In the fundamental investigation of Art. 114 we may suppose AD, CF, and BE to be the distances of A, B, and C, not from a given straight line but from a given plane; either perpendicular distances or distances measured parallel to a given straight line. Then, as in Arts. 114 and 115, if we know the distances of the points of application of the parallel forces from a given plane, we can obtain the distance of the centre of the parallel forces from that plane.

120. The weight of a body may be considered to be the aggregate of the weights of the particles which compose the body. The weights of these particles form a system of like parallel forces, and such a system always has a centre; see Art. 111. The centre of the parallel forces which consist of the weights of the particles of a body is called the centre of gravity of the body.

Thus the centre of gravity is a particular case of the centre of parallel forces; but it is found convenient to give especial attention to this particular case, and accordingly we shall consider it in the next Chapter. It will be observed that the theory of the centre of gravity is rather simpler than the general theory of the centre of parallel forces, because the weights of the particles of a body are all like forces, and thus we shall not have to consider the second case of Art. 114.

121. The following examples contain an interesting result.

A

(1) ABC is a triangle; parallel forces act at B and C, each proportional to the opposite side of the triangle: determine the position of the centre of the parallel forces. First let the forces be like. In BC take D so that BD is to DC as the force at C is to the force at B, that is as AB is to AC; then D is the centre of the parallel forces.

B

E

Hence, by Euclid, vi. 3, the point D is such that AD bisects the angle BAC.

Next let the forces be unlike. Suppose AB greater than AC. Then, proceeding as before, we find that the centre of the parallel forces is at E on BC produced such that AE bisects the angle between AC and BA produced. See Euclid, VI. A.

(2) Parallel forces act at the angular points of a triangle, each force being proportional to the opposite side of the triangle: determine the position of the centre of the parallel forces.

First let the forces be all like. By the preceding example D is the centre of the parallel forces at B and C; hence the centre of all the three parallel forces lies on the straight line AD which bisects the angle BAC. Similarly the centre lies on the straight line which bisects the angle ABC, and on the straight line which bisects the angle BČA、 Therefore the centre of all the parallel forces must coincide with the centre of the circle inscribed in the triangle ABC.

Next let the forces be not all like. Suppose that the forces at B and C are unlike, and the forces at A and C like. By example (1) the centre of all the three parallel forces must lie on the straight line which bisects the angle between CA and BA produced, and also on the straight line which bisects the angle ABC, and on the straight line which bisects the angle between AC and BC produced. Therefore the centre of all the parallel forces must coincide with the centre of the circle which touches AC, and BA and BC produced. See Notes on Euclid, Book IV.

EXAMPLES. VIII.

1. A body is acted on by two parallel forces 2P and 5P, applied in opposite directions, their lines of action being 6 inches apart: determine the magnitude and line of action of a third force which will be such as to keep the body at rest.

2. Parallel forces P and Q act at two adjacent corners of a parallelogram: determine the forces parallel to these which must act at the other corners, so that the

centre of the four parallel forces may be at the intersection of the diagonals of the parallelogram.

3. A rod without weight is a foot long; at one end a force of 2 lbs. acts, at the other end a force of 4 lbs., and at the middle point a force of 6lbs., and these forces are all parallel and like: find the magnitude and point of application of the single additional force which will keep the rod at rest.

4. Equal like parallel forces act at five of the angular points of a regular hexagon: determine the centre of the parallel forces.

5. Find the centre of like parallel forces of 7, 2, 8, 4, 6 lbs. which act in order at equal distances apart along a straight line.

6. The circumference of a circle is divided into n equal parts, and equal like parallel forces act at all the points of division except one: find their centre.

7. Like parallel forces of 1, 2, and 3 lbs. act on a bar at distances 4, 6, and 7 inches respectively from one end: find their centre.

8. ABC is a triangle; parallel forces Q and R act at B and C such that Q is to R as tan B is to tan C: shew that their centre is at the foot of the perpendicular from A on BC.

9. Parallel forces act at the angular points A, B, C of a triangle, proportional to tan A, tan B, tan C respectively: shew that their centre is at the intersection of the perpendiculars drawn from the angles of the triangle on the oppo

site sides.

10. Parallel forces P, Q, R act at the angular points A, B, C of a triangle: shew that the perpendicular distance of their centre from the side BC is

11. Parallel forces P, Q, R act at the angular points

A, B, C of a triangle: shew that the distance of the centre from BC measured parallel to AB is

Р× АВ
P+Q+R'

12. Parallel forces P, Q, R act at the angular points A, B, C of a triangle: determine the forces which must act at the middle points of BC, CA, AB, so that the second system may have the same centre and the same resultant as the first.

13. Like parallel forces of 3, 5, 7, 5lbs. act at A, B, C, D, which are the angular points of a quadrilateral figure taken in order: shew that the centre and the resultant will remain unchanged if instead of these forces we have acting at the middle points of AB, BC, CD, DA respectively P, 10-P, 4+P, 6-P lbs., where P may have any value.

14. Parallel forces act at the angular points A, B, C of a triangle, proportional to a cos A, b cos B, c cos C respectively: shew that their centre coincides with the centre of the circumscribed circle.

15. Parallel forces P, Q, R act at the angular points A, B, C of a triangle; and their centre is at 0: shew that

Q

R

P
area of BOC area of COA area of AOB*

=

16. Parallel forces P, Q, R, S act at A, B, C, D; and

P

area of BCD

area of CDA area of DAB

S

area of ABC shew that their centre is at the intersection of AB and CD.

17. Find the centre of equal like parallel forces acting at seven of the angular points of a cube.

18. Parallel forces P, Q, R, S act at the angular points A, B, C, D of a triangular pyramid: shew that the perpendicular distance of their centre from the face BCD is 3 volume of pyramid

P

P+Q+R+S

X

area of BCD

IX. Centre of Gravity.

122. We begin with the following definition:

The centre of gravity of a body or system of bodies is a point on which the body or system will balance in all positions, supposing the point to be supported, the body or system to be acted on only by gravity, and the parts of the body or system to be rigidly connected with the point.

123. To find the centre of gravity of two heavy particles.

Let A and B be the positions of the two particles whose weights are P and Q respectively.

Join AB and divide it at L, so that AL may be to LB as Q is to P; then L is the centre of gravity.

A

L

B

For, by Art. 60, the resultant of the weights P and Q acts through L; and therefore if A and B are connected by a rigid rod without weight the system will balance in every position when L is supported.

As the resultant of P and Q is P+Q the pressure on the point of support will be P+Q.

124. To find the centre of gravity of any number of heavy particles.