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Denote this resultant by R; then, since the forces X, Y, and Z are perpendicular to each other, we shall have,

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To find the Co-ordinates of the point of application of R. Consider each of the forces, X, Y, and Z, with respect to the axis whose name comes next in order, and denote the lever arm of X, with respect to the axis of Y, by z1; that of Y, with respect to the axis of Z, by x1; and that of Z, with respect to the axis of X, by y1 We shall have as in

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in which x1, y1, and z1, are the co-ordinates of the point of application of R.

Denoting the angles which R makes with the axes by a, b, and c, respectively, we have, as in the preceding article,

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The values of X, Y, and Z, may be computed by means of Equations (20), and these being substituted in (21), make known the value of the resultant. The co-ordinates of its point of application result from Equations (22), and its line of direction is shown by Equations (23). The intensity, direction, and point of application being known, the resultant is completely determined.

Measure of the tendency to Rotation about the Axes.

46. Let X, Y, and Z denote the components of the resultant of the system, as in

0

Prosy

X

Prosp

Pcosa

the last article, and denote, as before, the co-ordinates of the point of application of the resultant by x1, y1, and z1. To find the resultant moment, with respect to the axis of Z, it may be observed that the component Z, can produce no rotary effect, since it is parallel to the axis of Z; the moment of the component Y, with respect to the axis of Z, is Yx1; the moment of the component X, with respect to the same axis, is -Xy, the negative sign being taken because the force X tends to produce rotation in a negative direction. Hence, the resultant moment of the system, with respect to the axis of Z, is,

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Fig. 30.

or, substituting for X and Y their values, we have,

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In like manner for the resultant moment of the system, with respect to the axis X,

Zy1- Yz1 == Σ(Pcosy y- Pcosẞ z) (25.)

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And for the resultant moment, with respect to the axis of Y,

Xz1- Zx1 = (Pcosa 2 - Pcosy x) (26.)

Equilibrium of Forces in a Plane.

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47. In order that a system of forces lying in the same plane, and applied at points of a free solid, may be in equilibrium, two conditions must be fulfilled: First, the resultant of the system must have no tendency to produce

motion of translation; and, secondly, it must have no tendency to produce motion of rotation. Conversely, if these conditions are satisfied, the system will be in equilibrium.

The first condition will be fulfilled, and will only be fulfilled, when the resultant is equal to 0; but from Art. 45, we have,

R = √X2 + Y3.

The value of R can only be equal to 0 when X = 0, and =0; or, what is the same thing,

Y =

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The second condition will be fulfilled, and will only be fulfilled, when the moment of the resultant, with respect to any point of the plane, is equal to 0, whence,

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Hence, from Equations (27) and (28), in order that a system of forces, lying in the same plane, and applied at points of a free solid body, may be in equilibrium, we must have,

1st. The algebraic sum of the components of the forces in the direction of any two rectangular axes separately equal to 0.

2d. The algebraic sum of the moments of the forces, with respect to any point in the plane, equal to 0.

Equilibrium of Forces in Space.

48. In order that a system of forces situated in any manner in space, and applied at points of a free solid body, may be in equilibrium, two conditions must be fulfilled. First, the forces must have no tendency to produce motion of translation; and secondly, they must have no tendency to produce motion of rotation about either of the three rectangular axes. Conversely, when these conditions are fulfilled, the system will be in equilibrium. The first condition will be

fulfilled, and will only be fulfilled, when the resultant is equal to 0. But, from Equation (21),

R = √X + Y2 + Z2.

That this value of R may be 0, we must have, separately,

X = 0, Y = 0, and Z

or, what is the same thing,

(Pcosa) = 0, (Pcos)

==

=

= 0;

0, and (Pcosy) = 0. (29.)

The second condition will be fulfilled, and will only be fulfilled, when the moments, with respect to each of the three axes, are separately equal to 0. This gives (Art. 46),

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Hence (Equations 29 and 30), in order that a system of forces in space applied at points of a free solid may be in equilibrium :

1st. The algebraic sum of the components of the forces in the direction of any three rectangular axes must be separately equal to 0.

2d. The algebraic sum of the moments of the forces, with respect to any three rectangular axes, must be separately equal to 0.

Equilibrium of Forces applied to a Revolving Body.

49. If a body is restrained by a fixed axis, about which it is free to revolve, we may take this line as the axis of X. Since the axis is fixed, there can be no motion of translation, neither can there be any rotation about either of the other two axes of co-ordinates. All of Equations (29), and the first and third of Equations (30), will be satisfied by virtue of the connection of the body with the fixed axis.

The second of Equations (30) is, therefore, the only one that must be satisfied by the relation between the forces. We must have, therefore,

Σ(Pcosy y- Pcos3 z) = 0. (31.)

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That is, if a body is restrained by a fixed axis, the forces applied to it will be in equilibrium when the algebraic sum of the moments of the forces with respect to this axis is equal to 0.

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