If the components acting in the same direction along either of the two lines be given the plus sign and those acting in the other direction, the negative sign, then it follows from the foregoing that the condition of equilibrium for a concurrent system is that HIGH SCHOOL AT THREE RIVERS, MICH.

J. C. Llewellyn, Architect, Chicago, 111. Basement Plan Shown on Opposite Page. HIGH SCHOOL AT THREE RIVERS, MICH.

J. C. Llewellyn, Architect, Chicago, 111. Second-Floor Plan Shown on Opposite Page. For Exterior and Plan of First Floor. See Page 138 the algebraic sums of the components of the forces along each of two lines at rigid angles to each other must equal zero.

Examples. 1. It is required to determine the pull on the rope and the pressure on the plane in Example 1, Art. 16 (Fig. 12), it being given that the inclination of the plane to the horizontal is 30 degrees.

Let us denote the pull of the rope by F1 and the pressure of the plane by F2. The angles which these forces make the horizontal are 303 and 60° respectively; hence the horizontal component of F1 = F1 X cos 30° = 0.8660 F1 and " " " " F2 = F2 X cos 60° =0.5000 F2; also " " " " the weight = 0.

The angles which F1 and F2 make with the vertical are 60° and 30° respectively, hence the vertical component of F1 = F1 X cos 60o = 0.5000 F1 and the vertical component of F2 = F2 X cos 30° = 0.8660 F2; also the vertical component of the weight = 120.

Since the three forces are in equilibrium, the horizontal and the vertical components are balanced, and hence

0.866 Fl = 0.5 F2 and 0.5 F1 + 0.866 F2 = 120.

From these two equations F1 and F2 may be determined; thus from the first,

F2 = 0.866/0.5 F1= 1.732 F1 2 0.5

Substituting this value of F2 in the second equation we have

0.5 F1 + 0.866 X 1.732 F1 = 120, or 2 F1 = 120; hence, F1 = 120/2 = 60 pounds, and F2 = 1.732 X 60 = 103.92 pounds.

2. It is required to determine the pulls in the ropes of Fig. 13 by the algebraic method, it being given that the angles which the left- and right-hand ropes make with the ceiling are 30 and 70 degrees respectively and the body weighs 100 pounds.

Let us denote the pulls in the right- and left-hand ropes by F1 and F2 respectively. Then

 the horizontal component of F1 = F1 X cos 70o = 0.342 F1, the horizontal component of F2 = F2 X cos 30° = 0.866 F2, the horizontal component of the weight = o, the vertical component of F1 = F1 X cos 20o = 0.9397 F1, the vertical component of F2 = F2 X cos 60o = 0.500 F2, and the vertical component of the weight = 100.

Now since these three forces are in equilibrium, the horizontal and the vertical components balance; hence

0.342 F1 = 0.866 F2 and 0.9397 F1 + 0.5 F2 = 100. Fig. 15.

These equations may be solved for the unknown forces; thus from the first.

F1 = 0.866/0.342 F2 = 2.532 F2.

Substituting this value of F1 in the second equation, we get

0.9397 X 2.532 F2 + 0.5 F2 = 100, or, 2.88 F2 = 100 ; hence F2 = 100/2.88 = 34.72 pounds, and F1= 2.532 X 34.72 = 87.91 pounds.

## Examples For Practice

1. Solve Ex. 1, Art. 16 algebraically. (First determine the angle which the inclined rope makes with the horizontal; you should find it to be 63° 26'.)

2. Solve Ex. 2, Art. 10 algebraically.

3. Solve Ex. 3, Art 16 algebraically.