and suspended by a rod from the point L; LN is a lever having its fulcrum at M, and sustaining the piece D by a rod KH; O is a scale-pan suspended from the end N of the lever LN. The instrument is so constructed, that

EF: GF:: KM: LM;

and the distance KM is generally made equal to of MN. The parts are so arranged that the beam LN shall rest horizontally in equilibrium when no weight is placed on the platform.

If, now, a body be placed upon the platform, a part of its weight will be thrown upon the piece D, and, acting downwards, will produce an equal pressure at K. The remaining part will be thrown upon E, and, acting upon the lever FG, will produce a downward pressure at G, which will be transmitted to L; but, on account of the relation given by the above proportion, the effect of this pressure upon the lever LN will be the same as though the pressure thrown upon E had been applied directly at K. The final effect is, therefore, the same as though the weight of Q had been applied at K, and, to counterbalance it, a weight equal to of must be placed in the scale-pan O.

To weigh a body, then, by means of this scale, place it on the platform, and add weights to the scale-pan till the lever LN is horizontal, then 10 times the sum of the weight added will be equal to the weight required. By making other combinations of levers, or by combining the princi

ple of the steelyard with this balance, objects may be weighed by using a constant counterpoise.

EXAMPLES.

1. In a lever of the first class, the lever arm of the resistance is 22 inches, that of the power, 33, and the resistance 100 lbs. What is the power necessary to hold the resistance in equilibrium? Ans. 8 lbs.

2. Four weights of 1, 3, 5, and 7 lbs. respectively, are suspended from points of a straight lever, eight inches apart. How far from the point of application of the first weight must the fulcrum be situated, that the weights may be in equilibrium?

SOLUTION.

Let x denote the required distance. Then, from Art. (36)

1 x x + 3(x8) + 5(x — 16)+7(x24) = 0;

.. x 17 in. Ans.

3. A lever, of uniform thickness, and 12 feet long, is kept horizontal by a weight of 100 lbs. applied at one extremity, and a force P applied at the other extremity, so as to make an angle of 30° with the horizon. The fulcrum is 20 inches from the point of application of the weight, and the weight of the lever is 10 lbs. What is the value of P, and what is the pressure upon the fulcrum?

SOLUTION.

The lever arm of P is equal to 124 in. × sin 30° = 62 in., and the lever arm of the weight of the lever is 52 in. Hence,

20 × 100 = 10 × 52 + P × 62; .. P = 24 lbs. nearly.

We have, also,

R =

X2 + Y2 = √(110+24 sin 30°)+(24 cos 30°)3.

.. R = 123.8 lbs. ;

4. A heavy lever rests on a fulcrum which is 2 feet from one end, 8 feet from the other, and is kept horizontal by a weight of 100 lbs., applied at the first end, and a weight of 18 lbs., applied at the other end. What is the weight of the lever, supposed of uniform thickness throughout?

SOLUTION.

Denote the required weight by x; its arm of lever is 3 feet. We have, from the principle of the lever,

100 × 2 = x × 3 + 18 X 8; ... x = 183 lbs. Ans.

5. Two weights keep a horizontal lever at rest; the pressure on the fulcrum is 10 lbs., the difference of the weights is 3 lbs., and the difference of lever arms is 9 inches. What are the weights, and their lever arms?

Ans. The weights are 7 lbs. and 10 lbs.; their lever arms are 153 in., and 63 in.

6. The apparent weight of a body weighed in one pan of a false balance is 5 lbs., and in the other pan it is 6 lbs. What is the true weight?

7. In the preceding example, what is the ratio of the lever arms of the balance?

SOLUTION.

Denote the shorter arm by l, and the longer arm by nl. We shall have, from the principle of moments,

That is, the longer arm equals 1 times the shorter arm.

The Inclined Plane.

85. An inclined plane is a plane inclined to the horizon. In this machine, let the power be a force applied to a body either to prevent motion down the plane, or to produce motion up the plane, and let the resistance be the weight of the body acting vertically downwards. The power may be applied in any direction whatever; but we shall, for simplicity's sake, suppose it to be in a vertical plane, taken perpendicular to the inclined plane.

B

Let AB represent the inclined plane, O a body resting on it, R the weight of the body, and P the force applied to hold it in equilibrium. In order that these two forces may keep the body at rest, friction being neglected, their resultant must be perpendicular to AB (Art. 72).

When the direction of the force

a

R

Fig. 72.

P is given, its intensity may be found geometrically, as follows: draw OR to represent the weight, and OQ perpendicular to AB; through R draw RQ parallel to OP, and through draw QP parallel to OR; then will OP represent the required intensity, and OQ the pressure on the plane.

When the intensity of P is given, its direction may be found as follows: draw OR and OQ as before; with Ras a centre, and the given intensity as a radius, describe an are cutting OQ in 2; draw RQ, and through O draw OP parallel, and equal to RQ; it will represent the direction of the force P.

If we denote the angle between P and R by ¤, and the inclination of the plane by a, we shall have the angle ROQ equal to a, since OQ is perpendicular to AB, and OR to AC, and, consequently, the angle QOP — a, From =& the principle of Art. 35, we have,

PR

sina: sin(-a).. (38.)

From which, if either P or q be given, the other can be found.

If we suppose the power to be

applied parallel to the plane, we

B

That is, when the power is parallel to the plane, the power is to the resistance, as the height of the plane is to its length. If the power is parallel to the base of the plane, we shall have, o α = 90° — a; whence,

That is, the power is to the resistance as the height of the

plane is to its base.

From the last proportion we have,

PR

BC = AC

Rtana.

If we suppose a to increase, the value of P will increase, and when a becomes 90°, P will become infinite; that is, if friction be neglected, no finite horizontal force can sustain a body against a vertical wall.