may be conceived to be at A or B. Let B be the centre of motion, then we have a straight lever whose centre of motion is B, and the two forces A and A+B, acting perpendicularly upon it at the points A and C, sustain each other; also, À : B :: BC: AC; therefore A: A+B BC: BA.

105. Cor. 1. If two weights, or two forces, acting perpendicularly on the arms of a straight lever, keep each other in equilibrio, they are inversely as their distances from the centre of motion.

For the weights will balance when they are in that proportion, and if the proportion be altered by increasing or diminishing one of the weights, its effort to turn the lever round will be altered, or the equilibrium will be destroyed.

106. Cor. 2. Since A: B :: BC: AC when there is an equilibrium upon the lever AB, whose fulcrum is C, by multiplying extremes and means, A×AC= BX BC.

107. Cor. 3. When the power and weight act on the same side of the fulcrum, and keep each other in equilibrio, the weight sustained by the fulcrnm is equal to the difference between the power and the weight.

108. Cor. 4. In the common balance, the arms of the lever are equal; consequently, the power and weight, or two weights, which sustain each other, are equal. In the false balance, one arm is longer than the other; therefore the weight, which is suspended at this arm, is proportionally less than the weight which it sustains at the other.

109. Cor. 5. If the same body be weighed at the two ends of a false balance, its true weight is a mean proportional between the apparent weights.

Call the true weight x, and the apparent weights, when it is suspended at A and B, a and b respectively; then a :x :: AC: BC, and ≈ : b :: AC: BC; therefore a: x :: x : b.

110. Cor. 6. If a weight C, fig. 18, be placed upon a lever which is supported upon two props A and B in an horizontal position, the pressure upon A : the pressure upon B:: BC: AC.

For if B be conceived to be the fulcrum, we have this proportion, the weight sustained by A: the weight C :: BC: AB; in the same manner, if A be considered as the fulcrum, then the weight C : the weight sustained by B: AB CA; therefore, ex æquo, the weight sustained by A : the weight sustained by B: BC: AC.

111. Cor. 7. If a given weight P, fig. 19, be moved along the graduated arm of a straight lever, the weight W, which it will balance at A, is proportional to CD, the distance at which the given weight acts.

When there is an equilibrium, WXAC=PXDC (art. 106); and AC and P are invariable; therefore wαDC.

112. If two forces, acting upon the arms of any lever, keep it at rest, they are to each other inversely as the perpendiculars drawn from the centre of motion to the directions in which the forces act.

Case 1, Let two forces, A and B, fig. 20, act perpendicularly upon the arms CA, CB, of the lever ACB whose fulcrum is C, and keep each other at rest. Produce BC to D, and make CD=CA; then the effort of A to move the lever round C, will be the same, whether it be supposed to act perpendicularly at the extremity of the arm CA, or CD (art. 101, ax. 3); and on the latter suppositior, since there is an equilibrium, A : B :: CB : CD (art. 105); therefore A : B :: CB : CA.

Case 2. When the directions AD, BH, fig. 21, in which the forces act, are not perpendicular to the arms, take AD and BH, to represent

forces; draw CM and CN at right angles to those directions; also draw AF perpendicular, and DF parallel to AC, and complete the parallelogram GF; then the force AD is equivalent to the two AF, AG, of which AG acts in the direction of the arm, and therefore can have no effect in causing, or preventing any angular motion in the lever about C. Let BH be resolved, in the same manner, into the two BI, BK, of which BI is perpendicular to, and BK in the direction of the arm CB; then BK will have no effect in causing, or preventing any angular motion in the lever about C; and since the lever is kept at rest, AF and BI, which produce this effect, and act perpendicularly upon the arms, are to each other, by the first case, inversely as the arms; that is, AF: BI :: CB: CA, or AF × CA=BIX CB. Also, in the similar triangles ADF, ACM, AF : AD :: CM: CA, and AFX CA=AD x CM; in the same manner, BI×CB=BH× CN; therefore AD × CM=BH × CN, and AD : BH :: CN : CM.

113. Cor. 1. Let a body IK, fig. 22, be moveable about the centre C, and two forces act upon it at A and B, in the directions AD, BH, which coincide with the plane ACB; join AC, CB; then this body may be considered as a lever ACB, and drawing the perpendiculars CM, CN, there will be an equilibrium, when the force acting at A: the force acting at B: CN: CM.

114. Cor. 2. The effort of the force A, to turn the lever round, is the same, at whatever point in the direction MD it is applied; because the perpendicular CM remains the same.

115. Cor. 3. Since CA: CM :: rad. sin. CAM, CM =

CA X sin. CAM

rad.

; therefore, when there is an CBX sin. CBN CAX sin. CAM rad. rad.

:

equilibrium, the power at A: the weight at B:: :: CB X sin. CBN: CA x sin. CAM. 116. Cor. 4. If the lever ABC fig. 23, be straight, and the directions AD, BH, parallel, A: B :: BC: AC; because, in this case, sin. CAM―sin. CBH. Hence, also, AXAC = B× BC.

117. Cor. 5. If two weights balance each other upon a straight lever in any one position, they will balance each other in any other position of the lever; for the weights act in parallel directions, and the arms of the lever are invariable.

118. Cor. 6. If a man, balanced in a common pair of scales, press upwards by means of a rod, against any point in the beam, except that from which the scale is suspended, he will preponderate.

Let the action upwards take place at D, fig. 24, then the scale, by the re-action downwards, will be brought into the situation E; and the effect will be the same as if DA, AE, DE, constituted one mass; that is, drawing EF perpendicular to CA produced, as if the scale were applied at F (art. 114); consequently the weight, necessary to maintain the equilibrium, is greater than if the scale were suffered to hang freely from A, in the proportion of CF : CA. 119. Cor. 7. Let AD fig. 25, represent a wheel, bearing a weight at its centre C; AB an obstacle over which it is to be moved by a force acting in the direction CE; join CA, draw CD perpendicular to the horizon, and from A draw AG, AF, at right angles to CE, CD. Then CA may be considered as a lever whose centre of motion is A, CD the direction in which the weight acts, and CE the direction in which the power is applied; and there is an equilibrium on this lever when the power : the weight :: AF :: AG.

Supposing the wheel, the weight, and the obstacle given, the power is

least when AG is the greatest; that is, when CE is perpendicular to CA, or parallel to the tangent at A.

120. Cor. 8. Let two forces acting in the directions AD, BH, fig. 26, upon the arms of the lever ACB, keep each other in equilibrio; produce DA and HB: till they meet in P; join CP, and draw CL parallel to PB; then will PL, LC represent the two forces, and PC the pressure upon the fulcrum.

For, if PC be made the radius, CM and CN are the sines of the angles CPM, CPN, or CPL, PCL; and PL : LC :: sin. PCL : sin. LPC :: CN : CM, therefore PL, LC, represent the quantities and directions of the two forces, which may be supposed to be applied at P (art. 114), and which are sustained by the re-action of the fulcrum; consequently, CP represents the quantity and direction of that re-action (art. 58), or PC represents the pressure upon the fulcrum.

121. In a combination of straight levers, AB, CD, figs. 27, 28, whose centres of motion are E and F, if they act perpendicularly upon each other, and the directions in which the power and weight are applied be also perpendicular to the arms, there is an equilibrium when P: W:: EBX FD: EA x FC.

For the power at A: the weight at B, or C :: EB: EA; and the weight at C the weight at D :: FD: FC; therefore, P: W :: EB x FD: EA x FC.

By the same method we may find the proportion between the power and the weight, when there is an equilibrium, in any other combination of levers.

122, Cor. If E and F be considered as the power and weight, A and D the centres of motion, we have, as before, E: F::FDX BA: AEX CD. Hence the pressure upon E⚫ the pressure upon F :: FD×BA: AEX CD.

123. Any weights will keep each other in equilibrio on the arms of a straight lever, when the products, which arise from multiplying each weight by its distance from the fulcrum, are equal on each side of the fulcrum.

The weights A, B, D, and EF, fig. 29, will balance each other upon the lever AF whose fulcrum is C, if AxAC+B × BC + D × DC= EXEC+FX FC.

In CF take any point X, and let the weights r, s, t, placed at X, balance respectively, A, B, D; then Ax AC=rx XC; B× BC= sx XC; Dx DC=tx XC (art. 116); or, Ax AC+BxBC+Dx DC =r+s+tx XC. In the same manner, let p and 9, placed at Y, balance respectively, E and F; then p+-qxYC=ExEC+FX FC; but by the supposition Ax AC+B× BC+D× DC=EXEC+Fx FC; therefore r+s+txXC=p+qx YC, and the weights r, s, t, placed at X, balance the weights p, q, placed at Y; also A, B, D, balance the former weights, and E, F, the latter; consequently A, B. D, will balance E, and F.

124. Cor. 1. If the weights do not act in parallel directions, instead of the distances, we must substitute the perpendiculars, drawn from the centre of motion, upon the directions. (Art. 112.)

125. Cor. 2. In art. 111, the lever is supposed to be without weight, or the arms AC, CD to balance each other. In the formation of the common steel yard fig. 30, the longer arm CB is heavier than CA, and allowance must be made for