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 resultant 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 construction, 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.

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 P1, P2, P3, acting in the directions PO, PO, PO, 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 P2 and P3, which tend to turn the plane in one direction about A, equal the moment of P1, which tends to turn it in the opposite direction.

Through A draw DAB parallel to OP,, and produce OP2 to meet it in D. Take OD to represent P2, and take DB of 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.

2

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, A ODA+ AOAB=A ODB=A OBC, ... A ODA+A OAB=A OAC,

... OD × AN+ OB× AL= OC × AM,

1

2

3

3

Now P1 × AM, P2 × AN, P ̧ × AL, are the moments of P1, P2, P3, about A (Art. 4.)

6. If R be the resultant of P2 and P3, then since R is equal to P, and acts in the same straight line, m1R=m*P1, ...mt P2+mt P2=mt R. (8)

2

The sum of the moments therefore, about any point, of two pressures, P2 and P3 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 P1 and P3, which tend to turn the plane in opposite directions about A, &c. Then is R equal to P2, and in the same straight line with it, therefore moment Ris equal to moment P2.

1

But by equation (1) mt P1-m* P ̧=m* P2; ... m2 P1 - m* P ̧

=mt R.

2

Generally therefore, mt P, + m2 P2 = mt R ..... (2), the moment therefore of the resultant of any two pressures in the same plane is equal to the sum or difference of the oments 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, P3. . . . . 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 moments of those which tend to turn it in the opposite direction.

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

Therefore by the last proposition, it being understood that the moments of those of the pressures P1, P2, which tend to turn the plane to the left of A, are to be taken negatively, we have

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

1

3

m2 R1_1=m2 P1+ m2 P2+ m2 P ̧ + . . . . . + m2 P„. . . (3), where R is the resultant of all the pressures P1, P2,

. . P.

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

... m2 P1 + m2 P2 + m2 P2 + . . . . . mt P = 0.. . . (4).

3

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 P3, of which R, may be resolved into two, parallel and equal to P, and P2, so that R, may be resolved at any point of its direction into three pressures parallel and equal to P1, P2, P3: and in like manner, R, may be resolved into two pressures parallel and equal to R, and P1, and therefore into four pressures parallel and equal to P1, P2, P3, P41 and

so of the rest. Therefore R-1 may at any point of its direction be resolved into n pressures parallel and equal to P1, P2, P3, . . . . . P,; if, therefore, n such pressures were applied to that point, they would just be held in equilibrium by a pressure equal and opposite to R. But R1_=0; these n pressures would, therefore, be in equilibrium with one another if applied to this point.

Now it is evident, that if being thus applied to this point, they would be in equilibrium, they would be in equilibrium if similarly applied to any other point. Therefore, &c.

THE POLYGON OF PRESSURES.

9. The conditions of the equilibrium of any number of pressures applied to a point.

Let OP, OP, OP3, &c. represent in magnitude and direction pressures P1, P2, &c. applied to the same point O. Complete the parallelogram OP, AP2, and draw its diagonal OA; then will OA represent in magnitude and direction the resultant of P, and P2. Complete the parallelogram OABP3, then will OB represent in magnitude and direction the resultant of OA and P3; but OA is the resultant of P, and P2, therefore OB is the resultant of P1, P2, P3; similarly, if the parallelogram OBCP, be completed, its diagonal OC represents the resultant of OB and P1, that is, of P1, P2, P3, P1, and in like manner OD, the diagonal of the parallelogram OCDP, represents the resultant of P1, P2, P3, P4, P5.

Now let it be observed, that AP, is equal and parallel to OP2, AB to OP3, BC to OP, CD to OP, so that P,A, AB, BC, CD, represent P2, P3, P4, P, respectively in magnitude, and are parallel to their directions. Moreover OP, is in the direction of P1 and represents it in magnitude, so that the sides OP,, P,A, AB, BC, CD, of the polygon OP, ABCDO, represent the pressures P1, P2, P3, P4, P5, respectively in magnitude, and are parallel to their directions; whilst the side