To do this, we first find the centre of gravity of each body separately, by the rules already given. The weight of each body may then be regarded as a force applied at the centre of gravity of the body. The weights being parallel, we have a system of parallel forces, whose points of application are known. If these points are all in the same plane, we may find the lever arms of the resultant of all the weights, with respect to two lines, at right angles to each other in that plane; and these will make known the point of application of the resultant, or, what is the same thing, the centre of gravity of the system. If the points are not in the same plane, the lever arms of the resultant of all the weights may be found, with respect to three axes, at right angles to each other; these will make known the point of application of the resultant weight, or the required position of the centre of gravity.

1. Required the point of application of the resultant of three equal weights, applied at the three vertices of a plane triangle.

SOLUTION.

Let ABC (Fig. 34) represent the triangle. The resultant of the weights applied at B and C will be applied at D, the middle point of BC. The weight acting at D being double that at A, the total resultant will be applied at G, making GA 2GD; hence, the required point is at the centre of gravity of the triangle.

=

2. Required the point of application of the resultant of a system of equal parallel forces, applied at the vertices of any polygon?

Ans. At the centre of gravity of the polygon.

3. Parallel forces of 3, 4, 5, and 6 lbs., are applied at the successive vertices of a square, whose side is 12 inches. At what distance from the first vertex is the point of application of their resultant?

SOLUTION.

Take the sides of the square through the first vertex as axes of moments; call the side through the first and second vertex the axis of X, and that through the first and fourth the axis of Y. We shall have from Formulas (13),

Denoting the required distance by d, we have,

4. Seven equal forces are applied at seven of the vertices of a cube. What is the distance of the point of application of their resultant from the eighth vertex?

SOLUTION.

Take the eighth vertex as the origin of co-ordinates, and the three edges passing through it as axes of moments. We shall have from Equations (13), denoting one edge of the cube by a,

4a, y1 =
Y1 4a, and z = 4a.

Denoting the required distance by d, we have,

5. Two isosceles triangles are constructed on the same side of the base b, having altitudes respectively equal to h and h', h being greater than h'. Where is the centre of gravity of the space lying between the two triangles?

SOLUTION.

It must lie on the altitude of the greater triangle. Take the common base as an axis of moments; then will the moments of the triangles be, respectively, bh × 1h, and

bh' × }h'; and from the first of Formulas (13), we shall

That is, the required centre of gravity is on the altitude of the greater triangle, at a distance from the common base equal to one-third of the difference of the two altitudes.

included

6. When is the centre of gravity of the space between two circles tangent to each other internally?

SOLUTION.

Take their common tangent as an axis of moments. The centre of gravity will lie on the common normal, and its distance from the point of contact is given by the equation,

7. Let there be a square, and suppose it divided by its diagonals into four equal parts, one of which is removed. Required the distance of the centre of gravity of the remaining figure from the opposite side of the square.

7

Ans. of the side of the square.

8. To construct a triangle, having given its base and centre of gravity.

SOLUTION.

Draw through the middle of the base, and the centre of gravity, a straight line; lay off beyond the centre of gravity a distance equal to twice the distance from the middle of the base to the centre of gravity. The point thus found is the vertex.

9. Given the base and altitude of a triangle. Required the triangle, when its centre of gravity is perpendicularly over the extremity of the base.

10. Three men carry a cylindrical bar, one taking hold

of one end, and the others at a common point. Required the position of this point, in order that the three may sustain equal portions of the weight.

Pressure of one body upon another.

72. Let A be a movable body pressed against a fixed body B, and touching it at a single point. In order that A may be in equilibrium, the resultant of all the

forces acting upon it, including its weight, must pass through the point of contact, P'; otherwise there would be a tendency to rotation about P', which would be measured by the

P

moment of the resultant with respect to this point. Furthermore, the direction of the resultant must be normal to the surface of B at the point P', else the body A would have a tendency to slide along the body B, which tendency would be measured by the tangential component. The pressure upon B develops a latent force of reaction, which must be equal and directly opposed to it. The resultant of all the forces must be equal to zero (Art. 31). That is, when a body, resting upon another and acted upon by any number of forces is in equilibrium, the resultant of all the forces called into play is equal to 0.

If all the forces called into play are taken into account, the algebraic sums of their moments with respect to any three rectangular axes will be separately equal to 0.

Equations (29) and (30) are, then, perfectly general in every case of equilibrium, provided all of the forces called into play are taken into account.

Stable, Unstable, and Indifferent Equilibrium.

73. A body is in stable equilibrium when, on being slightly disturbed from its state of rest it has a tendency to

return to that state. This will, in general, be the case when the centre of gravity of the body is at its lowest point. Let A be a spherical body suspended from

Fig. 46.

an axis 0, about which it is free to turn. When the centre of gravity of A lies vertically below the axis, it is in equilibrium, for the weight of the body is exactly counterbalanced by the resistance of the axis. Moreover, the equilibrium is stable; for if the body be deflected to A', its weight tends to restore it to its position of rest, A. The measure of this tendency is W × OP, that is, the moment of the weight with respect to the axis 0. Under the action of the force W, the body will return to A, and, passing to the other side by virtue of its inertia, will finally come to rest and return again to A', and so on, till after a few vibrations, when it will come to rest at A.

A body is in unstable equilibrium when, being slightly disturbed from its state of rest, it tends to depart still farther from it. This will, in general, be the case when the centre of gravity of the body occupies its highest position. Let A be a sphere, connected by an

inflexible rod with the axis 0. When the centre of gravity of A lies verti cally above O, it will be in unstable equilibrium; for, if the sphere be deflected to the position A', its weight will act with the lever arm OP to increase this deflection. The motion will continue till, after a few vibrations, it comes to rest below the axis. In this last position, it will be in stable equilibrium.

Fig. 47.

A body is in indifferent, or neutral, equilibrium when it remains at rest wherever it may be placed.

This will, in

general, be the case when the centre of gravity continues in the same horizontal plane on being slightly disturbed.