82. Component of gravity effective in producing the motion of the pendulum. When a pendulum is drawn aside from its position of rest (Fig. 65), the force acting on the bob is its weight, and the direction of this force is vertical. Let it be represented by the line OR. The component of this force in the direction in which the bob is free to move is On, and the component at right angles to this direction is Om. The second component Om simply produces stretch in the string and pressure upon the point of suspension. The first component On is alone responsible for the motion of the bob. A consideration of the figure shows that this component becomes larger and larger the greater the dis- FIG. 65. Force acting on displaced pendulum placement of the bob. When the bob is directly beneath the point of support, the component producing motion is zero. Hence a pendulum can be permanently at rest only when its bob is directly beneath the point of suspension.* m R 51 QUESTIONS AND PROBLEMS 1. The engines of a steamer can drive it 12 mi. per hour. How fast can it go up a stream in which the current is 3 mi. per hour? How fast can it come down the same stream? 2. The wind drives a steamer east with a force which would carry it 12 mi. per hour, and its propeller is driving it south with a force which would carry it 15 mi. per hour. What distance will it actually travel in an hour? Draw a diagram to represent the exact path. * It is recommended that the study of the laws of the pendulum be introduced into the laboratory work at about this point (see Experiment 12, authors' Manual). 3. A barge is anchored in a river during a storm. If the wind acts eastward on it with a force of 3000 lb. and the tide northward with a force of 4000 lb., what is the direction and magnitude of the equilibrant ; that is, the pull of the anchor cable upon the barge? 4. A picture weighing 20 lb. hangs upon a cord whose parts make an angle of 120° with each other. Find the tension (pull) upon each part of the cord. 5. If the barrel of Fig. 66 weighs 200 lb., with what force must a man push parallel to the skid to keep the barrel in place if the skid is 9 ft. long and the platform 3 ft. high? 6. A cake of ice weighing 200 lb. is held at rest upon an inclined plane 12 ft. long and 3 ft. high. By the resolutionand-proportion method find FIG. 66. Force necessary to prevent a barrel from rolling down an inclined plane the component of its weight that tends to make the ice slide down the incline. With what force must one push to keep the ice at rest? How great is the component that tends to break the incline? 7. A tight-rope 20 ft. long is depressed 1 ft. at the center when a man weighing 120 lb. stands upon it. Determine graphically the tension in the rope. 8. The anchor rope of a kite balloon makes an angle of 60° with the surface of the earth. If the lifting power of the balloon is 1000 lb., find the pull of the balloon on the rope and the horizontal force of the wind against the balloon. 9. A canal boat and the engine towing it move in parallel paths which are 50 ft. apart. The tow rope is 130 ft. long, and the force (effort) applied to the end of the rope is 1300 lb. Find what component of the 1300 lb. acts parallel to the path of the boat. n m R FIG. 67. Forces acting on a kite 10. In Fig. 67 the line on represents the pull of gravity on a kite, and the line om represents the pull of the boy on the string. What is the name given to the force represented by the line oR? 11. If the force of the wind against the kite is represented by the line AB, and it is considered to be applied at o, what must be the relation between the force oR and the component of AB parallel to OR when the kite is in equilibrium under the action of the existing forces? 12. If the wind increases, why does the kite rise higher? 13. Show from Fig. 68 what force supports an aëroplane in flight. (Remember that oR, the component of the wind pressure AB perpendicular to the plane, is the only acting force out of which a support for the aeroplane can be derived.) (See frontispiece and opposite pp. 153, 316, and 317.) A Direction of Flight 18 B n R 0 m A B' FIG. 68. Forces acting on an aëroplane in flight GRAVITATION 83. Newton's law of universal gravitation. In order to account for the fact that the earth pulls bodies toward itself, and at the same time to account for the fact that the moon and planets are held in their respective orbits about the earth and the sun, Sir Isaac Newton (1642-1727) (see opposite p. 84) first announced the law which is now known as the law of universal gravitation. This law asserts first that every body in the universe attracts every other body with a force which varies inversely as the square of the distance between the two bodies. This means that if the distance between the two bodies considered is doubled, the force will become only one fourth as great; if the distance is made three, four, or five times as great, the force will be reduced to one ninth, one sixteenth, or one twenty-fifth of its original value; etc. The law further asserts that if the distance between two bodies remains the same, the force with which one body attracts the other is proportional to the product of the masses of the two bodies. Thus we know that the earth attracts 3 cubic centimeters of water with three times as much force as it attracts 1, that is, with a force of 3 grams. We know also, from the facts of astronomy, that if the mass of the earth were doubled, its diameter remaining the same, it would attract 3 cubic centimeters of water with twice as much force as it does at present, that is, with a force of 6 grams (multiplying the mass of one of the attracting bodies by 3 and that of the other by 2 multiplies the forces of attraction by 3 × 2, or 6). In brief, then, Newton's law of universal gravitation is as follows: Any two bodies in the universe attract each other with a force which is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. 1 Two masses of 1 gram each at a distance apart of 1 cm. attract each other with a force of about 15,000,000,000 gram. The masses of the sun and the earth are so great that even though 93,000,000 miles apart, they attract each other with a force of about 4,000,000,000,000,000,000 tons. A body weighing 100 pounds on the earth would weigh about 2700 pounds on the sun. A freely falling body on the earth drops 16 feet the first second, while on the sun it would fall 27 times that far in the first second, or 432 feet. On the moon we should weigh of what we do on the earth; we could jump 6 times as high and should fall as fast. 84. Variation of the force of gravity with distance above the earth's surface. If a body is spherical in shape and of uniform density, it attracts external bodies with the same force as though its mass were concentrated at its center. Since, therefore, the distance from the surface to the center of the earth is about 4000 miles, we learn from Newton's law that the earth's pull upon a body 4000 miles above its surface is but one fourth as much as it would be at the surface. It will be seen, then, that if a body be raised but a few feet or even a few miles above the earth's surface, the decrease in its weight must be a very small quantity, for the reason that a few feet or a few miles is a small distance compared with 4000 miles. As a matter of fact, at the top of a mountain 4 miles high 1000 grams of mass is attracted by the earth with 998 grams instead of 1000 grams of force. 85. Center of gravity. From the law of universal gravitation it follows that every particle of a body upon the earth's surface is pulled toward the earth. It is evident that the sum of all these little pulls on the particles of which the body is composed must be equal to the total pull of the earth upon the body. Now it is always possible to find one single point in a body at which a single force, equal in magnitude to the weight of the body and directed upward, can be applied so that the body will remain at rest in whatever position it is placed. This point is called the center of gravity of the body. Since this force counteracts entirely the earth's pull upon the body, it must be equal and opposite to the resultant of all the small forces which gravity is exerting upon the different particles of the body. Hence the center of gravity may be defined as the point of application of the resultant of all the little downward forces of gravity acting upon the parts of the body; that is, the center of gravity of a body is the point at which the entire weight of the body may be considered as concentrated. The earth's attraction for a body is therefore always considered not as a multitude of little forces but as one single force F (Fig. 69) equal to the pull of gravity upon the body and applied at its center of gravity G. It is evident, then, that under the influence of the earth's pull, every body tends to assume the position in which its center of gravity is as low as possible. 86. Method of finding center of gravity experimentally. From the above definition it will be seen that the most direct way of finding the center of gravity of any flat body, like that shown in Fig. 70, is to find the point upon which it will balance. FIG. 69. Center of gravity of an irreg ular body |