OD, OC must be equal and opposite, since OD represents the resultant of P and Q.

PQR0A: OB: OC=OA: AD: OD

= sin ODA sin AOD: sin OAD

19. The magnitude of the resultant R, of two forces P and Q, which act on a particle, and whose directions make an angle with one another, may be easily found. Let OA, OB represent the forces P, Q respectively. Complete the parallelogram OBCA, and join ‘OC: the latter represents R.

.._ R2 = P2 + Q2 + 2PQ cos 0.

Ex. 1. If forces of 3 lbs. and 4 lbs. have a resultant of 5 lbs., at what angle do they act? Ans. 90o.

Ex. 2. If one of two forces acting on a particle is 5 lbs., and the resultant is also 5 lbs., and at right angles to the known force, find the magnitude and direction of the other force.

Ans. 5/2 lbs., making an angle of 135° with the other force. Ex. 3. At what angle must forces P and 2P act on a particle in order that their resultant may be at right angles to one of them? Ans. 120o.

X

Ex. 4. If three forces, whose magnitudes are expressed by the numbers 3, 6, 9, act on a particle, and keep it at rest, shew that they must all act in the same straight line.

Ex. 5. Shew that three forces cannot maintain a particle in equilibrium if one of them be greater than the sum of the other two.

Ex. 6. Find the magnitude of the resultant of (i) 3 lbs. and 4 lbs. at 45o to one another; (ii) 2lbs. and 3 lbs. at 105o; (iii) 16 lbs. and 21 lbs. at 120o. Ans. (i) 6.56 lbs. (ii) 3.15 lbs. (iii) 19.04 lbs. Ex. 7. If the three forces in Ex. 4 act in directions represented by the sides of an equilateral triangle, taken in order: determine their resultant. Ans. A force 3√3, acting at right angles to the force 6.

Ex. 8. Three forces acting on a particle keep it in equilibrium: the greatest force is 5 lbs., and the least is 3 lbs., and the angle between two of the forces is a right angle: find the other force. Ans. 4 lbs.

Ex. 9. Two equal forces act at a certain angle on a particle, and have a certain resultant: also if the direction of one of the forces be reversed and its magnitude be doubled, the resultant is of the same magnitude as before: shew that the two equal forces are inclined at an angle of 60o.

Ex. 10. Determine the resultant of four forces of 5, 6, 9, 10 lbs. acting on a particle and represented in direction by OA, OB, OC, `OD, respectively, where O is the point of intersection of the diagonals of a square ABCD.

Ans. 4/2 lbs., in the direction bisecting the angle COD. Ex. 11. Forces P, P√3, and 2P act on a particle: find the angles between their respective directions that there may be equilibrium.

Ans. Between P and P√3, 90°; between P and 2P, 120°; between P√3 and 2P, 150o.

Ex. 12. Five equal forces act on a particle, in directions parallel to five consecutive sides of a regular hexagon taken in order; find the magnitude and direction of their resultant.

Ans. The direction is parallel to the third force, and the magnitude equal that of any one force.

20. The following proposition is sometimes useful.

If two forces acting on a particle be represented by m times the line OA, and n times the line OB, respectively,

their resultant is represented by (m +n) times the line OG, where G is the point between A and B, such that mAG= nBG.

By the triangle of forces mOA is equivalent to mGA and mOG, and the force nOB to nGB and nOG. But since mAG nBG, and they are opposite, these two forces counterbalance one another, so that we

=

with (m+n) OG only.

Fig.9

G

B

3

21*. Def. Let A,, A,, A,... A, be a series of points; join A,A,, and take B between them, so that

1

1

join BA, and take B, between them, so that

2

join BA, and take B, between them, so that

and so on until we arrive at B-1: this point is called the Centroid of A,, A„,  ̧........ „

2' 3

The centroid of the n points A1, A2...An is sometimes defined as the point whose distance from any plane is one nth the sum of the distances of A1, A2...A from that plane. We can easily shew that the definition of the centroid we have already given leads to this definition also.

Draw A1M1, AM, &c. B1N1, B2N2 &c. perpendicular to any given plane. Draw Anm parallel to M ̧Ñ12, and B1nm, parallel to N1NM.

This proves the statement for two and three points, and by the method of induction the proof can easily be extended to any number of points.

Note. The distances from the plane must be considered positive when they are on one side of it, negative when they are on the other. We may suppose that any number of the points become coincident: for instance, if A, and A, coincide with A1, B1 and B., will also coincide with A1, and Bg will be in the line A144, and such that B44=3B41. We may extend the idea of the centroid by supposing that some of the points are negative, in which case the process of finding the centroid will be somewhat modified: for instance, if A, be a negative point, B, will be in AB1, but beyond B1 not between B, and A ̧, and such that BA=2B,B1: as B2 is the centroid of two positive and one negative point B, will divide the line B1⁄4 équally. Also the distance of a negative point from a plane must be taken of opposite sign to what it would be if the point were a positive one, and in estimating the number of points, we must take the difference between the numbers of positive and negative points.

1

29

22*. Prop. If ОA,, ОA, OA, &c. ...... OA, represent a number of forces acting on a particle, their resultant will

r-1'

-1

be represented by r times the line OВ, where В is the centroid of A,, A...A..

2

For, by the last proposition, putting_m=n= 1, the resultant of OA, and OA, is 20B,; putting m = 1, n = 2, that of OA, and 20B, is 30B,, and so on, until we obtain rOB, as the final resultant.

After reading Chap. IV. it will be obvious that the centroid of a number of points is the Centre of Mass of equal particles situate one at each point. As a direct result of this proposition we see, that the resultant attraction or repulsion on a particle of any mass of which each particle attracts or repels with a force varying as its distance and its mass conjointly, is the same as the attraction or repulsion of the whole mass collected at its Centre of Mass.

Ex. 1. Find a point such that, if it be acted on by forces represented by the lines joining it to the vertices of a triangle, it will be in equilibrium.

The required point must be the centroid of the three points, i.e. (Art. 21) the point of intersection of the lines drawn from the vertices to the middle points of the opposite sides.

Ex. 2. O is any point in the plane of a triangle ABC, and D, E, F are the middle points of the sides. Shew that the system of forces OA, OB, OC is equivalent to the system OD, OE, OF.

It can be shewn that the centroid of the points A, B, C is also that of D, E, F.

: Ex. 3. The circumference of a circle is divided into a given number of equal parts, and forces acting on a particle are represented by straight lines drawn from any point to the points of division: shew that their resultant passes through the centre of the circle, and that its magnitude varies as the distance of the point from the centre.

The centre of the circle is clearly the centroid of the points.

Ex. 4. AOB and COD are chords of an ellipse parallel to conjugate diameters: forces are represented in magnitude and direction by OA, OB, OC, OD: shew that their resultant is represented in direction by the straight line which joins O to the centre of the ellipse, and in magnitude by twice this line.

The centroid of the points A, B, C, D is midway between O and the

centre.