Let ABCD, fig. (6), be the quadrilateral, O being the intersection of the diagonals, and let the forces act along AB, DC, or along BA, CD. Produce AB, DC, to meet in T, and draw OE, OF, parallel to DC, AB, respectively.

L

The angle OBE is equal to ¿AOB+¿BAO, and therefore to DOC+BAC, and therefore to DOC+BDC, or to DOC+ CDO, and therefore to OCF. Also BEO=CFO. Hence the triangles OBE, OCF, have the angles OBE, BEO, equal respectively to the angles OCF, CFO, and are therefore similar triangles. Again the triangles AOB, COD, are similar, since the angle BAO is equal to the angle CDO, and the angle AOB to the angle COD.

:: force along DC: force along AB.

Hence the resultant of these two forces must pass through 0.

(7) The resultant of two equal forces, applied at a given point, is represented in magnitude and direction by a given straight line drawn from the point: prove that the locus of the extremity of the straight line which represents either force is an indefinite perpendicular erected at the middle point of the given straight line.

(8) If ABC be a right-angled triangle and ABDE, ACFG, be the squares on the sides, constructed as in Euclid 1. 47; prove that the resultant of forces represented by CD, BF, is parallel to a diagonal of the rectangle the sides of which are AE, AG.

(9) Having given a line representing in length the difference between the resultant and one of two equal forces, which act upon a point at right angles to each other, shew how the lines representing the forces may be found by a geometrical construc

tion.

(10) Two forces act along opposite sides of a quadrilateral in a circle, towards opposite parts, and are respectively proportional to these sides; prove that their resultant passes through the intersection of the two other sides.

SECT. 2. Trigonometrical Method.

Let two forces P, Q, act upon a point, and let a be the angle between their directions: then, if R be their resultant, and if R's direction be inclined to the directions of P, Q, at angles o, Y, respectively,

The first and either of the last two of these formulæ determine the magnitude and the direction of the resultant.

If two forces X, Y, act at right angles to each other, and if their resultant R make angles λ, p, with X, Y, respectively, then

(1) If two forces, each equal to P, act upon a point, the angle between their directions being 60°, to find the magnitude of their resultant.

Let R denote their resultant: then

This question may be treated also as follows.

Let C, fig. (7), be the point at which the two forces act: draw Ca to bisect the angle between their directions, and draw y Cy' at right angles to Cx. Then one of the forces is equal to

Pcos 30° along Cx and Psin 30° along Cy, the other force being equal to Pcos 30° along Cx and Psin 30° along Cy': the components along Cy, Cy', destroy one another, and there remains for the resultant a force along Cx equal to 2P cos 30° = P√3.

(2) A cord PAQ is tied round a pin at the fixed point A, and its two ends are drawn in different directions by the forces P and Q to determine the angle between these directions, supposing the pressure on the pin to be equal to (P+ Q).

Let be the required angle: then, (P+ Q) being by the hypothesis the resultant of the two forces,

} (P+ Q)2 = P2 + Q2 + 2PQ cos 0,

which determines the value of 0.

(3) A string ACP, fig. (8), with a weight P at the end of it, is attached to a point A in a wall AB: from a point B of the wall a thin rod BC projects horizontally, pushing out the string: to find the pressure of the string on the rod.

The end C of the rod is acted on by the tension P of the portion CA of the string, in the direction CA, and by the tension P of the portion CP of the string vertically downwards.

Let ACB = a: then the angle between the two equal forces P, P, acting at C, is equal to +a: hence the pressure on the end of the rod is equal to

(4) A string is wrapped round a regular polygon, the tension of the string being given: to find the sum of the pressures on the angles of the polygon, and thence to determine the whole pressure on the circumference of a circle under like circum

stances.

Let n be the number of the sides of the polygon; 0, fig. (9), its centre; AB, BC, any two consecutive sides: join AO, BO,

CO. Let P be the tension of the string. Let R be the pressure on the angle B, being the resultant of the two equal forces P, P, acting along BA, BC. Then

Consequently the sum of the pressures upon the n angles is equal to

2 n.P sin.

Let no then the whole

n

pressure upon the circumference of the circle, into which the polygonal periphery degenerates,

(5) The directions of two forces, equivalent to 3 lbs. and 4 lbs. respectively, are inclined to each other at an angle of 60°: to determine the magnitude and direction of their resultant.

R=√(37), sin &=

2√3
√(37)'

sin

=

31/3 2√(37)

(6) If two equal forces are inclined to each other at an angle of 120°, shew that their resultant is equal to either of them.

(7) Two forces, acting on a particle, are inclined to each other at an angle of 30°, one force being equal to two pounds and the other to three pounds: to find the magnitude of their resultant.

The resultant is equal to

(13+ 6/3).

(8) If two forces, each equal to P, act at an angle of 135° to each other, shew that their resultant is equal to (2/2). P.

(9) Two forces of 9 and 12 pounds act upon a point in directions at right angles to each other: to find the magnitude of their resultant.

The resultant is equivalent to a force of 15 pounds.

(10) Two forces act at right angles to each other, one of them being 4 pounds: to find the other when the resultant is 5 pounds.

The other is a force of 3 pounds.

(11) If two forces of 5 and 12 pounds act at the same point at right angles to each other, what single force will produce the same result?

The single force required = 13 pounds.

(12) If the magnitudes of two forces are 6 and 11, and the angle between their directions 30°, shew that the magnitude of their resultant is 16.47 nearly.

(13) Shew that, in the preceding question, the resultant makes with the force 6 an angle the sine of which is approximately 3339, and with the force 11 an angle of which the sine is approximately 1821.

(14) Two forces of 1 and 2 pounds act upon a point: to find the angle between the forces, supposing the resultant to be /7.

The required angle is 60o.

(15) Apply two pressures of P pounds and P√/3 pounds so as to produce the same effect as a pressure of 2P pounds applied in a given direction.

The two forces P and P√3 must be at right angles to each other, their inclinations to the force 2P being 60° and 30° respectively.