3. If the length of a lever be 10 feet, the power 170 lb., and the weight to be balanced 1530 lb., where must be the fulcrum? Ans. 9 feet from the power.

A pulley consists of a wheel movable about its axis, and so arranged as to be put in motion by means of a cord passing over it. The movable pulley is the only one by which a gain in power is effected. A double movable pulley is represented at fig. 2. By such a pulley an equilibrium is produced, when the power is to the weight, as one to the number of ropes sustaining the weight. If a single movable pulley be employed, the weight is sustained by two ropes; and the power will be to the weight, as 1 to 2. If two movable pulleys be employed, the weight is sustained by four ropes, and the power will be to the weight, as 1 to 4, &c.; that is, each movable pulley is sustained by two ropes, and consequently doubles the effect of the power employed.

Prob. 23. Given the number of movable pulleys and the power, to find the weight,

RULE.-Multiply the power applied by twice the number of movable pulleys.

Ex. 1. If in a system of pulleys there be three movable pulleys, what weight will a power of 72 lb. balance? 72 × 6-432 lb. Ans.

2. What weight will a power of 15 lb. balance by a system of 6 movable pulleys? Ans. 180 lb.

3. By the aid of a system of 12 movable pulleys, how many pounds will a man sustain who is capable of applying a power of 150 lb.? Ans. 3600.

Prob. 24. Given the number of movable pulleys, and the weight to be balanced, to find the required power.

RULE. Divide the weight by twice the number of movable pulleys.

Ex. 1. How many pounds would be required by the aid of two movable pulleys, to sustain 800 lb.? Ans. 200 lb.

2. By the aid of 10 movable pulleys, how many lb. would be required to balance 2000 lb. ? Ans. 100 lb.

3. What weight would balance 10 pounds of silver by the aid of a system of 100 pulleys?, Ans, 12 pwt.

WHEEL AND AXLE.

The wheel and axle form a kind of perpetual lever, the long arm of which is the semidiameter of the wheel, and the short arm, the semi-diameter of the axle. Consequently, the power of the wheel and axle is increased either by making the wheel larger, while the axle remains unaltered, or by making the axle smaller, while the wheel remains the same. The power must always be applied to the wheel, and the weight to the axle, when an increase of action is to be gained. In the

use of the wheel and axle, an equilibrium is produced, when the power is to the weight as the semi-diameter of the axle is to the semi-diameter of the wheel,

Prob. 25. Given the diameter of the wheel, the diameter of the axle, and the power, to find the weight.

RULE. Multiply the diameter of the wheel by the power applied, and divide the product by the diameter of the axle.

Ex. 1. If the diameter of the axle be six inches, and that of the wheel 6 feet, what weight attached to the axle, will 16 lb. attached to the wheel, balance? Ans. 192 lb.

3. If the diameter of a wheel be 20 feet, and that of the axle 2 feet, and a power of 400 lb. be applied to the wheel, what weight will be balanced at the axle ? Ans. 4000 lb.

Prob. 26. Given the diameter of the wheel, the diameter of the axle, and the weight to be balanced, to find the required power.

RULE.-Multiply the diameter of the axle by the weight, and divide the product by the diameter of the wheel.

Ex. 1. If the diameter of the axle be 6 inches, and the diameter of the wheel 12 feet, what power will balance a weight of 360 lb? Ans. 15 lb.

2. If the diameter of the axle be 2 feet, and the diameter of the wheel 20 feet, what power will balance a weight of 4000 lb.? Ans. 400 lb.

THE INCLINED PLANE.

An inclined plane is a plane that makes with the ground or floor some certain angle less than a right angle. In the application of this instrument, the ratio of the power and weight is always the same as that of the height and length of the plane.

Prob. 27. Given the length and height of the plane, and also the power, to determine the weight..

RULE.-Multiply the power by the length of the plane, and ivide the product by its perpendicular height.

Ex. 1. If the length of a plane be 16 feet, and its height 4 feet, what weight will a power of 32 lb. sustain? Ans. 128 lb. 2. What weight will 164 lb. sustain on a plane 112 feet in length and 3 feet in height? Ans. 61224 lb.

Prob. 28. Given the length and height of the plane, and also the weight, to find what power will sustain it.

RULE.-Multiply the weight by the height of the plane, and divide the product by the length..

Ex. 1. What power will balance 128 lb. on an inclined plane, the length of which is 16 feet, and the height 4 feet? Ans. 32 lb.

2. Suppose on a railroad there is an inclined plane 150 rods in length, and rising to the perpendicular height of 50 feet; what power will be required to sustain a weight of 84000 lb. Ans. 28000 lb.

THE WEDGE.

The wedge is composed of two inclined planes, whose bases unite and form the base of the wedge. The power applied to the wedge is to the effect produced at the side of the wedge, as the thickness of the head to the length of the wedge, no allowance being made for friction.

Prob. 29. Given the thickness of the head, the length of the side, and the power acting upon the head of a wedge, to determine the force produced on the side.

RULE.-Multiply the length of the wedge by the power applied, and divide the product by the thickness of the head.

Ex. 1. If the length of a wedge be 12 inches, the thickness of the head 3 inches, and the force applied, 64 lb., what will be the resistance at the side? Ans. 256 lb.

2. If the length of a wedge be 32 inches, the thickness of the head 2 inches, and the force applied 1600 lb., what will be the resistance at the side? Ans. 25600 lb.

Prob. 30. Given the length of the side, the thickness of the head, and the resistance upon the side of a wedge, to find the force acting upon the head.

RULE. Multiply the resistance at the side by the thickness of the head, and divide the product by the length of the side of the wedge.

Ex. 1. If the length of the wedge be 32 inches, the thickness of the head 2 inches, and the resistance at the side be 25600 lb. what must be the force upon the head, no allowance being made for friction? Ans. 1600 lb.

2. If the resistance at the side of a wedge be 20000 lb., the length of the wedge 20 inches, and the thickness of the head 3 inches, what force is required to be applied to counteract the resistance at the sides? Ans. 3000 lb.

THE SCREW.

The screw is a cylindrical piece of wood, or metal, with a spiral thread running round it with a gradual and uniform inclination. The thread, therefore, forms an inclined plane of the same length as the thread itself, and whose height equals the length of the screw.

It is then obvious that the power of the screw depends in part upon the fineness of the threads, since the finer they are, the greater will be the length of the plane, while its height remains the same. The power of the screw also depends in part on the length of the lever employed to move it. It is, therefore, a compound power, involving in its action both the principle of the lever and the inclined plane, and is consequently more efficient than either of the other mechanical powers. In the action of this instrument, an equilibrium between the power and weight is produced, when the power is to the weight as the distance between two contiguous threads of the screw is to the circumference of the circle described by the power in one revolution.

Prob. 31. Given the distance between the threads of the screw, the length of the lever, and the power applied to find the weight.

RULE. Multiply the circumference of the circle described by one revolution of the lever by the power applied, and divide the product by the distance between the threads of the screw.

Note.-The lever is the semi-diameter of the circle it describes; the circumference of the circle may therefore be found by Prob. 6th.