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Proof: Let a, b, c,... be the resultant masses of the partial systems and a, ẞ, y their position vectors from 0. Then the mass-centre of the partial systems is given

by

=

μ (a+b+c+...) aa + bẞ+cy +.........(1). But if a1, a2, aз... be the masses of the first partial system, and α1, a, a... the position vectors, then

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Similarly if b1, b2, b... be the masses of the second, C1, C2, C3... the masses of the third &c. and if B1, B2, B3... V1, V2, Y3... &c. be the position vectors, then

Ιβ = Σοβι

γ= Σαγι

therefore substituting in the equation (1) we get

μ (α1 + α2+... b1 + b2 + ... C1 + C2+.....)

2

= (α2α1 + α2α2+ ... b1ß1 + b2ß2 + ... C171 + C2Y2 + ...), which is the equation for determining the mass-centre of the original system.

The theorem really says nothing more than that the associative law holds for the addition of mass-vectors.

74. A few of the many geometrical theorems which are immediate consequences of the properties just proved for the mass-centre are given below.

(i) The three median lines of a triangle meet in a point and divide each other in the ratio I: 2.

Suppose equal masses placed at A, B, C the vertices of the triangle, then the mass-centre of B and C lies at A' their mid-point, then the mass-centre of the three is that of A and A' with masses 1:2. The mass-centre of the three is therefore at M which divides AA' so that A'M: MA = 1:2. The same is true for the other medians, these therefore pass through the mass-centre.

(ii) In a quadrilateral the three lines which join the mid-points of opposite sides and of the two diagonals respectively meet in a point and bisect each other.

The point of intersection of the three lines is the masscentre of four equal masses at the vertices of the quadrilateral; on dividing these four points in groups of two the above follows.

(iii) If we suppose masses placed at the vertices of a triangle proportional to the lengths of opposite sides, then by finding the position vector of the mass-centre of these points we can shew at once that the three bisectors of the angles meet at a point.

(iv) If A, B, C, D are four points in space, they form the vertices of a tetrahedron. Equal masses at these points can be divided three times into two pairs. Thus the masses at A and D can be concentrated at E,

(1)A

E

D (1)

M/

B(1)
Fig 48.

the mid-point of AD,
the masses at B and C
can be concentrated at
F, the mid-point of BC;
hence the mass-centre
of the whole is at M the

mid-point of EF. This
process can
be gone
through in three ways,
hence the theorem :-

The three lines joining the mid-points of opposite edges of a tetrahedron meet in a point and bisect each other.

If we take three points in one group, we get:

The four lines which join the vertices of a tetrahedron each to the mass-centre of the three points of the opposite face, meet in a point and divide one another in the ratio 1 : 3.

If we consider the planes in which the mass-centre must lie, we get:

The three planes through concurrent edges of a tetrahedron which bisect each the opposite edge meet in a line; the four lines thus obtained meet in a point.

75. Theorem: The mass-centre of a number of mass-points in a plane lies in that plane.

Proof: Take the origin in the plane, then the vector polygon and the resultant vector will lie in that plane.

76.

Scalar Equations for the determination of the mass-centre.

...

Let m, m2, M3, be the masses, a, B, y, ... their position vectors and μ the position vector of the masscentre. Then each vector may be decomposed in two fixed directions in the plane given by the orts and λ (§ 46), so

that

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and y being the coordinates of the mass-centre.

Hence the equation

becomes

Ση = m1a + m2ß + M3Y + ...

...

Σm (ãк + ÿλ) = m1 (x ̧ñ + î1λ) + M2 (X2K+Y2λ) + which, since and λ are independent, splits up into the two scalar equations

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Definition. If the axes are perpendicular, then mix is called the moment of the mass m1 about the axis of y, and Σm the moment of the resultant mass about the same axis.

=

The equation aΣm Σmx, says-multiply each mass by its distance from a given line (measured in a given direction), add all these products together and their sum is equal to the product of the resultant mass and its distance from the line.

Also:-The sum of the moments about any axis of all the masses equals the moment of the resultant mass, for :If the axis passes through the mass-centre then ≈= 0. Hence for every line through the mass-centre the sum of the moments of all masses vanishes, or

Ση1 = 0.

77. If the points are not in a plane, then the vectors

Fig. 50.

N

then the equation

becomes

y

a, B, y can be decomposed in 3 fixed directions given. by 3 orts κ, λ, μ, see § 51. In figure

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three scalar equations for determining the mass-centre.

If the orts are mutually perpendicular, then x, y, z, are the perpendiculars from P on the three planes yОz, zОx, xOy, and mz, is called the moment of my about the plane xOy.

The equation mž = Σm1z1 says—multiply each mass by its distance (measured in a given direction) from any plane, add all these products together and their sum is equal to the product of the resultant mass and its distance from that plane.

≈ thus determines a plane parallel to xOy in which the mass-centre must lie, x and y determine two other planes parallel to yOz and zOx respectively, the mass-centre is therefore the point of intersection of these three planes.

EXERCISES V.

(1) Taking six points at random on a sheet of paper, and supposing them to have masses 1, 2, 3, 4, 5, 6, verify by drawing that the position of the mass-centre is independent of the origin chosen for finding it.

(2) Starting from the mass-centre found in (1), shew thát μ=0. (3) Verify the theorem of article, § 68, by dividing the masses in (1) into three partial systems.

(4) A, B, C, D are any four points; shew that their mass-centre is the point of intersection of the lines joining the mass-centres of DBC, DAB, and DAC to A, C, and B respectively.

A, B

(5) A, B, C, D, E, F are any six points; prove that, if P, Q, R, S be the mass-centres of A, B, C; C, D, E; D, E, F; F, tively, then PQRS is a parallelogram.

respec

(6) The mass and the position of a number of mass-points in a line being as given in table, find the distance of the mass-centre from the origin.

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(7) Four mass-points having masses 2, 5, 7, 10 are in a line, the distances between them in the above order being 12, 3, and 8 inches; find the mass-centre.

(8) The distances of a number of mass-points in a line from a fixed point in the line are 3, 3, 4, 2, and −1 ft., the masses being 1, 8, 3, 5, 7; find the mass-centre.

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