EXERCISE XXXV. 1. The French "force de cheval" is 4,500 kilogrammetres per minute; express it in terms of the horse-power. 2. Given 1 horse-power = 550 foot-pounds per sec., and 1 "force de cheval" 75 kilogrammetres per sec.; deduce the relation of the "force de cheval" to the horse-power. It is supposed that the intensity of gravity is the same in the two definitions. 3. Find the value of the horse-power in terms of kilogrammetres per minute. 4. Determine the horse-power of a machine capable of raising 10 tons through a height of 20 feet in 2 minutes. 5. If an engine consumes 2 pounds of coal per horse-power per hour, how many foot-pounds of work will it perform when consuming 112 pounds of coal? 6. How many foot-pounds of work are required to raise 30,000 lbs. of water from a depth of a furlong; and how many horse-power to do it in five minutes? 7. If a pressure of 1 ton is exerted through 10 yards, how many foot-pounds of work are done; and at what horse-power does an engine work which does the work in half a minute? 8. A pumping engine is partly worked by a weight of 2 tons, which at each stroke of the pump falls through 4 ft.; the pump makes 10 strokes a minute; how many gallons of water are lifted per minute by the weight from a depth of 200 ft. ? 9. Calculate the horse-power of an engine from the following data:-stroke 24 in., diameter of piston 16 in., 100 revolutions per minute, average effective pressure in cylinder 60 lbs. per square inch. 10. In the transmission of power by a rope, the wheel carrying the rope is 14 feet in diameter and makes 30 revolutions per minute, the tension of the rope being 100 lbs. Find the amount of power transmitted estimated as horse-power. 11. What diameter of cylinder will develop 50 horse-power with a four-foot stroke, 40 revolutions per minute, and a mean effective steam pressure of 30 lbs. per square inch above the atmosphere, the engine being non-condensing? 12. The cylinder of an engine is 12 inches diameter by 20 inches long; with an average pressure of 60 lbs. per square inch it has a power of 40 horse-power. Find the rate of revolution of the engine. SECTION XXXVI.-MECHANICAL ADVANTAGE. ART. 163.-Virtual Velocity. Since 1 W = F resistance by L displacement, 1 W per T = (F resistance by L displt.) per T, = F resistance by (L displt. per T); for the force is supposed to be constant during the displacement. The velocity indicated is that of the force in its own direction, that is, the virtual velocity. ART. 164.--Velocity Ratio and Mechanical Advantage. The scholium to Newton's third law of motion asserts that in the case of any machine, the rate at which work is done by the agent is equal to the rate at which work is done against the resistance, when there is no acceleration of the parts of the machine, and no loss of energy through friction. Suppose that the speed of the agent is v L per T, and that of the resistance v' L per T; then v/v L displt. of agent = L displt. of resistance. But, as the value of F by L is the same for the agent and for the resistance, The former is called the velocity ratio of the machine, and the latter the mechanical advantage. The value of the one is the reciprocal of the value of the other. EXAMPLES. Er. 1. If the thread of a screw makes 25 turns in 3 inches, and the arm is 24 inches; what force must be applied to sustain a weight of 1 cwt. ? 25 turns of screw 3 inch, resistance ; 2 × 24 × 25 inch, agent = 3 inch, resistance; Ex. 2. In a crane the numbers of the teeth in the two wheels are 45 and 60 respectively; the numbers of leaves in the two pinions 9 and 15; the arm of the winch is 3 feet, and the radius of the weight-bearing axle 4 inches. Find the mechanical advantage of the crane. = 6 ft., agent revolution of 1st pinion, Ec. 3. In a small hydraulic press the ram is 2 in. and the plunger in. in diameter; the length of the lever handle is 2 ft., and the distance from the fulcrum to the plunger is 14 in. Find the power exerted on the ram when a 10-lb. weight is hung at the end of the lever. = 1 sq. inch plunger sq. inch ram; .. 16 in. displacement of plunger = in. displacement of ram, and 24 in. displacement of agent = in. displacement of plunger ; 24 x 16 x 2 3 in. displacement of agent = in. displacement of ram; 1. Find the steepest incline up which a force of 5 cwt. can just move a weight of 2 tons, friction being left out of account. 2. A man weighing 12.5 stones is lowered into a well by means of a windlass, the arm and axle of which are 30 inches and 8 in diameter. must be applied to let him down with uniform velocity. Find the force which 3. The wire-rope working a railway signal passes over and has its end attached to the rim of a fixed grooved pulley; the pulley is 1 foot in diameter, and is turned by a lever 3 feet long. Find the tension transmitted along the rope when a man pulls with a force of 37 lbs. at right angles to the lever and at its extremity. 4. Twelve sailors, each exerting a force of 48 pounds, work a capstan with levers 7 feet 6 inches long; the radius of the capstan is 16 inches; what resistance can they unitedly sustain? 5. A driving wheel has 64 teeth, and the driven wheel 33 teeth; how many revolutions per second will the latter make when the former makes 10 revolutions per second? 6. A screw having 4 threads to the inch is worked by an arm 18 inches long; what is the force exerted by the screw when a force of 15 lbs. is applied at the end of the arm? 7. A screw, the pitch of which is a quarter inch, is turned by means of a lever 4 feet long; find the force which will raise 15 cwt. 8. The screw of a steamer has a pitch of p feet, and makes n revolutions per minute; deduce the theoretical velocity of the steamer in knots per hour. 9. If the thread of a screw be inclined at an angle of 30 degrees to the horizontal, the radius of the horizontal section of the cylinder 9 inches and the length of the lever 4 feet; find what power will sustain 15 cwt. on it. 10. In a screw-jack, where a worm wheel is used, the pitch of the screw is inch, the number of teeth in the worm wheel is 16, and the length of the lever handle is 10 inches; find the gain in power. 11. A wheel and axle is used to raise a bucket from a well. The radius of the wheel is 15 inches; and while it makes 7 revolutions the bucket, which weighs 30 lbs., rises 5 feet. Show what is the smallest force that can be employed to turn the wheel. 12. Suppose that we have four weightless pulleys, three moveable and one fixed, forming an example of the first system, and that the weight is a man weighing 160 lbs.; find what pull the man must exert on the power end of the rope in order to raise himself thereby. 13. A weight of 3 cwt. is raised 3 feet by means of a single moveable pulley, the block of which has three sheaves; determine the power and the space through which it has acted. 14. In the second system of pulleys, if there be three pulleys in the lower block, which weighs 8 lbs., and the string be fastened to the upper block; find the weight which a power of 20 lbs. can support. 15. Find the ratio of the power to the weight in a system of three pulleys, in which all the strings are attached to the weight, neglecting the weight of the pulleys. 16. In a compound wheel and axle the diameters of the two parts of the axle are 5 and 6 inches respectively. The weight raised hangs from a single moveable pulley in the usual manner, and is supported by a pressure applied perpendicularly to a lever handle 15 inches in length. Find the ratio of the agent to the resistance. 17. In a lifting crab the lever handle is 14 inches long, the diameter of the drum is 6 inches, and the wheel and pinion have 57 and 11 teeth respectively. Find the weight in pounds which could be raised by a force of 50 lbs. applied to the lever handle, friction being neglected. 18. The thread of a screw makes 12 turns in a foot of its length; the power is applied at the end of an arm 2 feet long; it is found that when the power is 30 lbs. it can just raise a weight of 1,200 lbs. ; what portion of the power is used in overcoming friction, and how many foot-pounds of work are done by the power when the weight is raised 2 feet? SECTION XXXVII.-MOMENT OF A FORCE. ART. 165. Moment of a Force. When the virtual velocity of an agent is round a circle, F by L per T can be expressed as F by radius by (radian per T). The angular velocities of the agent and the resistance being the same, the condition that the value of W per T must be the same for both reduces to the condition that the value of F by L radius be the same. This idea F by L radius is called the moment of a force. Special units are poundal by foot, lb. by weight by foot, kgm. by weight by metre, etc. Let r L and r' L be the measures of the arms, then, since the angular velocity is the same for both, the velocity-ratio becomes r/r' L agent L resistance; and therefore the mechanical advantage becomes r'r F agent F resistance. ART. 166.-Horizontal Lever. In the case of a vertical force acting at the end of a horizontal lever, the displacement of the end of the lever is at the beginning in the direction of the force. Hence the condition for equilibrium is that the value of F by L arm should be the same for the two opposing forces. When the |