of Z by 7, 7', 7", &c. Resolving each force into components, respectively parallel to the three co-ordinate axes, and denoting the resultants of the groups in the directions of the respective axes by X, Y, and Z, we shall have, as in the preceding article, X = Σ (P cos a), Y = Σ (P cos 3), Z = Σ (P cos y.) If we denote the resultant of the system by R, and the angles which it makes with the axes by a, b, and c, we shall have, as in Article 31, The application of these formulas is entirely analogous to that of the formulas in the preceding article. Expression for the Resultant of two Forces. P' 35. Let us consider two forces, P and P', situated in the same plane. Since the position of the co-ordinate axes is perfectly arbitrary, let the axis of X be so taken as to coincide with the force P; a will then be equal to 0, and we shall have sin a = 0, and cos a = 1. The value of X (Equation 5), will Fig. 12. become P+P' cos a', and the value of Y will be Squaring these values, substituting (6), and reducing by the relation 1, we have, √P2 + P'2 + 2PP' cos a' (7.) The angle a' is the angle included between the given forces. Hence, The resultant of any two forces, applied at the same point, is equal to the square root of the sum of the squares of the two forces, plus twice the product of the forces into the cosine of their included angle. If we make a' greater than 90°, and less than 270°, its cosine will be negative, and we shall have, R VP2 + P2 2PP' cos a'. = 0, its cosine will be 1, and we shall RPP'. If we make a' = 90°, its cosine will be equal to 0, and we shall have, R = √P2 + P”. - If we make a' = 180°, its cosine will be 1, and we shall have, R P'. R The last three results conform to principles already deduced. Let P and Q be two forces, and R their resultant. The figure QP being a parallelogram, the side PR is equal to Q. From the triangle ORP we have, in accordance with the principles of trigonometry, Fig. 13. R P: QR:: sin ORP : sin ROP: sin OPR. (8.) If we apply a force R' equal and directly opposed to R, the forces P, Q, and R', will be in equilibrium. The angles ORP, and QOR', being opposite exterior and interior angles, are supplements of each other; hence, sin ORP sin QOR'. The angles ROP, and POR', are adjacent, and, consequently, supple R Fig 14. mentary; hence, sin ROP = sin POR'. The angles OPR, and POQ, are interior angles on the same side, and, consequently, supplementary; hence, sin OPR = sin POQ. We have also R = R'. Making these substitutions in the preceding proportion, we have, Fig 14. P: Q: R': sin QOR': sin POR': sin POQ. R Hence, if three forces are in equilibrium, each is proportional to the sine of the angle between the other two. 1. EXAMPLES. Two forces, P and Q, are equal in intensity to 24 and 30, respectively, and the angle between them is 105°. What is the intensity of their resultant ? R = √242 + 302 + 2 × 24 × 30 cos 105° = 33.21. 2. Two forces, P and Q, whose intensities are respectively equal to 5 and 12, have a resultant whose intensity is 13. Required the angle between them. 13 = √25 + 144 + 2 × 5 × 13 cos a. COS α = 0, or a 90°. Ans. 3. A boat is impelled by the current at the rate of 4 miles per hour, and by the wind at the rate of 7 miles per hour. What will be her rate per hour when the direction of the wind makes an angle of 45° with that of the current? R = √16 +49 + 2 × 4 × 7 cos 45° 10.2m. Ans. 4. A weight of 50 lbs., suspended by a string, is drawn aside by a horizontal force until the string makes an angle of 30° with the vertical. Required the value of the horizontal force, and the tension of the string. Ans. 28.8675 lbs., and 57.735 lbs. 5. Two forces, and their resultant, are all equal. What is the value of the angle between the two forces? 120°. 6. A point is kept at rest by three forces of 6, 8, and 11 lbs., respectively. Required the angles which they make with each other. SOLUTION. We have P = 8, Q 6, and R' = = 11. Since the forces are in equilibrium, we shall have R' = R = 11; hence from the preceding article, 11 = √64 +36 + 96 cos QOP; cos QOP; or, QOP = 77° 21′ 52". 36. The moment of a force, with respect to a point, is the product obtained by multiplying the intensity of the force by the perpendicular distance from the point to the line of direction of the force. The fixed point is called the centre of moments; the perpendicular distance is called the lever arm of the force; and the moment itself measures the tendency of the force to produce rotation about the centre of moments. pectively by p, q, and . Then will PP, Q, and Rr, be the moments of the forces P, Q, and R. Draw CO, and from P let fall the perpendicular PS, upon OR. Denote the angle ROP, by a, the angle ROQ, or its equal, ORP, by B, and the angle ROC by q. Since PR Q, we have from the right-angled triangles OPS and PRS, the equations, Multiplying both members of the first equation by sin, and both members of the second by cos q, then adding the resulting equations, we find, Substituting in the preceding equation, and reducing, we have, RrQqPp. When the point C falls within the angle POR, & α becomes negative, and the equation just deduced becomes |