a_,-:1....1;.;. and m1+m2 ’”1+m2° 3 1 il , ::3 ·-— F° 7;-*- ° ZZ-—· , Solution. Force F will cause the entire system to move with an acriielvgation a. The equations of Newton’s second law for each block W1 e F-fi= vw. fi--1‘z= ma. fz—f¤= ma. fs= ma where ji, fz, is are the tensions in the threads (Fig. 204). I is ’ n · t · fa f2 J9 777. 772 772 . nI I I F é Fig. 204 By solving these equations it is possible to calculate all the ten- sions as well as the acceleration a with which the system will move. 68. Solution. It is difficult to start a heavy railway train when the couplings between the freight-cars are tensioned. In this case the traction of the locomotive has to impart an acceleration to the whole train at once. lf the train is iirst backed up, the couplings between the cars will be slackened and the locomotive can with the same {ull impart much larger accelerations first to the nearest car and t en, successively, to all the other cars. 69. Solution. lf before the motion begins all the couplings in the train are tensioned, the break may occur in the couplings of the cars closest to the locomotive. The tension in these couplings should be greatest since it is intended to produce an acceleration for the greater mass of the oars behind all at once (see Problem 68L. If all the couplings between the cars are first slac ened the break may occur in any Fart of the train depending on the ratios of the ten- sions in the coup ings between separate cars. 70. The dynamometer will show a force: (1) in ~f= 1 1121: (2) fn iv F= 2 kefi ¤¤d (3) ;n—.:£?]:=1.5 kgf. Solution. In all three cases the system will move with some acce- leration a in the direction of the larger force and the dynamometer will show the bonding force fn acting between the weights. To hud fn, it is necessary to write the equation of IjIewton’s second law separately for each weight. For the first case (Fig. 205). F··fn= Ma. fn-—f= ma
Hence, a—---L -M ·-|- m and Mfn ·— F-·· HQFE (F ··}) Since m < M and 1 it may be assumed that M+ m fn W f The other cases can he considered in a similar manner utilizing J, the equations of Newton’s second _, mL,. F law and the given ratios of masses. i il ’ 71. a=!tB?. g. D ' Q P +0 ° Tn, • • • • • • M P =·——- 1 k . F' 205 I P4-Q ( + ) 1 . g Solution. The bodies P and Q move with accelerations of equa magnitude a. The body P is acted uson by the force of gravit and the tension in the thread f, and the bo y Q by the tension in the thread and the force of friction j, = kQ. The equation of N ewton’s second law for the motion of each body will be P Q P- =—— , —-k =-—- I g 4 f Q g a The solution of these equations determines the values of a and j __P-kQ __ PQ 1 4- ,,+0 z and 1‘—————·,,_,_Q( +k> .. mg . 72. a-. 2M +m , _ 2M (M -l-m) . T- 2M -|—m. g' 2Mm { :.50-[$7; g, F - 2T Solution. All the bodies in the system move with accelerations of equal magnitude a (Fig. 206). T e left-hand weight M is acted upon by the force of gravity Mg and the tension in the thread T, and the 1·1ght·hand weight by the force
CHAPTER 1. MECHANICS 15{ of gravity Mg, the pressure of the small weight f and the tension_in the thread T. The small weight m is acted upon by the force of gravity mg and the pressure f exerted by the /, / J, weight M. For each of the three masses the equa- tions of N ewton’s second law will be T — Mg = Ma Mg + f — T = M a mg — f = ma The solutions of these equations deter- T la mine the values of a, T and f. _ . The pressure on the axis of the pulley ` will be equal to twice the tension in the Mg M threads F = 2T. T mi]: 1;-.. .:1/f‘&.fri’2?a.»o.oS. _ l (P 2-P 1] 8 _ mg Note. The acceleration of the weights ,- M can be found from the equations of New- My ton’s second law (see the solution to Prob- _ . P2-P, Fig. 206 lem 72) and will be oz -17:*3- g, and the 4 z . time of motion from the kinematic equations of uniformly acce- lerated motions from rest will be 1/ W t= — a 74. The centre of gravity moves down with an acceleration j:(Pr-—Pz)2 g (P1+P2)z Solution. After a time t each weight will be dgsplaged from the t ... initial position by a distance S = Lg- Wh61‘6 a = ·pi·j|_·Ii· 8 (SBB the solution to Problem 72). _ , In this case the centre of gravity of the system should obviously move down a certain distance L from the_1n1t1al position towards the larger weight (Fig. 207). Upon determining the centre of gravity, the distances of the centre of gravity from the weights should be inver- sely proportional to the magnitudes of these weights, 1.e., S -4- L ___ lh S -· L __ P2 11-1218
162 ANSWERS AND SOLUTIONS or E-$-1 P2 P,—P2 1 Pl+P2 P2 . atz Inserting the value S=-2- , we get _·Pi—P2 (liz L*Pi+PzX 2 Comparing this result with the formula for the path of uniformly accelerated motion and then inserting the values of the acceleration of the weights we find that the centre of gra- * - · f vity Should move down with an acceleration i=%E ¤= (i$%)2g # The magnitude of the acceleration of the I centre of gravity is less than that of each · weight separately. _ . 75. In the first case the system moves w1th | an acceleration a == 9 cm/s“ and the force of Q M fricggn petween the block and the cart is f= <»l;| In the second case the acceleration of the · , block is a, = 7.5 m/S2 and that of the cart az: 0_""‘ _iL ‘_0 = 0.25 m/S2; the force of friction isi = 0.5 kgf. T ‘ Solution. First case. The maximum force of ;+°·` p friction at rest, if,. = kmg = 0.5 kgf, is larger -L B than the flqrcp F applied to tllige blpcki)1Folr 1 T * this reason t e orce F cannot ma e t e oc _ Slide along the cart. The entire System will Fig. 207 move as a single whole with a total accelera- tion a, and the force of friction should be deter- mined as the magnitude of the bonding force from the equation of Newton’s second law. The second law equations for the block and the cart will be F'*'fjr.; may ff'--; hence p MF a=·-—-:9 cm/sz, f;r=·-———=l80 gf M —\- m M -{-m Second CQSB. The maximum force. of friction fj]- is less than t.llB force F. For this reason the force F will cause the block to slide along the cart. The block and the cart will have different accelerations ai and az, The force of friction will have its maximum value jh.: kmg: = 0.5 kgf during motion.
CHAPTER I. MECHANICS 163