which is attached to the anchor or other weight to be raised. The force W is divided between the two parts of the rope pulling on S, so and large drums, respectively, the moments of the forces exerted by the W rope on the drum are r and -R and the difference between these W two moments must be balanced by the moment of the force P acting on the end of the capstan bar of length 7. Hence we have in case of equilibrium The advantage of such an arrangement is evidently the same as if one end of the rope were fixed and the other, after passing around S, were wound up on an axle whose radius was R-r. But such an axle being of small diameter would not have the strength of the larger axle with two drums. The differential pulley is a similar device used for raising heavy weights. There is an upper pulley having a single sheave with two grooves of different diameters like the two drums of the Chinese capstan. An endless chain passes over one groove in the upper pulley then around a pulley attached to the weight to be raised, and then around the second groove of the upper or fixed pulley. The grooves of the upper pulley have notches to receive the chain so that it cannot slip, and the chain is passed over it in such a way that it is wound up on one groove at the same time that it unwinds from the other. If the difference in diameters of the two grooves in the upper sheave is small, a small pull on the chain may suffice to support a large weight. Problems. 1. A 200-lb. barrel is rolled up an inclined plane 10 ft. long to a platform 4 ft. above the ground. How much force must be exerted along the plane, and how much work is done? Find also force and work when the plane is 20 ft. long, the height being the same. 2. How much force parallel to the plane is required to support a 20 kilogram weight on a smooth inclined plane 10 meters long and 6 meters high? Also find the pressure of the weight against the plane. 3. If the coefficient of friction between weight and plane in the last question is 0.20, find how much more force must be exerted in drawing the weight up the plane than in lowering it. 4. A force of 50 lbs. applied to a jack screw by means of a lever arm 2 ft. long will raise what weight, if the screw has 3 threads to the inch, neglecting friction? 5. Find the tension on a bicycle chain when the pedal is pressed down with a force of 100 lbs., the crank arm being 6 inches long and the sprocket wheel 8 inches in diameter. 6. How much force must be exerted on the crank of a windlass to raise a weight of 300 lbs., if the crank arm is 18 inches long and the drum on which the rope is wound is 8 inches in diameter. 7. Find the direction and amount of the force on the bearings of the windlass in the previous question, first, when the crank is in a horizontal position and being pressed down; second, when the crank arm is vertical. 8. A man weighing 150 lbs. raises himself in a sling by means of a rope passing over a movable pulley attached to the sling and a fixed pulley overhead. With how much force must he pull? Show also how to obtain your result by the principle to work. 9. A man weighing 150 lbs. runs up 20 steps, each 7 inches high, in 5 seconds. How much work does he do, and what H. P. does he expend? 10. A locomotive drawing a train along a level track at 30 miles per hour expends 75 horse-power; find the total air and frictional resistance overcome. Ans. 937.3 lbs. II. What load can two horses draw along a level road at the rate of 3 miles an hour if they spend two H. P. in pulling the load? Coefficient of friction of wagon on road being. Ans. 2500 lbs. 12. A locomotive draws a 300-ton train along a level track at the rate of 20 miles per hour, while working at the same rate it draws it up a per cent. grade at 15 miles per hour, what H. P. is expended, supposing the frictional and air resistances the same in both cases, and what is the resistance in pounds? Ans. Resistance = 4500 lbs.; H. P. =240. III. KINETICS OF A PARTICLE RECTILINEAR MOTION OF A MASS. 93. Introductory.-Up to this point we have studied especially cases of equilibrium, where the forces acting are so balanced that there is no acceleration. We must now examine in some detail the various forms of motion where forces are involved in such a way as to cause acceleration. This part of mechanics, as Mach says, "is a wholly modern science. All that the Greeks achieved in mechanics belongs to the realm of statics. Dynamics was first founded by Galileo." Before 1638, when Galileo first published the results of his experiments, so little progress had been made in this direction that it was currently held that heavy bodies fell faster than light ones. Galileo carefully studied the motion of falling bodies, and of bodies rolling down inclined planes, and showed that in each of these cases the motion was with constant acceleration. As pendulum clocks had not been invented at that time, he made use of a simple water clock to measure short intervals of time in his experiments. This consisted of a large vessel of water having a jet closed by the finger, and the water was allowed to escape during the time interval to be measured. Thus the weight of water escaping while a body rolled down an inclined plane served to measure the time of descent. 94. Dyne and Poundal.-In dealing with cases of equilibrium we have used the ordinary gravitation measures of force, the weight of a pound or gram, but in studying the accelerating effect of forces a new unit will be found more convenient. It has been shown in §37 that when a force acts on a mass there is acceleration, and the rate of acceleration is proportional to the force when the mass is constant, and that to give a large mass the same acceleration as a small one the forces must be proportional to the masses. These two laws are both really included in Newton's second law of motion and are expressed by the formula where F represents the force, m the mass, and a the acceleration, and k is an absolute constant which depends on the particular units employed. It is convenient to adopt as the unit a force which will make the constant k equal to unity in the above expression, so that we may write simply Defined in this way, unit force is one which will give unit acceleration to unit mass, or unit force acting for unit time on unit mass will change its velocity by unity. When the centimeter gram and second are the fundamental units as in the C.G.S. system, the unit force is called the dyne, from the Greek word for force. It is a force which, acting on a mass of one gram for one second, will change its velocity by one centimeter per second. Hence to find the force in dynes in a given case of motion it is only necessary to multiply the mass in grams by its rate of acceleration measured in centimeter-seconds. Thus a force of 100 dynes acting on a mass of 10 grams will give it an acceleration 10, or in one second will give it an increase in velocity of 10 cm. per second. A unit of force similarly based on the foot, pound, and second as units of length, mass, and time, respectively, is sometimes used and is called the poundal, it is the force which acting on a mass of one pound will increase its velocity one foot per second for every second that it acts. 95. Motion in a Straight Line with Constant Velocity.-When a body moves in a straight line with constant velocity the acceleration is zero and therefore the force must be zero according to the formula F=ma. The moving mass is therefore in equilibrium. This is the case considered in Newton's first law of motion. A railway train while running at constant speed is in a state of equilibrium. The force of the locomotive urging it on is exactly balanced by the resistance of the air and friction of the wheels. So when a bucket is drawn up out of a well with constant speed it is in equilibrium and the upward pull on the rope is exactly equal to the weight of the bucket of water. 96. Motion in a Straight Line with Constant Acceleration. When a body moves in a straight line with a velocity which is increasing or diminishing at a constant rate, it has a constant acceleration in the direction of the motion in one case and opposite to it in the other. When the acceleration a is constant, the change in velocity of the moving body in t seconds is at. And if the velocity at the beginning of the time t is u, and that at the end of the time is v, then v=u+at when the speed is increasing; (1) The space passed over in t seconds will be found by multiplying the average velocity during the interval by the time t. But since the acceleration is constant the velocity increases uniformly with the time, and therefore the average velocity is the v + u arithmetical mean of the initial and final velocities, or The space traversed in time t may therefore be expressed by the formula 2 |