the string. The string may then be described as the means of communicating the two portions of a stress which act one on each of the masses; in this way, by comparing the actual motion of the masses with the motions given by our calculations, we have a practical test of the truth of the Third Law of Motion. H Example i. In an Atwood's machine the masses suspended A and B, each contain M lbs. ; a small mass C of m lbs. is placed on the mass A; find the resulting acceleration and the tension of the string. I. Consider the motion of the masses (A and C), treating them as one mass. The external forces acting are (i) the weight of (M + m) lbs.; that is, (M + m) 8 poundals, (ii) the tension of the string; let this be T poundals. The weight acts vertically downwards and the tension acts vertically upwards ; hence, the resultant external force is {(M+ m)g- T} poundals downwards, Let (A and C) have a celos downwards; then it has (M+ m2) a pound-celos; so that the force acting upon it must be (M + m) a poundals downwards. Therefore, (M+ m) 8- T=(M+ m) a (i). II. Consider the motion of the mass B. The string being inextensible, whatever motion A may have, B must have an equal motion in the opposite direction; therefore B has an acceleration - a celos downwards. The mass acceleration of B is therefore - Ma pound-celos, downwards. So that it is acted on by Ma poundals. The resultant external force is (Mg - 7) poundals, downwards. For the tension of the string is a stress which acts equally on A and B in opposite directions. B (ii). Therefore Mg- T= - Ma Hence, by subtraction of (ii) from (i), mg=(2M+m) a, that is, 2M + m 78 2M (M + m) Also from (ii) T= Mg + Ma, m a = (iii). T 2M +m 8 Example ii. In an Atwood's machine the masses A and B are each 15} oz. and a mass C of 1 oz. is placed gently on the mass A when at rest; after 2 seconds C is removed ; discuss the resulting motion. I. Consider the motion of the masses A and C, considering them as one mass. or a = I The external forces are its weight, which is 33 x 32 poundals, and the tension, T poundals; let a celos downwards be its acceleration; then its mass-acceleration is fixa pound-celos; whence, as in Ex. i, 33 - T=33a (i). II. Consider the motion of the mass B ; then as before and as in Example i, we have 31 - T=-31 a (ii). Whence by subtraction 2= 44 a=2a, (iii). Thus, as long as the mass C forms part of the mass A each mass has I celo, the mass B upwards and the masses A and C downwards. Hence at the end of 2 seconds each mass will have 2 velos and will have moved over (i ata), that is, 2 feet. Forming the equations (i) and (ii) on the supposition that C is not put upon A, we have, 31 – Tria (i)? and, 31 - T=-31a whence a=0. So that when the i oz. mass is removed the acceleration vanishes; that is, the velocity is uniform ; and the masses each move with 2 velos, A downwards and B upwards. Example iii. Find the stress between the small mass C and the mass A in Example ii of page 47. The mass A and the mass C each descends with i celo. The weight of the mass C, viz, 2 poundals acting on to lb. would produce in it 32 celos. Hence the stress is such that the upward part of it acting on ' lb. destroys in it 31 celos. The upward action of the stress is therefore if poundals. Example iv. A horizontal platform is made to descend vertically with 4 celos; a man stands on the platform with 56 lbs, in his hand; shew that this mass appears to weigh only 49 pounds. The force acting on the 56 lbs. (besides the pressure of the hand) is its weight downwards, that is, 56 x 32 poundals, This would produce in it 32 celos, if the upward force of the man's hand did not prevent. The 56 lbs. actually has 4 celos, which requires 56 x 4 poundals to produce it. Thus, of the weight (56 x 32 poundals) the part 56 x 4 poundals are taken up in producing the 4 celos; so that the upward push of the man's hand must counteract the remainder. In other words this push is 56 x 28 poundals, which is the weight of 49 lbs. 57. To find the acceleration in an Atwood's machine we may proceed as follows. Let the masses be m lbs. and m' lbs. Then, considering the string as the means of preventing relative motion between the masses, and therefore the mass moved to be the sum of the masses m lbs. and m' lbs., the tension of the thread becomes an internal stress and has no effect on the mass-acceleration produced. The resultant external force in the direction of the motion of the particles is (the weight of the greater mass 1 – that of the less m'), hence we have mg – m'g=(m + m') a. 58. The problem of Atwood's machine is similar to the following Two particles A and B of masses m lbs. and m' lbs., connected by a light inextensible string, are placed with the string stretched, on a smooth horizontal plane; A is acted on by a horizontal force mg poundals in the direction BA, and B is acted on by a horizontal force m'g poundals in the direction AB; find the tension of the string and the acceleration of the masses. 2mm' It will be found that the tension is m+me'8 poundals; and that the acceleration is m - m' m + m's celos. EXAMPLES. XVII. Find the tension of the string and the acceleration of the masses in an Atwood's machine, when the masses A and B have the following values : 1. Mass of A, 1 lb. oz.; of B, 15 oz. 5. In an Atwood's machine the masses A and B are 3 lbs. 15 oz. each; a mass of 2 oz. is gently placed on A and is removed by a catch after A has descended 1 ft.; shew that A will take 4 seconds to descend the next 4 ft. 6. In an Atwood's machine the masses are each 7 lbs. 15 oz.; a mass of 2 oz. is gently placed on one of them and is removed by a catch after descending 8 ft.; how long will it take to pass over the next foot? 7. In an Atwood's machine it is observed that after descending 8 ft. from rest the masses have a velocity of 4 ft. per second; the larger mass is 32} oz.; find the other mass. 8. In an Atwood's machine the masses are equal and it is observed that a weight of 1 oz. added to one of them produces i celo; what are the masses? 9. In an Atwood's machine the pulley is suspended by a hook from a nail; shew that (with the usual notation) the force on the 4mm'g nail is (m2*yon poundals + the weight of the pulley). 10. Shew that in an Atwood's machine the tension of the string diminishes as the ratio of the larger mass to the smaller increases, their sum remaining unaltered. 11. Prove that in an Atwood's machine when the tension of the string is one fourth of the sum of the weights m : m' =3+2N2 : 1. L. D. 4 12. Two masses m lbs. and m' lbs. connected by an inextensible thread, are placed with the thread straight on a smooth horizontal plane; the mass m is acted on in the direction of the string away from m' by a force f poundals, and m' is acted on by a force fi poundals in the opposite direction ; find the acceleration of the masses. 13. Prove that if in Ex. 12 the forces are equal to the weights of the masses on which they act, the acceleration is m-m' m+m8 celos. 14. A bullet fired directly into a block of wood will penetrate 3 in.; what part of its velocity will it lose in passing through a board of the same wood i in. thick supposing the resistance uniform? 15. A mass of i cwt. is placed on a lift which descends vertically with a uniform acceleration of 10 celos; find the stress between the lift and the mass. 16. If the lift in Question 16 is ascending vertically with an acceleration 16 celos, what is then the pressure ? 17. A man can raise a sack of corn weighing i} cwt.; shew that when on a lift which is made to descend vertically with 8 celos, he can raise 2 cwt. 18. A balloon ascends vertically with a uniformly accelerated motion so that a weight of 1 lb. produces, on the hand of the aeronaut sustaining it, a downward pressure equal to that which 17 oz. produces when at rest ; find the height which the balloon attains in i min. from rest. 19. Find the pressure on the mass of 2 oz. in Question 5. 20. Find the magnitude of the stress between the 2 oz. and the mass on which it is placed in Question 6. |