to one another. But if P, be equal to P, their resultant will manifestly have its direction as much towards one of these pressures as the other; that is, it will have its direction midway between them, and it will bisect the angle BAC: but the diagonal AF in this case also bisects the angle BAC, since P, being equal to P,, AC is equal to AB; so that in this particular case the direction of the resultant is the direction of the diagonal, and the proposition is true, and similarly it is true of P, and P,, since these pressures are equal. Since then it is true of P, and P, when they are equal, and also of P, and P,, therefore it is true in this case of P, and P, + P,, that is of P, and 2 P,. And since it is true of P, and P,, and also of P, and 2 P,, therefore it is true of P, and P, +2 P,, that is of P, and 3 P,; and so of P, and m P,, if m be any whole number; and similarly since it is true of m P, and P,, therefore it is true of m P, and 2 P,, &c., and of m P, and n P, where n is any whole number. Therefore the proposition is true of any two pressures m P, and n P, which are commensurable.

It is moreover true when the pressures are incommensurable. For let AC and AB represent any two such pressures P, and P, in magnitude

and direction, and complete the parallelogram ABDC, then will the direction of the resultant of P, and P, be in AD; for if not, let its direction be AE, and draw EG parallel to CD. Divide AB into equal parts, each less than GC, and set off on AC parts equal to those from A towards C. One of the divisions of these will manifestly fall in GC. Let it be II, and complete the parallelogram AHFB. Then the pressure P, being conceived to be divided into as many equal units of pressure as there are equal parts in the line AB, AH may be taken to represent a pressure P, containing as many of these units of pressure as there are equal parts in AII, and these pressures P, and P, will be commensurable, being measured in terms of the same unit. Their resultant is therefore in the direction AF, and this resultant of P, and P, has its direction nearer to AC than the resultant AE of P, and P, has; which is absurd, since P, is greater than P,.

3

1

Therefore AE is not in the direction of the resultant of P, and P,; and in the same manner it may be shown that no other than AD is in that direction. Therefore, &c.

3. The resultant of two pressures applied in any directions to a point, is represented in magnitude as well as in direction by the diagonal of the parallelogram whose adjacent sides represent those pressures in magnitude and in direc

tion.

Let BA and CA represent, in magnitude and direction, any two pressures applied to the point A. Complete the parallelogram BC. Then by the last proposition AD will represent the resultant of these pressures in direction. It will also represent it in magnitude; for, produce DA to G, and conceive a pressure to be applied in GA equal to the resultant of BA and CA, and opposite to it, and let this pressure be represented in magnitude by the line GA. Then will the pressures represented by the lines BA, CA, and GA, manifestly be pressures in equilibrium. Complete the parallelogram BG, then is the resul.ant of GA and BA in the direction FA; also since GA and BA are in equilibrium with CA, therefore this resultant is in equilibrium with CA, but when two pressures are in equilibrium, their directions are in the same straight line; therefore FAC is a straight line. But AC is parallel to BD, therefore FA is parallel to BD, and FB is, by construc.ion, parallel to GD, therefore AFBD is a parallelogram, and AD is equal to FB and therefore to AG. But AG represents the resultant of CA and BA in magnitude, AD therefore represents it in magnitude. Therefore, &c.*

THE PRINCIPLE OF THE EQUALITY OF MOMENTS.

4. DEFINITION. If any number of pressures act in the same plane, and any point be taken in that plane, and perpendiculars be drawn from it upon the directions of all these pressures, produced if necessary, and if the number of units in each pressure be then multiplied by the number of units in the corresponding perpendicular, then this product is called the moment of that pressure about the point from which the perpendiculars are drawn, and these moments are said to be measured from that point.

*Note (d) Ed. App.

5. If three pressures be in equilibrium, and their moments be taken about any point in the plane in which they act, then the sum of the moments of those two pressures which tend to turn the plane in one direction about the point from which the moments are measured, is equal to the moment of that pressure which tends to turn it in the opposite direction.

Let P., P2, P., acting in the directions P,O, P,O, P,O, be any three pressures in equilibrium. Take any point A in the plane. in which they act, and measure their moments from A, then will the sum of the moments of P, and P,, which tend to turn the plane in one direction about A, equal the moment of P,, which tends to turn it in the opposite direction.

Through A draw DAB parallel to OP,, and produce OP, to meet it in D. Take OD to represent P, and take DB such a length that OD may have the same proportion to DB that P, has to P. Complete the parallelogram ODBC, then will OD and OC represent P, and P, in magnitude and direction. Therefore OB will represent P, in magnitude and direction.

Draw AM, AN, AL, perpendiculars on OC, OD, OB, and join AO, AC. Now the triangle OBC is equal to the triangle OAC, since these triangles are upon the same base and between the same parallels.

Also, AODA+AОAB=▲OBD = AOBC, ...A ODA+AOAB=AOAC,

···÷OD× AN++ OB×AL = { OC × AM,

... P2 × AN+P‚×AL=P,×AM.

1

2

2

Now P, XAM, P, × AN, P,× AL, are the moments of P1, P1, P, about A (Art. 4.)

... m2 P2 + m2 P, = mt P,

Therefore, &c. &c.

6. If R be the resultant of P, and P,, then since R is equal to P, and acts in the same straight line, m'R= m.P1, ...m'P,+m'P1 = m1R.

The sum of the moments therefore, about any point, of two pressures, P, and P, in the same plane, which tend to

turn it in the same direction about that point, is equal to the moment of their resultant about that point.

If they had tended to turn it in opposite directions, then the difference of their moments would have equalled the moment of their resultant. For let R be the resultant of P, and P, which tend to turn the plane in opposite directions about A, &c. Then is R equal to P,, and in the same straight line with it, therefore moment R is equal to moment P,. But by equation (1) m'P,-m1P ̧ = m1P ̧; ... m'P-m'P, = m1R.

2

=

(2),

Generally, therefore, mt P, + mt P, mt R. . . . the moment, therefore, of the resultant of any two pressures in the same plane is equal to the sum or difference of the moments of its components, according as they act to turn the plane in the same direction about the point from which the moments are measured, or in opposite directions.*

7. If any number of pressures in the same plane be in equilibrium, and any point be taken, in that plane, from which their moments are measured, then the sum of the moments of those pressures which tend to turn the plane in one direction about that point is equal to the sum of the moments of those which tend to turn it in the opposite direction.

Let P1, P2, P,

.

n

P be any number of pressures in the same plane which are in equilibrium, and A any point in the plane from which their moments are measured, then will the sum of the moments of those pressures which tend to turn the plane in one direction about A equal the sum of the inoments of those which tend to turn it in the opposite direction.

Let R, be the resultant of P, and P2,

R

R

&c.

R2-1

R, and P,

R, and P,
&c.

Rn-2 and Pn.

Therefore, by the last proposition, it being understood that the moments of those of the pressures P,, P, which tend to turn the plane to the left of A, are to be taken nega. tively, we have

Note (c) Ed. App.

Rn-1

m2

mt Rant R-2+mt P2.

Adding these equations together, and striking out the terms common to both sides, we have

m2 R2-1

P 22

=

mt P, +mt P, + m2 P, +

2

+ m2 Pr

(3), where R-1 is the resultant of all the pressures P1, . Pr.

But these pressures are in equilibrium; they have, therefore, no resultant.

.·. Re—1 = 0.. mt R-1 = 0, ...mt P, + m' P2 + m2 P„, + . . . . . m2 P2 = 0.

(4).

Now, in this equation the moments of those pressures which tend to turn the system to the left hand are to be taken negatively. Moreover, the sum of the negative terms must equal the sum of the positive terms, otherwise the whole sum could not equal zero. It follows, therefore, that the sum of the moments of those pressures which tend to turn the system to the right must equal the sum of the moments of those which tend to turn it to the left. Therefore, &c. &c.

8. If any number of pressures acting in the same plane be in equilibrium, and they be imagined to be moved parallel to their existing directions, and all applied to the same point, so as all to act upon that point in directions parallel to those in which they before acted upon different points, then will they be in equilibrium about that point.

For (see the preceding figure) the pressure R, at whatever point in its direction it be conceived to be applied, may be resolved at that point into two pressures parallel and equal to P, and P,: similarly, R, may be resolved, at any point in its direction, into two pressures parallel and equal to R, and P,, of which R, may be resolved into two, parallel and equal to P, and P,, so that R, may be resolved at any point of its direction into three pressures parallel and equal tô P1, P., P ̧: and, in like manner, R, may be resolved into two pressures parallel and equal to R, and P,, and therefore into four pres sures parallel and equal to P,, P., P., P., and so of the rest.