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statical couple, and all tendency to rotatory motion can be referred to forces forming such couples. If the forces are inclined at an angle a to AB, then P. AB. sin. a is the moment of the couple.

16. PROP. To find the resultant of any number of parallel forces which act at any points in one plane.

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Draw the lines AF, BG, CH, DK, parallel to Oy; and Ac, Cb, parallel to Ox. The triangles ACc, CBb, are similar, and

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(P1+P2) × CH = P1× AF+P2× BG

or, R1 × CH = P1× AF+P2×BG

Taking another force, P3, and compounding with R1 acting at C, we find the second resultant, R2 = R1+P3= P1+P2+P3, and R2x EL R1 × CH+P3 × DK

= P1x AF+ P2 × BG + P2 × DK

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and so onwards for any number of forces.

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the points A, B, D, &c., the points of application of the n forces, the above formula becomes

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and if a, ÿ, were the co-ordinates of the point of application of R, the resultant of the n forces, we should have

R=P1+P2+P3+ &c. . . . + Pn

and R.y = P1.y1+P2• Y2+P3• Y3+&c. ... +Pn•Yn

If we had drawn lines parallel to Ox, we should have found, in the same way,

R. P1.x1+P2. x2+P3.X3+ &c. . . . +Pn.n

These formulæ are often written more concisely by using the Greek letter as the sign of summation; and P being any one of the forces, a, g, being the co-ordinates of its point of application, then

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The point whose co-ordinates are π, ÿ, is called the center of parallel forces. It depends on the magnitude of the forces and their points of application; but is independent of the angle which their direction makes with any given line.

Secondly. When some of the forces act in opposite directions, they must be taken negative; and so also when the coordinates of the points of application are negative, they must be applied with their proper signs; and then the above formulæ will apply to all cases.

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Drawing Ce, Ed, parallel to Ox and meeting DK in d, EL produced in e, the triangles ECe, EDd, are similar, and

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or, R1(CH-EL) = P3(DK—EL)

or, (R1-P3). EL = R1 × CH-P2× DK

or, R, × EL = P1 × AF+P2 × BG−P3 × DK

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or, R1.ÿ = P1.y1-P2.y2, as we should have found by putting y2 with its proper sign in the general formula, which thus applies generally to all cases of parallel forces.

When some of the forces are negative and others positive, we may have the sum Σ(P) = 0;

or, R = 0

and the system of forces may be equivalent to a couple.

17. PROP. The algebraic sum of the moments of any number of parallel forces which act in one plane about any point in the plane, equals the moment of their resultant about that point.

Let P1, P2, P3, be parallel forces acting at the points A, B,

D, respectively; R1 and R2 the resultants, as before; G the

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=(P1+P2)Ge+P2 x cb-P1Ac... since cb=ef, de=Ac =R1 × Ge+ (P2 × CB−P1 × AC) cos. BAb

=R1x Ge=moment of the resultant about G

since P1× AC=P2× CB when C is the point of application of the resultant.

1

In the second figure, the algebraic sum of the moments about G

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=P2(Ge+ef)-P1(de-Ge)

=(P1+P2) Ge+P2 × ef-P1 × de

=R1x Ge... as before, since P2 xef=P1x de

If we take another force, P3, we find, in the same way, the moment of the second resultant, R2, equals the algebraic sum of

the moments of R1 and P3, or of P1, P2, and P3,

number of forces.

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EXAMPLES.

1. Two parallel forces acting in the same direction have their magnitudes 5 and 13, and their points of application 6 feet distant; shew that their resultant acts at a point 43 feet from the point of application of the force 5, and 1 feet from that of the force 13. What is the magnitude of the resultant?

2. If the forces in the last question act in opposite directions, shew that the point of application of their resultant is distant 3 feet from the point of application of force 13, and 93 feet from that of force 5. What is the magnitude of the resultant?

3. If two parallel forces, P and Q, act in the same direction at the points A and B, and make an angle ✪ with the line AB, shew that the moment of each of them about the point of appli

cation of their resultant:

=

P.Q
P+Q

AB sin. 0.

4. If three forces which act at a point be represented in direction and magnitude by the sides of a triangle taken in order, they will make equilibrium; shew that if, instead of acting at one point, they each act in the line which is the side of the triangle representing it, they are equivalent to a statical couple.

5. If three equal parallel forces act at the three angles of an equilateral triangle, shew that their center, or point of application of their resultant, is in the line drawn from any angle to the middle of the opposite side, and at the length of the line measured from the angle: being independent of the angle which the forces make with the plane of the triangle.

6. In question 4, if a, b, c, be the sides of the triangle opposite to the angles A, B, C, respectively, then the moment of the couple equals ab sin. C=ac sin. B=bc sin. A.

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