82. DEF. The resultant of two distances from a given point is the one distance which is the combined effect or sum of the two distances, taking into account their direction as well as their magnitude. 83. PROP. The parallelogram of Distances. The resultant of two given distances OA, OB is OR, the diagonal of the parallelogram AOB. This follows because the opposite sides of a parallelogram are equal. In the figure below the distance OP is the resultant of the distances OM and MP; that is, of OM and ON, for MP is equal and parallel to ON. 84. When the two distances are in the same direction their resultant coincides with the two distances placed end and end, so that its magnitude is their sum. When the two distances are in opposite directions the magnitude of their resultant is their difference. 85. It is often convenient to express the distance of one point from another as the resultant of two distances in given directions. Example. Express the distance OP as the resultant of two distances in directions parallel to OA and OB respectively. N M Through P draw PM and PN parallel to OB and OA respectively. 86. When the one distance OP is expressed as the sum of two others OM and ON, the two distances OM and ON are called the components of the distance OP. The distance OP is said to be resolved into its two components OM and ON in those directions. The distance OP is clearly the resultant of its two components OM and ON. It is usually convenient to take the two component directions at right angles to each other. When the two components are mutually at right angles, each component is called the resolved part of the resultant in its own direction. Example i. The road from A to B runs in a straight line for 2 miles from A to H; then after turning through an angle of 30°, it runs in a straight line for 3 miles from H to K; then after turning through an angle of 15°, runs in a straight line for i mile from K to B; how far is it from A to B as the crow flies? Resolve the distance HK into its two resolved parts HN and HI.=NK along and perpendicular to AH. Resolve the distance KB into its two resolved parts KR =NM and NK=HK sin 30=1 x 3 miles, V2*1 mile, I I RB= KB sin 45o= x 1 mile. N2 Hence, the number of miles in AM is 2+1 3+3+1/2. The number of miles in MB is ** 3+2. Therefore, the number of miles in AB is {(2+1x343+1/2) + ( +52)} ={(2+ * 3 * 1*7320... + 1 x 14142...)" + (1'5+1 * 1'4142...)"} = {(5*305...)2 + (2-207...)"}=5'746... nearly. Example ii. The distance of B from A is the resultant of the three distances AP, AP', AP" of 2 miles, 3 miles and 1 mile respectively where the angle PAP'=30°, and the angle P'AP” = 15°; what is the distance of B from A? It will be found that this is only another way of stating Example i. EXAMPLES. XXIV. 1. Resolve a distance of 100 yards into two components mutually at right angles of which makes an angle of 30° with it. 2. Resolve a distance of 25 yards into two components mutually at right angles one of which makes an angle whose sine is with it. 3. A man walks 200 yards in a straight line, then turns through an angle of 45° and walks 150 yards further in a straight line; find his resultant distance from the starting point. 4. Find the resultant of two distances of 300 yards and 450 yards making an angle of 45° with each other. 5. I walk 1 mile towards the north-east; what is the resolved part of this distance in the northerly direction? what is the resolved part of this distance towards the north-west ? 6. Find the resolved part of a distance of 360 yards in the direction which makes an angle of 60° with it. 87. When a point has a given uniform velocity u velos in a given direction, then after an interval t seconds the point is at a certain distance from its initial position; this distance is ut feet, and is in the direction of the velocity. This distance is called the distance due to the velocity. 88. A point is said to have simultaneous velocities and accelerations when its distance after an interval from its initial position is the resultant of the distances due to those velocities and accelerations separately. Example. A point has vertically – 32 celos, and 200 velos, and it has 40 velos in a horizontal direction; find its position after 3 seconds. Let O be its initial position. From N draw NQ vertically upwards to represent 3 x 200 ft. and from Q draw QP vertically downwards to represent * x 32 x 9 ft. Then P represents the position of the point after 5 seconds. EXAMPLES. XXV. Find the position of a point having the following; 4. 20 velos northwards, 15 velos eastwards and 15 velos northwards after 8 secs. 5. 20 velos and – 3 celos westwards after ro seconds. 6. Three velocities of 20 velos each making an angle of 120° with cach other, after 10 seconds. 7. 20 velos northwards and 32 celos westwards after 2 secs. 8. Three accelerations, 15 celos westwards, 20 celos northwards and 15 celos eastwards after i second. 89. The motion of a projectile affords an important example of a point which after any interval has two simultaneous distances from its initial position. Example. A particle is projected with u velos in a direction making an angle a with the horizontal plane; find its position after t seconds. The only force acting on the particle is its own weight, which acts vertically downwards. By Art. 33 this force produces in the mass the same acceleration whether the particle has other velocity or not. Hence, the distance of the particle after t seconds from its initial position is the resultant of two distances, one due to the u velos, the other due to the acceleration g celos vertically downwards. Let O be the initial position of the particle P; let OH be a horizontal line; then we may find the position of P after t seconds thus : Drawing OQ, making an angle a with OH, so that OQ represents a distance of ut feet; from Q draw QP vertically downwards, so that QP represents £gt2 feet. Then P is the position of the particle after t seconds. For OQ is the distance due to the velocity u velos, and QP is the distance due to the acceleration g celos. |