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RESULTANT OF TWO FORCES ACTING ON A POINT.
SECT. 1. Geometrical Method.
IF two forces P, Q, act upon a point O, fig. (1), in the directions OA, OB, respectively, and if
then OC, the diagonal of the parallelogram OA CB, will coincide with the direction of the resultant R of the two forces, and
OA OB OC :: P: Q: R.
(1) Two forces of 6 and 8 lbs., acting on a point, have a resultant 10 lbs.: to find the angle between the two former forces.
Let O, fig. (1), be the point, OA, OB, the directions of the forces 6, 8, respectively: take OA=6 units of length, and OB=8 units of length: complete the parallelogram OACB, and join OC. Then OC must be equal to 10 units of length.
Since 62 +82 = 102, the square on OC must be equal to the square on OA together with that on AC: hence, by Euclid, the angle OAC, and therefore the angle AOB, that is, the angle between the two forces, must be a right angle.
(2) If the angle between two equal forces acting on a point be 120°, what is their resultant?
Let CA, CB, fig. (2), represent the two forces: complete the parallelogram CBDA, and join CD: then CD will represent the
and therefore, since ACB = 120°, ‹ DCA = 60° CDA: but the sum of the angles of the triangle ACD is 180°: hence also < CAD.= 60°, and the triangle CAD is equilateral. Thus CD is equal to CA or CB and bisects the angle ACB, that is, the resultant is equal to either of the component forces and bisects the angle between them.
(3) Two forces act on a point at right angles to each other and the resultant is double of the less: shew that the angle which the resultant makes with the less is double the angle which it makes with the greater.
Let CA, CB, fig. (3), represent the two forces: complete the rectangle BCAD and draw the diagonal CD, which will represent the resultant of CA, CB.
Bisect CD in O and join OA: then, since a circle can be described about the triangle CAD, having CD for its diameter, it is plain that OD, OA, which are radii, must be equal to each other: but AD = OD, by the hypothesis: hence the triangle AOD is equilateral and therefore equiangular.
Again, OA, OC, being radii, the triangle AOC is isosceles, and ACO= ▲ CAO.
The exterior angle AOD of the triangle ACO is equal to the angle ACO together with the angle CAO, and therefore to twice the angle ACO:
(4) At any point of a parabola forces are applied, represented in magnitude and direction by the tangent and normal at the point: to prove that, the forces being supposed to act both towards or both from the axis, the resultant will pass through the focus.
Let PT, PG, fig. (4), be the tangent and normal at any point P of the parabola. Then the force PT acting on P is equivalent to forces PS, ST, acting on P; and the force PG acting on Pis equivalent to forces PS, SG, acting on P, or, since the line SG is equal to the line ST, to forces PS, TS, acting on P.
Hence the forces PT, PG, acting on P, are equivalent to the force 2PS acting on P in the direction PS.
(5) Two forces, which act at a point of a curve, are represented in magnitude and direction by chords passing through the point; if the position of one of the chords and the direction of the resultant be given, to determine the positions of the other chord.
Let BAC, fig. (5), be the curve; let the chord AP, given in position, represent the force; and let the indefinite line AS represent the given direction of the resultant of the force AP and the required force. Produce PA to T, making AT equal to AP. Through T draw the line TQ parallel to AS, cutting the curve in any point Q. Draw QR parallel to TA, cutting AS in R: join PR. Then QR, AP, are equal to one another, being each equal to AT; they have also been drawn parallel to each other. Hence, by Euclid, AQ= PR. Thus the chord AQ represents a force which, combined with the force AP, corresponds to a resultant AR.
If the line TQ, produced if necessary, cuts the curve in any other points Q', Q", Q"",...; there will be other chords AQ', AQ", AQ",... which satisfy the conditions of the problem.
(6) Two forces act along opposite sides of a quadrilateral in a circle, towards the same parts, and are respectively proportional to these sides to prove that the resultant will pass through the intersection of the diagonals.