Page images
PDF
EPUB

but its effect on the body in that plane must be a force

X along ON.

But if S be the tension of the rod, its effect along ON is S cos AON = S sin 0,

[blocks in formation]

Also the ball moves horizontally, and is therefore in equilibrium vertically.

Whence

[blocks in formation]

[The rod is the only body touching the ball, therefore S is the only force acting on it besides W.]

[blocks in formation]

h is called the height of the simple governor, or the height due to the revolutions, and the result shows that it is independent of weight of ball or of the length of the rods.

The method we have used is not that usually given, nor is it a very convenient method. Usually the ball is shown, as in Fig. 170, in equilibrium under the forces S, X, and W.

h

N

W

Fig. 170.

Then we have at once

S sin =X, S cos 0=W,

and hence the same results as before. But there is in this method a danger that the student may think of X as a force acting on the revolving body, whereas, as drawn in Fig. 170, it is the

force exerted by the ball on the rod.

The forces, as drawn in Fig. 170, do not represent forces acting on the ball, but on the end of the rod

where it meets the ball. Suppose now the rod to extend to O, and to be connected to the ball by a pin passing through its centre (Fig.

[merged small][ocr errors][merged small][merged small]

then as from Fig. 170 are quite correct, always bearing in mind what we have just been saying, and hence we shall always put the forces in that form.

The next question is-If the speed change from A to A' what pull is caused on the rods EE, and hence

N

Fig. 172.

w

on the slider G? As the speed increases the balls tend to rise, but they cannot actually do so until they exert a sufficient pull on the rods EE (Fig. 167); we are now then going to consider the governor revolving at A', but at the proper height for A, i.e. before the slider has moved.

Let T be the tension of NC (we x' consider one only), C being the middle point of AO, and let the other forces be as usual. Then the forces acting on the rod and pin at

O are T, W, X', and the action at the point A.

To avoid considering this latter, take moments about A, and we have

Th sin + Wr=X'h.

X' is the centrifugal force due to A',

.. Th sin +Wr=

WA'2r.h
g

But g/A'2', the proper height for the new revolutions

[merged small][merged small][ocr errors]
[ocr errors]
[blocks in formation]

Next consider the slider; it is acted on, taking now the two balls, by the two tensions T, there

fore the total pull on the slider, say F, equals

AT 2T cos 0,

[blocks in formation]

Fig. 173.

If h and h' be found, this gives at once the value of F.

Now the resistance to motion of the slider, say R, will prevent motion taking place till F equals R; so that if we know R we can find the change of speed necessary before the slider begins to move.

We shall

have for this

[blocks in formation]

d

or this will give the necessary weight of the balls in order that a given change of velocity may be sufficient to move the slider. The smaller the change of speed necessary the greater is said to be the sensitiveness of the Governor.

Bursting Effect of Centrifugal Force. This effect, mentioned on page 228, we can now explain.

tk

The figure represents the rim of a fly-wheel, rotating with angular velocity A.

a

Fig. 174.

Let

r-radius of rim,

b=thickness or width perpendicular to the paper,
t=width in plane of paper.

Consider now the forces acting on I inch in length of the rim, contained between the two radial lines ab, cd. If

w=weight of 1 c. in. in lbs.

w.t.b.weight of 1 inch length of rim.

Then, to make this piece rotate in the circle, there must be applied to it a force

w.t.b.A2r
g

This force can only be applied by the action on the 1-inch length, shown shaded, of the remainder of the wheel; which action is applied over the plane ends ab and cd; and if A be so great that the strength of the metal is not sufficient to supply such a force, then the shaded piece will no longer travel in the circle, i.e. the rim will burst. For the completion of this investigation we must refer to page 274; where the nature of the actions between the small piece which we have picked out for consideration, and the rest of the rim will be better understood.

EXAMPLES.

I. In testing the power of an engine by the transmission dynamometer of Fig. 163, the revolutions of B were observed to be 300 per minute and the thrust P 10 lbs. The dimensions were, radius of B 24 ins., centres of C and D 44 ins. apart, arm E 30 ins. long. Find the brake H. P.

2. The power of an engine is tested by a tail-rope meter. The wheel is 5 ft. diameter, the weight 300 lbs. horse power when the spring balance shows a pull of and the fly-wheel makes 150 revolutions per minute.

Ans. 1.56

dynamo

Find the 180 lbs., Ans. I.

3. The coefficient of axle friction is sometimes determined by hanging a heavy pendulum on a shaft so that it can revolve freely on the shaft. The shaft is then rotated at a given speed and the pendulum takes a position of equilibrium at an angle to

the vertical. If W=weight of pendulum, / the distance of its C. G. from the centre of the shaft, N the number of revolutions per minute, prove that the horse power lost in friction is WIN sin 0/5255. Deduce the value of ƒ' (page 69).

Ans. f'=

/ sin @

r

r-radius of shaft.

4. In question 2 find the effect, Ist, of a fall-off of 3 per cent in the power of the engine; 2d, of a decrease of 10 per cent in the friction due to increased lubrication.

Ans. Ist, Leaving the weight unaltered, the revolutions will drop 3 per cent, and the spring balance be unaltered;

2d, the balance will show 190 lbs., and the revolutions rise to 164 nearly.

5. In testing an engine by a Prony brake, the arm was vertical and connected to a spring balance. In order to support the weight of the brake, 720 lbs., a stiff spring was fitted on which the brake rested. This spring being supposed to be directly under the centre of the shaft; if the engine were running at 140 revolutions, show that a displacement of 1 ft. in the position of the spring will falsify the results by 19 H. P.

6. Find the height of a simple governor revolving at 75 revolutions. Ans. 6 ins.

7. The balls of a governor weigh each 3 lbs., and are each hinged to a pair of equal rods, one of each pair connected to the spindle, and the other to a slider which moves the throttle. The speed suddenly increases from 75 to 77 revolutions. the pull on the slider.

Find Ans. .162 lbs.

8. Solve the preceding when the governor is constructed as in Fig. 167, page 246. Ans. .324 lbs.

9. If a plate rotate about any axis perpendicular to itself, prove that the centrifugal force is the same as if the whole mass were concentrated at the C. G., revolving with the same angular velocity.

Ans. Let P be any point, w the weight of a small particle at

P, O the centre of rotation, G the C. G. The centrifugal force of w is w/g. A2OP along OP. This is by statics equivalent to two forces w/gA2. OG, and w/gA2. GP. Adding up all these the first set gives W/gA2. OG, and the second set vanish because G is the C. G. Hence the result.

10. A fly-wheel, diameter 12 ft., weighing 5 tons, is bored out

« PreviousContinue »