Choosing the 40-pound force next, lay off CD to represent the magnitude and direction of that force and mark its line of action ed. Then OD (direction 0 to D) represents the magnitude and direction of the resultant of 00 and CD, and od (parallel to OD and passing through the point of concurrence of the forces 00 and CD) is the line of action of it.
Next lay off a line DE representing the magnitude and direction of the 60-pound force and mark the line of action de. Then OE (direction 0 to E) represents the magnitUde and direc tion of the resultant of OD and DE, and oe (parallel to OE and passing through the point of concurrence of the forces OD and DE) is the line of action of it.
It remains now to compound the last resultant (OE) and the first component (AO). AE represents the magnitude and direc tion of their resultant, and ae (parallel to AE and passing through the point of concurrence of the forces OE and AO) is the line of action.
32. Definitions and Rule for Composition. The point 0 (Fig. 31) is called a pole, and the lines drawn to it are called rays. The lines oa, ob, oc, etc., are called strings and collectively they are called a string polygon. The string parallel to the ray drawh to the beginning of the force polygon (A) is called the first string, and the one parallel to the ray drawn to the end of the force polygon is called the last string.
The method of construction may now be described as follows: 1. Draw a force polygon for the given forces. The line drawn from the beginning to the end of the polygon represents the magnitude and direction of the resultant.
2. Select a pole, draw the rays and then the string polygon. The line through the intersection of the first and last strings parallel to the direction of the resultant is the line of action of the resultant. (In constructing the string polygon, observe carefully that the two strings intersecting on the line of action of any one of the given forces are parallel to the two rays which are drawn to the ends of the line representing that force in the force polygon.) 1. Determine the resultant of the 50-, 70-, 80- and 120 pound forces of Fig. 5.
Ans. 260 pounds acting upwards 1.8 and 0.1 feet to the right of A and D respectively.
2. Determine the resultant of the 40-, 10-, 30- and 20 pound forces of Fig. 32.
Ans: end.
33. acting down 1g feet from left 33. Algebraic Composition. The algebraic method of com position is best adapted to parallel forces and is herein explained only for that case.
If the plus sign is given to the forces acting in one direction, and the minus sign to those acting in the opposite direction, the magnitude and sense of the resultant is given by the algebraic sum of the forces; the magnitude of the resultant equals the value of The algebraic sum; the direction of the resultant is given by the sign of the sum, thus the resultant acts in the direction which has been called plus or minus according as the sign of the sum is plus or minus.
If, for example, we call up plus and down minus, the alge braic sum of the forces represented in Fig. 32 is
40+ 10 30 20 + 50 15 = 45; hence the resultant equals 45 pounds and acts downward.
The line of action of the resultant is found by means of the principle of moments which is (as explained'in "Strength of Materials") that the moment of the resultant of any number of forces about any origin equals the algebraic sum of the moments of the forces. It follows from the principle that the arm of the resultant with respect to any origin equals the quotient of the algebraic sum of the moments of the forces divided by the result ant; also the line of action of the resultant is on such a side of the origin that the sign of the moment of the resultant is the same as that of the algebraic sum of the moments of the given. forces.
For example, choosing 0 as origin of moments in Fig. 32, the moments of the forces taking them in their order from left to right are Hence the algebraic sum equals 200 + 40 90 20 100 + 45 = 325 foot-pounds.
The sign of the sum being negative, the moment of the resultant about 0 must also be negative, and since the resultant acts down, its line of action must be on the left side of 0. Its actual distance from 0 equals 1. Make a sketch representing five parallel forces, 200, 150 100, 225, and 75 pounds, all acting in the same direction and 2 feet apart. Determine their resultant.
( Resultant = 750 pounds, and acts in the same Ans. ) direction as the given forces and 4.47 feet to the ( left of the 75-pound force.
2. Solve the preceding example, supposing that the first three forces act in one direction and the last two in the opposite direction.
(Resultant --=- 150 pounds, and acts in the same Ans. ) direction with the first three forces and 16.3 feet ( to the left of the 75-pound force.
Two parallel forces acting in the same direction can be com pounded by the methods explained in the foregoing, but it is sometimes convenient to remember that the resultant equals the sum of the forces, acts in the same direction as that of the two forces and between them so that the line of action of the resultant divides the distance between the forces inversely as their magni tudes. For example, let and F2 (Fig. 33) be two parallel forces. Then if R denotes the resultant and a and b its distances to F, and as shown in the figure, 34. Couples. Two parallel forces which are equal and act in opposite directions are called a couple. The forces of a couple cannot be compounded, that is, no single force can produce the same effect as a couple. The perpendicular distance between the lines of action of the two forces is called the arm, and the product of one of the forces and the arm is called the moment of the couple.
A plus or minus sign is given to the moment of a couple according as the couple turns or tends to turn the body on which it acts in the clockwise or counter-clockwise direction.