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33. To find the resultant of any number of forces concurring in one point P, and acting in the same plane.

Let the forces be represented by PA, P B, P C, PD (Fig. 9), and from the point A draw A b parallel and equal to P B, and complete the parallelogram A B, and draw the diagonal P b. Then complete the parallelogram b C in the same manner, and draw the diagonal P c.

Also complete the parallelogram c D, and draw the diagonal P d; and so on for any number of forces whatever PA and P B are equivalent to P b.

.. PA, PB, P C, are equivalent to P b, P C, that is to P. c.

.. P A, P B, P C, P D, are equivalent to P c, P D, that is to P d.

That is, Pd is the resultant of the forces PA, PB, PC, P D.

Cor. Hence it follows that if any number of forces be represented by the sides of a polygon taken in order, as P A, A b, bc, cd, their resultant will be represented by the line P d, that completes the polygon.

THE MECHANICAL POWERS.

34. The Mechanical Powers are contrivances by which we are enabled to sustain a great weight, or overcome a great resistance, by a small force.

The mechanical powers are usually accounted six in number, viz. the Lever, the Wheel and Axle, the Pulley, the Inclined Plane, the Wedge, and the Screw.

THE LEVER.

35. A Lever is an inflexible rod, moveable about a centre of motion, or fulcrum, and having forces applied to two or more points in it. There are three kinds or orders

of levers.

36. A lever of the first order has the fulcrum C between the weight W and the power P (Fig. 12). And of this order are balances, crows, handspikes, scissors, &c.

37. A lever of the second order has the weight between the power and the fulcrum; such as oars, rudders, &c. 38. A lever of the third order has the power between the weight and the fulcrum; such as tongs, the bones and muscles of animals, a man raising a ladder, &c.

39. The bended lever is sometimes considered as a fourth kind; such as a hammer drawing a nail. But it evidently has the same properties as that of the first order.

40. When the power and weight keep the lever in equilibrio, they are to each other reciprocally as the distances of their lines of direction from the fulcrum. That is, P: W:: CD: CE (Fig. 10); where CD and CE are perpendicular to W O and A O, the directions of the two weights, or the power A and weight W.

For draw C F parallel to A O, and C B parallel to WO: also join C O, which will be the direction of the pressure on the fulcrum C; for there cannot be an equilibrium, unless the directions of the three forces all meet in, or tend to the same point, as O. Then, because these three forces keep each other in equilibrio, they are proportional to the sides of the triangle C B O or C F O, drawn in the direction of these forces; therefore,

P: W: CF: FO or B C.

But since E B is parallel to C F, and B C to F D, the triangles C E B and C D F are similar; therefore C D ; CE CFC B.

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Hence, by equality, P: W:: CD: CE.

That is, each force is reciprocally proportional to the distance of its direction from the fulcrum.

Cor. 1. When the two forces act perpendicularly on the lever, as two weights, then, in case of an equilibrium, D coincides with W, and E with P; therefore the above proportion becomes

P: W:: CW: CA.

Cor. 2. Since the product of the extremes is equal to the product of the means, P x CE=WxCD; or if P× the forces act perpendicularly on the lever, we have, by the last Cor. PxAC W x C W.

Cor. 3. If any force P act at A in the direction A E, its effect on the lever to turn it about the centre of motion C, is as the length of the lever A C, and the sine of the angle of direction C A E. For the perpendicular C E is as A C x sine of the angle C A E.

Cor. 4. And in the bended lever A C W, we have P × A Cx sine CAD=WxCW x sine C W G. (Fig. 11.) -For C D and C G are perpendicular to the direction of the forces P and W respectively, and are therefore the distances of their respective directions from the fulcrum; consequently Px CDW x C G, but C D = AC × sin. C A B, and C G C W x sine C W G.

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..PX AC x sin. CADW x CW x sin. C W G. Cor. 5. It also follows, from Cor. 2, that in a straight lever of the first order, P x AC-WXC W, and the pressure on the fulcrum is P + W. (Fig. 12.)

Cor. 6. In a straight lever of the second order P x AC WXCW; but the pressure on the fulcrum is, in this case, WP. (Fig. 13.)

Cor. 7. In a straight lever of the third order, P x AC = W × CW, and the pressure on the fulcrum is PW. (Fig. 14.)

Cor. 8. If a straight lever be kept in equilibrio, by several weights P, Q, R, S, T, acting perpendicularly, (Fig. 15,) then,

PXAC+QxBC+ RXDC-SXEC+ Tx

F C.

Cor. 9. Since P: W:: CW: A C, we have, by composition, (Fig. 16)

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This corollary is useful in finding the fulcrum, when the power and weight are both given, together with the whole length of the lever.

Or it may be done very simply by Algebra. Thus, let A W the whole length of the lever a, and C W = x;

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41. In the compound lever, or where several levers act perpendicularly upon one another, as A B, B C, C D (Fig. 17), the fulcrums of which are F, G, I; then P: W :: BF × CGXDI: AFX BG XI C.

For the power P acting at A: weight at B:: BF: AF, and the force or power at B : the weight at C :: CG: BG. Also the power at C; the weight W:: DI: IC.

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Therefore, ex equo, P: W:: BFXCGXDI: AF x B G x IC.

The pressure on the fulcrum F

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P+Q=

Cor. PXAB

by Cor. 9 to the last Art.

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42. The following rule holds, whether the lever be of the first, second, or third order,

Multiply the power by its distance from the fulcrum, and this product will be equal to the weight multiplied by its distance from the fulcrum, the weight of the lever not being considered.

43. If the weight upon each end of the lever be given, to find the fulcrum:

Then the sum of the weights, is to either of the weights, as the sum of the distances, or whole length of the lever, is to the distance of the other.

THE BALANCE.

44. The Balance is a lever with equal arms, and, when well constructed, must have the following properties:

1. The points of suspension of the scales, and the centre of motion of the beam, A, C, B, (Fig. 18) should be in a straight line. 2. The arms, A C, B C, must be of an equal length. 3. That the centre of gravity be in the centre of motion C, or a little below it; for when these centres coincide, and the weights are equal, the addition of the smallest weight would overset the balance, and place it in a vertical situation, from which it would have no tendency to return. The sensibility, in this case, would be the greatest possible; but the requisites of level and stability would be entirely lost. 4. That they be in

* If a be the length of the arm of the balance, and b the distance between the centre of suspension and centre of gravity, P the load in either scale, and W the weight of the beam, the sensibility of the balance is as It is therefore

α

b (2 P+W)

greater, the greater the length of the arm, the less the distance between the two centres, and the less the weight with which the balance is loaded.

The stability, or the force with which the state of equilibrium is recovered, is proportional to (2 P + W) b, the denominator of the preceding fraction. The diminution of b, therefore,

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