Parallelogram of Forces.-If two forces, acting simultaneously towards or from a point, be represented in direction and magnitude by the adjacent sides of a parallelogram, then the resultant of these forces will be represented in direction and magnitude by the diagonal of the parallelogram which passes through their point of intersection.

Component

B

R

Resultant

Component

A

For example, let any two forces, P and Q, act from the point O at any convenient angle, say 60°, then, if OA and OB be plotted to scale to represent these forces in direction and magnitude, the diagonal OD of the parallelogram OADB will represent in direction and to the same scale their resultant R. But the resultant R is equal and opposite in direction to a force E, which would exactly balance the effect of P and Q, or to a force represented in direction and in magnitude by the line DO. Further, since the side AD is equal and parallel to the side OB, it may be taken to represent Q in direction and magnitude. Hence we have the three sides of the triangle OAD taken in the order OA, AD, DO, representing in direction and magnitude three forces, P, Q, E, in equilibrium, acting from the point O. Hence we have a general proposition termed the "triangle of forces," or a deduction from the "parallelogram of forces."

PARALLELOGRAM OF FORCES.

E

TRIANGLE OF FORCES.

Triangle of Forces.-If three forces acting towards or from a point are in equilibrium, and a triangle be drawn with its sides

TRIANGLE OF FORCES.

67

respectively parallel to those forces taken in due order, then the forces will be represented to scale by the sides of the triangle.

CONVERSELY:-If three forces acting towards or from a point are represented in direction and to scale by the sides of a triangle taken in due order, these three forces are in equilibrium.

For example, let the three forces P, Q and E act from the point O, and be in equilibrium. Draw a triangle with its sides, P, Q, E, respectively parallel to these forces; then the sides of this triangle, taken in that order, represent to the same scale these forces. Or, if the triangle, whose sides are respectively P, Q and E, represent in direction and to scale the three forces P, Q and E, as they act from a point O, these forces are in equilibrium. We have shown by a dotted line the resultant R, and its direction as opposed to E, by the same side of the triangle.

It is quite evident that if the forces P, Q and E acted towards the point O, instead of from it, the triangle P, Q, E would still represent these forces in magnitude, but the direction of all the arrows would have to be pointed the opposite way.

SPECIAL CASES.—Three Equal Forces in Equilibrium.—It can easily be proved by the apparatus used for Experiment I., or by construction, that if you have three equal forces in equilibrium they must act at 120° from each other, and that the triangle representing their directions and magnitudes will be an equilateral triangle, or a triangle whose angles are each equal to 60°.

Two Forces acting at Right Angles.--In this case it can be proved by the same apparatus, or by Euclid, Book I. Prop. 47, that any two forces P and Q, acting at right angles to each other, have a resultant R, or are balanced by a third force E, of such magnitude that

E2 R2 P2+ Q2

Consequently, if you have any two forces in the proportion of 3 to 4 acting at right angles to each other, their resultant must have a value of 5.

Conversely, if any two forces in the proportion of 3 to 4 units are balanced by a third force proportionally of 5 units, the forces 3 and 4 must be acting at right angles to each other.

Resolution of a Force into Two Components at Right Angles to each other.*-Let R be the force to be resolved, P and Q the components, and let R make an angle, 0, with the force Q.

Resultant of Two Forces acting at any Angle on a Point.The proof of this general case must be left to the Author's Advanced Treatise on Applied Mechanics, but the formula may be given here, viz. : R2 P2 + Q2 + 2P × Q Cos a

=

where P and Q are any two forces, R their resultant, and a the angle between the directions of the forces P and Q.

Resultant of any Number of Forces Acting at a Point.-Let P1, P2, P3, &c., be any number of forces acting at a point; then, by the

A 3

5

B

parallelogram of forces find a resultant, R, for P, and P.; and a resultant, R, for R, and P1; and so on. The last resultant will be the resultant of all the forces.

Example I.-Forces 3, 5 and 7 units act from a central point O at equal angles. Find the resultant.

ANSWER. Let OA, OB and OC represent the forces in direction and magnitude. Then you can follow out the above rule and find a resultant for, say, 3 and 5-call this R, ; and finally find a resultant for R, and 7. But it is obvious that you may subtract 3 units from each of them without affecting the result, since the forces are acting at equal angles from each other. This will destroy one of them, and leave OB, to represent 2 units, and OC, to represent 4 units. Then, by the parallelogram of forces you find the resultant R=3'5 units.

*The reverse of this may be applied to the composition of two or more forces acting at a point in one plane, but we will leave the demonstration of such problems, as well as that of the polygon of forces, to our Advanced Book on Applied Mechanics.

STRESSES IN JIB CRANES.

69

Stresses in Jib Cranes.-As a practical example of the application of the "triangle of forces," take the case of an ordinary hand-worked jib crane. The load is suspended from the hook H of the snatch-block SB; or, in the case of a crane for lifting light loads quickly, to a simple hook with a swivel attached directly

to the end shackle of the chain C, as it comes down from the jib pulley JP, instead of the chain passing round a snatch-block pulley, and up to an eyebolt near the point of the jib.

(1) The load produces a tension on the chain C.

(2) A thrust along the jib J, from the jib pulley to the eye-bolt connecting the shoe of the jib to the bottom of the framing F.

(3) A tension in the tie-rods from the top of the framing to their connection with the top of the jib.

(4) This tension on the tie-rods produces a horizontal pull, tending to bend and break the crane-post CP, where it leaves the upper foundation plate-bearing and joins the framing. Cranes for heavy lifts require a back balance weight to counteract this force. (See the third figure in Lecture XIII.)

(5) It also causes an upward stress, tending to unship or lift up the crane-post from its bearings in the upper and lower foundation plates.

The directions and values of these stresses will be better understood by the student after considering a particular example.

EXAMPLE II. In a hand-worked jib crane of the form shown by the above figure, the length of the jib is 30 feet, the lengths of the tie rods are 25 feet each, and the vertical distance between the attachments of the tie-rods and of the jib to the framing, is 12 feet. Find the stresses produced on these parts of the crane