The following table shows the spaces fallen through, and the Velocities acquired at the end of each 20 seconds. To find the space descended by a body in 7" and the velocity acquired. 16 1×49=788 1 of space, 32 2x7" 225 2 of velocity. Look into the table at 7" and you have the answers. EXAMPLE II. To find the time of generating a velocity of 100 feet per second, and the whole × 12 space descended. 100 32 2 =3" 23 time. 155, space descended. EXAMPLE III. To find the time of descending 400 feet, and the velocity at the end of that time. Or these answers can be found from the Table by Proportion. PENDULUM. The vibrations of pendulums are as the square roots of their lengths; and as it has been found by many accurate experiments, that the pendulum vibrating seconds in the latitude of London, is 39 inches long nearly, the length of any other pendulum may be found by the following rule, viz.-As the number of vibrations given is to 60, so is the square root of the length of the pendulum that vibrates seconds, to the square root of the length of the pendulum that will oscillate the given number of vibrations; or, as the square root of the length of the pendulum given, is to the square root of the length of the pendulum that vibrates seconds, so is 60 to the number of vibrations of the given pendulum. Since the pendulum that vibrates seconds, or 60, is 391 inches long, the calculation is rendered simple; for √39 ×60=375, a constant number, therefore 375, divided by the square root of the pendulum's length, gives the vibrations per minute, and divided by the vibrations per minute, gives the square root of the length of the pendulum. EXAMPLE I. many vibrations will a pendulum of 49 inches long How make in a minute? What length of a pendulum will it require to make 90 vibrations in a minute? What is the length of a pendulum, whose vibrations will be the same number as the inches in its length? ✔(375)=52 inches long, and 52 vibrations. It is proposed to determine the length of a pendulum vibrating seconds, in the latitude of London, where a heavy body falls through 16 feet in the first second of time? 3.1416 circumference, the diameter being 1. 16 feet 193 inches fall in the 1" of time. 12 193 × 2=386.00000000 3.14162 9.86965056 or 39.11 inches. = 39.109 inches. } By experiment this length is found to be 39 inches. What is the length of a pendulum vibrating in 2 seconds, and another in half a second? 375 ✓396.25 x 60-375. 30 375 120 12.5 squared-156.25 inches the length of a 2 se[conds' pendulum. =3.125 squared 9.765625 inches the length of a second's pendulum. MECHANICAL POWERS, &c. The Science of Mechanics is simply the application of Weight and Power, or Force and Resistance. The weight is the resistance to be overcome; the power is the force requisite to overcome that resistance. When the force is equal to the resistance, they are in a state of equilibrium, and no motion can take place; but when the force becomes greater than the resistance, they are not in a state of equilibrium, and motion takes place; consequently, the greater the force is to the resistance, the greater is the motion or velocity. The Science of Equilibrium is called Statics; the Science of Motion is called Dynamics. Mechanical Powers are the most simple of mechanical applications to increase force and overcome resistance. They are usually accounted six in number, viz. The Lever -The Wheel and Axle-The Pulley-The Inclined Plane -The Wedge-and the Screw. LEVER. To make the principle easily understood, we must suppose the lever an inflexible rod without weight; when this is done, the rule to find the equilibrium between the power and the weight, is, Multiply the weight by its distance from the fulcrum, prop, or centre of motion, and the power by its distance from the same point: if the products are equal, the weight and power are in equilibrio, if not, they are to each other as their products. EXAMPLE I. A weight of 100 lbs. on one end of a lever, is 6 inches from the prop, and the weight of 20 lbs. at the other end, is 25 inches from the prop-What additional weight must be added to the 20 lbs. to make it balance the 100 lbs. ? 100 × 6 25 =24—20=4 lbs. weight to be added. EXAMPLE II. A block of 960 lbs. is to be lifted by a lever 30 feet long, and the power to be applied is 60 lbs.- -on what part of the lever must the fulcrum be placed? 960 60 =16, that is, the weight is to the power as 16 is to 1; therefore the whole length 30 16+1 = 11%, the distance from the block, and 30-11-28, the distance from the power. EXAMPLE III. A beam 32 feet long, and supported at both ends, bears a weight of 6 tons, 12 feet from one end.-What proportion of weight does each of the supports bear? 12 × 6 32 weight. 20 × 6 32 2 tons, support at end farthest from the -3 tons, support at end nearest the weight. · EXAMPLE IV. A beam supported at both ends, and 16 feet long, carries a weight of 6 tons, 3 feet from one end, and another weight of 4 tons, 2 feet from the other end. What proportion of weight does each of the supports bear? 3×6 14x4 74 + = 16 16 -418 tons, end at the 4 tons. WHEEL AND AXLE. The nature of this machine is suggested by its name. To it may be referred all turning or wheel machines of different radii: as well-rollers and handles, cranes, capstans, windlasses, &c. The mechanical property is the same as in the lever: that is, the product of the weight into the distance at which it acts is equal to the product of the power into the distance at which it acts, the distances being estimated in directions perpendicular to those in which the weight and power act respectively, because the wheel and axle is only a kind of perpetual lever. And hence also this property: The product of the power applied, multiplied by its velocity, is equal to that of the weight to be raised into its velocity. When a series of wheels and axles act upon each other |