Let AB be the given line, and C the given point; μ = the density

C

A

PQ

B

at a point in AB, whose distance from C = 1; a= CA, b = CB, ∞ = CP, Sx = PQ. Since a physical line is of uniform thickness throughout, we may take the length of any portion of it as the measure of the volume of that portion; hence the volume of PQ, and as the density varies as (distance from C)";

=

... 1" : " :: μ: μη".

Wherefore the density at P is ua", and PQ is ultimately of uniform density, therefore the mass of PQ is

Again, the moment of the mass of PQ about C,

n

dx

.. moment of AB about C = μfa+1dæ

between the same limits as before.

Wherefore being the distance of the centre of gravity

of the line from C, we have

Ex. 2. To find the centre of gravity of a triangular plate, of uniform thickness, the density of which at any point varies as the nth power of its distance from a line through the vertex parallel to the base.

Let ABC (fig. 37) be the triangle, CD a line through its vertex parallel to its base; μ the density at a point in the triangle at the distance 1 from CD; P, Q two points in AC very near each other, through which draw Pp, Qq parallel to the base; b = AC, c = AB, x = CP, dx = PQ, 0 = L CAB = ACD.

Then the density at every point in the line Pp = (a sin ()", which may be ultimately taken as the density at every point of the element Pq. We may regard Pq as a parallelogram, whose base Pp

=

and whose altitude is PQ sin 0 So. sin 0; its area, which we may take as the measure of its volume, is therefore

(b sin 0)2+1

Wherefore, if a line passing through the centre of gravity of the triangle, parallel to the base, cut AC at a distance from C, the distance of the centre of gravity from CD will be a sin 0, and

And if CE be drawn bisecting the base, the centre of gravity must be in that line; hence we have two lines passing through the centre of gravity, and consequently it is the point of their intersection.

Ex. 3. To find the centre of gravity of a quadrant of a circle, the density at any point of which varies as the nth power of its distance from the centre.

=

Let ABC (fig. 38) be the quadrant; CD, Cd two radii making angles with CA respectively equal to 0, 0+80; AC a, CP Cp = r, PQ = pq = dr; μ = density at the = distance 1 from the centre; therefore the density at Por p = μur". Now we may ultimately consider Pq as a parallelogram, whose sides are PQ and Pp, or dr and rde, and its area = rdr. 80, which may be taken as the measure of its volume; and its mass

.. moment of quad. about CB = √, fr (u m2 +2 cos 0).