9. Graphical Composition of Two Concurrent Forces. If two forces are represented in magnitude and direction by AB and BC {Fig. 5), the magnitude and direction of their resultant is represented by AC. This is known as the "triangle law."

The line of action of the resultant is parallel to AC and passes through the point of concurrence of the two given forces; thus the line of action of the resultant is ac.

The law can be proved experimentally by means of two spring balances, a drawing board, and a few cords arranged as shown in Fig. 4. The drawing board (not shown) is set up vertically, then from two nails in it the spring balances are hung, and these in turn support by means of two cords a small ring A from which a heavy body (not shown) is suspended. The ring A is in equilibrium under the action of three forces, a downward force equal to the weight of the suspended body, and two forces exerted by the upper cords whose values or magnitudes can be read from the spring balances. The first force is the equilibrant of the other two. Knowing the weight of the suspended body and the readings of the balances, lay off AB equal to the pull of the right-hand upper string according to some convenient scale, and BC parallel to the left-hand upper string and equal to the force exerted by it. It will then be found that the line joining A and C is vertical, and equals (by scale) the weight of the suspended body. Hence AC, with arrowhead pointing down, represents the equilibrant of the two upward pulls on the ring; and with arrowhead pointing up, it represents the resultant of those two forces. Fig. 3. Fig. 4.

Notice especially bow the arrowheads are related in the triangle (Fig. 3), and be certain that you understand this law before proceeding far, as it is the basis of most of this subject.

Examples. Fig. 5 represents a board 3 feet square to which forces are applied as shown. It is required to compound or find the resultant of the 100- and 80-pound forces.

First we make a drawing of the board and mark upon it the lines of action of the two forces whose resultant is to be found, as in Fig. 6. Then by some convenient scale, as 100 pounds to the inch, lay off from any convenient point A, a line AB in the direction of the 100-pound force, and make AB one inch long, representing 100 pounds by the scale. Then from B lay off a line BC in the direction of the second force and make BC, 0.8 of an inch long representing 80 pounds by the scale. Then the line AC, with the arrow pointing from A to C, represents the magnitude and direction of the resultant. Since AC equals 1.06 inch, the resultant equals Fig. 5.  Fig. 6.

1.06 X 100 = 106 pounds.

The line of action of the resultant is ac, parallel to AC and passing through the intersection of the lines of action (the point of concurrence) of the given forces. To complete the notation, we mark these lines of action ab and ho as in the figure.

Examples For Practice

1. Determine the resultant of the 100- and the 120-pound forces represented in Fig. 5.

 Ans. The magnitude is 188 pounds; the force acts upward through A and a point 1.62 feet to the right of D.

2. Determine the resultant of the 120- and the 160-pound forces represented in Fig. 5.

 Ans. The magnitude is 200 pounds; the force acts upward through A and a point 9 inches below C.

10. Algebraic Composition of Two Concurrent Forces. If the angle between the lines of action of the two forces is not 90 degrees, the algebraic method is not simple, and the graphical is usually preferable. If the angle is 90 degrees, the algebraic meth-od is usually the shorter, and this is the only case herein explained. Let F1 and F2 be two forces acting through some point of a body as represented in Fig. 7a. AB and BC represent the magnitudes and direction of F1 and F2 respectively; then, according to the triangle law (Art. 9), AG represents the magnitude and direction of the resultant of F1 and F2, and the line marked R (parallel to AC) is the line of action of that resultant. Since ABC is a right triangle,

(AC)2 = (AB)2 +(BC)2 and, tan CAB = BC/AB

* Use sheets of paper not smaller than large letter size, and devote a full sheet to each example. In reading the answers to these examples, remember that the board on which the forces act was stated to be 3 feet square. Fig. 7.

Now let R denote the resultant. Since AC, AB, and BC represent R, F1, and F2 respectively, and angle CAB = x,

R2 = F12 + F22; or R = and, tan x = F2 ÷ F1.

By the help of these two equations we compute the magnitude of the resultant and inclination of its line of action to the force F1.

Example. It is required to determine the resultant of the 120- and the 100-pound forces represented in Fig. 5.

Let us call the 160-pound force F1 ; then,

R= = = = 200 pounds; and, tan x = 120/160 = ¾ ; hence x = 36° 52'.

The resultant therefore is 200 pounds in magnitude, acts through A (Fig. 5) upward and to the right, making an angle of 36° 52' with the horizontal.

Examples For Practice

1. Determine the resultant of the 50- and 70-pound forces represented in Fig. 5.

 Ans. R = 8b pounds; angle between R and 70-pound force - 35° 32'.

2. Determine the resultant of the 00- and 70-pound forces represented in Fig 5.

 Ans. R = 92.2 pounds ; angle between R and 70-pound force = 40° 36'.

11. Force Polygon. If lines representing the magnitudes and directions of any number of forces be drawn continuous and so that the arrowheads on the lines point the same way around on the series of lines, the figure so formed is called the force polygon for the forces. Thus ABCD (Fig. 8) is a force polygon for the 80-, 90-, and 100-pound forces of Fig. 5, for AB, BC, and CD represent the magnitudes and directions of those forces respectively, and the arrowheads point in the same way around, from A to D.