember that

the unit mass is i lb., while

the weight of 1 lb. is 32:2.... poundals.

;

48. The two following examples should be noticed and the results compared.

Example i. A mass of 1 ton lies on a smooth horizontal plane; a man by means of a windlass and string applies continuously to the mass a horizontal force equal to the weight of 28 lbs.; how long will he take to move the mass 5 ft. from rest.

Let the force produce a celos in the ton; that is, in 2240 lbs.
The mass acceleration is therefore 2240a pound-celos.
To produce 22400 pound-celos the force required is

22402 poundals.
The force actually applied is the weight of 28 lbs. ;;
which is,

28 x 32 poundals. Hence

28 x 32 poundals=2240 a poundals; whence,

a=. Let s ft. be the distance moved from rest in time t seconds; then,

s={ata; here,

5=x x 2;

= 25; therefore, t=5. The time required is 5 seconds.

Example ii. A mass of 1 ton lies on a smooth horizontal table, and is attached by a light string which passes over the edge of the table, to a mass of 28 lbs., which hangs freely under the action of its own weight; how long will the masses take to move 5 ft. from rest?

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[NOTE: Here, the force acting on the ton mass is not equal to the weight of 28 lbs.; for

The force acting on the ton is the pull or tension of the string; and the tension of the string is not equal to the weight of the 28 lbs.; for if it were, the 28 lbs. mass would have acting on it (i) a pull equal to the weight of 28 lbs. upwards and (ii) its own weight (which is that of 28 lbs.) downwards; that is, it would have no resultant external force acting on it, and would therefore have no acceleration.

But, if the ton has acceleration, the 28 lbs. must have acceleration also.

Hence, the force acting on the ton, i.e. the tension of the string, is less than the weight of 28 lbs.; as will presently appear.]

(i).

I. Consider the external forces acting on the 28 lbs.

Let T poundals be the tension of the string; then the resultant external forces acting on the mass of 28 lbs. are its weight directly downwards and the pull of the string directly upwards, that is,

28 x 32 poundals – T poundals; Let this resultant force produce a celos in the 28 lbs.

It produces therefore 28a pound-celos; the force necessary to do this is 28a poundals; hence we see that

(28 x 32 - 7) poundals=28a poundals II. Consider the external force acting on the ton.

The weight of the ton does not concern the question; it is supposed to be entirely obliterated by the upward support of the smooth horizontal plane.

The only other force acting on the ton is the tension of the string. The string is supposed to be light, and therefore of no mass ; so that i exerts the same pull on the ton as it does on the 28 lbs. ; which pull is

T poundals. Also, since the two masses are connected by an inextensible string, the acceleration of the ton is the same as that of the 28 lbs. viz., a celos. The force necessary to produce a celos in 2240 lbs. is

2240 a poundals; but, the force which produces the a celos is T poundals; hence we see that T poundals= 20 X 11 2a poundals;

(ii) whence, adding (ii) to (i) 28 x 32 = 81 X 28a; therefore,

a=ji. Let t seconds be the time in which the masses pass over 5 ft. from rest; then,

5= $xiA; or,

t=5= 4 x 2'236...

=5'031... The time required is 5'031... seconds.

NOTE. Comparing these two Examples we observe that in Example i, the acceleration of the ton is ši celos, in Example ii the acceleration is less, viz. şi celos. In Example i the weight of 28 lbs. acts on a ton, in Example ii, the weight of 28 lbs. acts on (a ton + 28 lbs.).

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