1. A mass of 200 grammes is acted on by a force equal to the weight of 10 grammes for 20 seconds. What distance will the mass have passed through, and what velocity will it have acquired? 2. A body whose mass is 108 lbs. is placed on a smooth horizontal plane, and under the action of a certain force describes from rest a distance of 111 feet in 5 secs.; what is the force in British absolute units? 3. A body resting on a smooth horizontal table is acted on by a horizontal force equal to the weight of 2 ounces, and moves on the table over a distance of 10 feet in 2 seconds; find the mass of the body. 4. It is found that a body, considered as a point, has its velocity increased by 7 feet per second in any second of its motion; it is known that the body weighs 23 lbs.; what is the magnitude of the force producing the acceleration? How many pounds of matter would this force support against gravity in a place where g=32.2? 5. How long must a force of 14 pounds act on a mass of 1,000 tons to give it a velocity of one foot per second? 6. Calculate in pounds the moving force which, acting for a minute upon the mass of a ton, will get up in it the velocity of 30 miles an hour. 7. A force equal to the weight of one lb. acts on a mass of 2 lbs. for one second; if the value of g be 32, find the velocity of the mass and the space which it has travelled over. At the end of the first second the force ceases to act; how much farther will the body move during the next minute ? 8. How far will a lateral pressure of an ounce move a pound on a smooth horizontal plane in five minutes? 9. Taking 10 pounds as the unit of mass, a minute as the unit of time, and a yard as the unit of length; compare the resulting systematic unit of force with that of the ft. -lb.-sec. system. 10. Determine the unit of time in order that the foot being the unit of length, the value of the intensity of gravity may be expressed by 1 instead of 32·2. 11. A fine string, carrying two unequal masses at its extremities, is hung over a smooth pulley. It is observed that at the end of 5 seconds the heavier weight is descending with a velocity of g feet per second. Find the ratio of the masses. 12. Two weights of 5 pounds and 7 pounds are connected by a string passing over a fixed smooth pulley; the system having no initial motion, find the velocity after three seconds. 13. Two masses of 48 and 50 grammes respectively are attached to the string of an Attwood's machine; and starting from rest, the larger mass passes through 10 centimetres in one second. Determine from these data the value of the acceleration due to gravity, the units being the centimetre and the second. 14. In Attwood's machine one of the boxes is heavier than the other by half an What must be the load of each in order that the over-weighted box may fall through one foot during the first second? ounce. 15. In an Attwood's machine-neglecting friction and the inertia of the wheels -if the two weights attached to the string be each 5 ounces, what must the moving weight be, in order that at the end of one second these weights may be moving with a velocity of one foot per second? In this case, how far have the weights moved from rest in the first second? 16. In Attwood's machine, where the weights are 17 oz. and 16 oz., find the acceleration and the tension of the cord. 17. The two ends of a string passing over the pulley of an Attwood's machine are loaded as follows:-A with 16 and B with 15 ounces. Find the tension at A, when it is in motion downwards. 18. The weights at the extremities of a string which passes over the pulley of an Attwood's machine are 500 and 502 grammes. The larger weight is allowed to descend; and 3 seconds after motion has begun 3 grammes are removed from the descending weight. What time will elapse before the weights are again at rest? 19. A smooth inclined plane, whose height is one half of its length, has a small pulley at the top over which a string passes. To one end of the string is attached a mass of 12 lbs., which rests on the plane; while from the other end, which hangs vertically, is suspended a mass of 8 lbs.; and the masses are left free to Find the acceleration and the distance traversed from rest by either mass in 5 seconds. move. 20. A mass of 63 lbs. is put on a smooth horizontal table and connected by a fine thread to a mass of 14 lbs., which hangs over the edge of the table; if the latter body is allowed to fall, dragging the former after it, what force does it exert on the former body, and what is its accelerative effect on the velocity of that body? SECTION XXVIII.---COMPOSITION OF FORCES. ART. 145.-Composition of Forces. Force involves mass-vector in its idea (Art. 141), but as the mass acted on is constant, the different forces can be represented as simple vectors. The magnitudes of the vectors must be made proportional to the magnitudes of the forces, and be drawn in corresponding directions. Suppose that the particle at A (Fig. 19) is simultaneously AC represents their resultant on the same scale. Hence the magnitude of the resultant is angle which has the tangent p2 + q2 F, and its direction is the qpLopp. per L adj. When the two forces are not inclined at a right angle, the construction leads to a parallelogram instead of a rectangle. ART. 146.-Coefficient of Friction. Friction is a force which opposes the sliding of one body over another. Its utmost amount is proportional to the force with which the two surfaces press against each other, provided the surfaces are plane. Hence μ F resistance F normal pressure. = The constant μ is greater when motion does not take place, than when it does take place. In the former case it is called the coefficient of statical friction, in the latter the coefficient of kinetic friction. The weight of a body resting on a plane inclined to the horizon at a is mg F downwards. The components of this force along and normal to the plane are my sin a F down the plane, and mg cos a F normal. Hence the frictional resistance is μmg cos a F up the plane. Motion will be about to ensue when mg sin a - μmg cos a = 0, The limiting angle at which sliding just does not commence is called the angle of repose; its tangent is equal to the coefficient of statical friction. EXAMPLES. Ex. 1. A weight of 20 lbs. rests on a horizontal plate which is made to ascend first with a constant velocity of one foot per second, second with a velocity constantly increasing at the rate of one foot per second per second; find in each case the pressure on the plate. Take the acceleration due to gravity at 32 feet per second per second. X In the first case there is a force on the weight of 20 × 32 poundals downwards and no force on the plate, since the velocity is constant; therefore the pressure on the plate is 640 poundals. In the second case there is as before a force of 20 × 32 poundals downwards on the weight, and there is in addition its resistance due to the acceleration imparted by the ascending plate, namely 20 × 1 poundals, hence the whole pressure on the plate is 20 × (32 + 1) poundals, that is, 660 poundals. Ex. 2. Forces equal to the weights of 2 lbs., 3 lbs., 4 lbs., 5 lbs. act on a particle in the directions north, east, south, west respectively; find the magnitude and direction of the resultant force. Here the intensity of gravity enters as a factor in each of the forces, hence the vectors representing the forces may be taken as 2 L north, 3 L east, 4 L south, 5 L west. The first and third. compounded give 2 L south, and similarly the second and fourth compounded give 2 L west; hence the resultant vector is 2√2 L south-west, and the resultant force is 2√2 lbs. in the direction south-west. Ex. 3. Two rafters, making an angle of 120°, support a gasalier weighing one cwt.; what is the pressure along each rafter? Draw the vector AB representing the force of 112 lbs. (Fig. 20). Draw BC parallel to AD and BD parallel to AC. It may be shown that ACB is an equilateral triangle; therefore the vector AC has the same length as AB; and so has AD. Hence the pressure along each rafter is 112 lb. weight. Ex. 4. The slope of a plane is 45°; find the time in which a body would slide from rest down 100 feet of its length, the angle of friction being 15°. The intensity of the weight is 32 poundals vertical per lb., i.e., 32 sin 45° poundals downwards per lb. + 32 cos 45° poundals normal per lb. But tan 15° poundals up per lb. = poundal normal per lb., 32 cos 45° tan 15° poundals up per lb., .*. (32 sin 45° – 32 cos 45° tan 15°) poundals down per lb., 1 foot per second per second = poundal per lb. ; 32 √/2 |