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58. By a combination of wheels, we may multiply the power to any extent whatever, by making the lesser wheels to turn the greater.

A combination of wheels and axles is the very same in principle as the combination of levers, given in Art. 41; therefore P: W:: the product of the radii : the product of the radii of all the wheels.

of all the axles (See Fig. 26.) Hence P WX the product of the radii of all the axles the product of the radii of all the wheels

W PX the product of the radii of all the wheels the product of the radii of all the axles

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(14)

(13)

Or, since the number of teeth in wheels are as their radii,

P: W:: the product of the number of teeth in all the pinions the product of the number of teeth in all the wheels.

..P=

W =

W X the product of the number of teeth in all the pinions. (15)

the product of the number of teeth in all the wheels

PX the product of the number of teeth in all the wheels
the product of the number of teeth in all the pinions

Example 1.

(16)

A weight of 1000 lbs. is sustained by a rope of 2 inches diameter, going round an axle the diameter of which is 6 inches; what weight must be suspended at the circumference of the wheel, by a rope of the same thickness, to obtain an equilibrium, the diameter of the wheel being 1 feet?

By form. 1, we have P =

Here W = 1000, r=3, R t= 1.

W (r + & t)
R+ t

= 9, and t2; therefore

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If the thickness of the rope had not been taken into the account, then,

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A weight of 40 lbs. is suspended by a rope of one inch diameter, going round an axle of 4 inches diameter; what power, acting at the circumference of a wheel of 20 inches diameter, will support the equilibrium ?

In this case, we must take form. 5, for the power acts the wheel without a rope.

upon

P

W (r+ & t)

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For here W

40 lbs. r = 2 inches, R

40 (2+)
10

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In a combination of wheels and axles, given the radií of the wheels 24, 26, and 20 inches respectively, and the radii of the pinions 8, 5, and 4 inches; and if a power of 56 lbs. be applied to the extremity of the first wheel, what weight will it be able to sustain at the extremity of the last pinion?

By form. 14, W=

39 cwt.

56 x 24 x 26 x 20

8 x 5 x 4

=

4368 lbs.

And, in this case, the power will move 78 times as fast as the weight.

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THE PULLEY.

59. A pulley is a small wheel which turns about an axis passing through its centre. The centre may be either

fixed or moveable. The pulley is either single, or combined together to increase the power.

60. When the power supports the weight by means of a fixed pulley, the power and weight are equal. For, through the centre C draw A B, which will represent a lever of the first order, the fulcrum of which is C, and AC BC; therefore, since the distance of the power from the fulcrum is equal to the distance of the weight from the fulcrum, to obtain an equilibrium the power must be equal to the weight. (Fig. 27.)

61. When the power sustains the weight by means of a single moveable pulley, the power is but half the weight, if the portions of the sustaining cord are parallel to each other.

For A B may be considered as a lever of the second order, the power acting at A, the weight at C, and the fulcrum or fixed point is at B; therefore P: W :: B C : AB; but BCA B, consequently PW. (Fig. 28.)

62. The same principle may be applied to a combination of pulleys, all drawn by one cord going over all the pulleys. (Fig. 29.) Then P: W:: 1: the number of parts of the cord going round the moveable block.

For the whole weight W is supported by the number of the parts of the cord going round the moveable block; therefore each of these parts must bear an equal portion of the weight; and if n the number of these parts, then each part must bear the nth part of the weight, or P: W :: 1: n. The pressure upon the hook A or B P+W or (n + 1) P.

63. If, instead of the same cord going round all the pulleys, each pulley hangs by a separate cord, then, to obtain an equilibrium, P; W:: 1: 2", n being the number of moveable pulleys. (Fig. 30.)

P: weight at B :: 1 : 2 weight at B weight at C:: 1: 2 weight at C: weight at D:: 1:2

... P : W :: 1 : 2 × 2 × 2, &c. :: 1 : 2n

Cor. Hence 2n P ==

W, or P =W ÷ 2n.

64. When the strings or cords are not parallel to each other, but form the angle A D E (see Fig. 31), then P: W radius: twice the cosine of the angle of inclination of the direction of the power to the direction of the weight.

For produce A B, the direction of the power, to D; and from C, the centre of the moveable pulley, draw CB perpendicular to C D, the direction of the weight; then let D B represent the force in the direction D B, and resolve it into D C, B C, and D C will be that part of it which is effective in sustaining the weight; and since the cord E F sustains the same weight that the cord A B sustains, the whole weight sustained by the cord E F BA will be represented by 2 C D, consequently P: W :: DB : 2 CD :: radius: twice the cosine of the angle B D C.

Example.

If a weight be sustained by a power which is attached to a rope going round a moveable pulley (see Fig. 31), and making an angle of 30° with a vertical line passing through the centre of the pulley, what proportion does the power bear to the weight?

By Art. 64, P: W:: 1: 2 cos. 30°.

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If the angle B DC 45°, then since cos. 45° = √ √2, we have P: W:: 1:

√2.

If the angle B DC 60°, its cosine is .

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.. P: W:: 1 : 1.

Hence the power and weight are equal.

THE INCLINED PLANE.

65. The Inclined Plane is a plane inclined to the horizon, or a plane which makes any angle whatever with an horizontal plane. (Fig. 32.)

66. If a weight W be sustained upon an inclined plane by a power P, acting in a direction parallel to that plane, then the power P is in proportion to the weight W as the height of the plane is to its length; that is, P: W :: BC A C.

Draw B D perpendicular to A C. Now the weight W is sustained by three forces, viz. the force of gravity, or its own weight W, acting in the direction B C, perpendicular to the horizon; the power P, acting in the direction D C, parallel to the plane; and the pressure against the plane, acting in the direction D B, perpendicular to the plane. Therefore, by Art. 29, these forces will be respectively proportional to the three sides B C, D C, and D B, of the triangle B D C. But the triangles B DC and A B C are similar, therefore their like sides are proportional; that is, BC, DC, and D B, are to each other respectively as A C, B C, and A B, which are therefore as the three forces W, P, and P', where P' represents the pressure against the plane.

That is, the power P is to the weight W:: BC: A C. And the power P is to the pressure against the plane P: BC: A B.

Also the weight W is to the pressure against the plane P: AC: A B.

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Let the length of the plane A C = 10 feet, the height BC 6 feet, and the weight 600 lbs.; then 10: 6:: 600 : 360 lbs. the power necessary to sustain the weight; and

=

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