5 he ension in each sring is w (= mg) Answers, Even-Numbered Problems, Chaper 5 54 (a) 540 N (b) The θ = 0 58 (a) (b) 4 53 0 N 4 336 0 N 50 (a) The free-body diagram for he car is given in Figure 50 The verical weigh w and he ension T in he cable have each been replaced by heir x and y componens (b) 5460 N (c 70 N Figure 50 5 (a) 849 N (b) 600 N Figure 5a, b 54 (a) T = wsin α (b) T = wsinα (c) n = n = w cos α A B
(d) For α 0, T = T 0 and n = n w For α 90, T A B n A = n = 0 B = w, T = w and 56 a) 396 m/s (b) 7 N 58 (a) 08 m/s (b T = 04 N; T = 64 N A B 50 (a) 08 m/s φ = 3 (b) 59 m/s 5 (a) 60 m (b) 6000 N 54 (a) (b) 78 m/s downward 4 m/s (c) n = 0 means ay (d) In par (a), 680 N In par (c), T = 0 3 56 (a) A = 50 m/s B = 050 m/s (b) A, 550 m/s (c) 56w = 56 mg (d) 4 87 0 N = g The suden should worry; he elevaor is in free-fall 58 (a) The fricion is saic for P = 0o P = 750 N The fricion is ineic for P > 750 N (b μ = 0556 μ = 0370 s (c) When he bloc is moving he fricion is ineic and has he consan value f = μ n, independen of P This is why he graph is horizonal for P > 750 N When he bloc is a res, f s = P since his prevens relaive moion This is why he graph for P < 750 N has slope + 530 (a) If here is no applied force, no fricion force is needed o eep he box a res (b) f s = 60 N in he opposie direcion
(c) 60 N (d) 80 N is required o eep i moving a consan velociy 53 μ = 05 534 (a) 540 m (b) 63 m/s 536 (a) μ = 0556 (b) 3 m/s The acceleraion is upward and bloc B slows down 538 Low pressure, 0059 High pressure, 000505 540 30 N 54 (a) 93 (b) 09 m s (c) 3 m/s μ mg 544 (a) F = cos θ + μ sin θ (b) 90 N 548 (a) (5/4)g, down (b) (3/4)g, down 550 (a) 090 (b), 44 m/s 55 (a) 69 s (b) No 554 (a) 069 (b) 0067 m 556 (a) 54 m/s (b) 833 N ; 93 N (c) 4 s 3
(d) mg = 760 N, wice his rue weigh 558 (a) 30 m (b) 450 N 560 (a) 464 m/s, upward (b) 05 N 56 The ension in he lower chain is equal o w he ension in he rope, w T he ension in he upper chain is also w mx 564 Tx ( ) = F LM ( m) + 566 (a) 0 N (b 50 N 568 (a) 69 N (b) 0 N 5 570 (a) 57 0 N (b) 470 m 57 50 m s 574 8 m s 576 (a646 m, which is greaer han 40 m You don sop before you reach he hole, so you fall ino i (b) 6 m s μs 578 The fracion ha hangs over is + μ 580 93 m/s = 655 mi/h He was guily 58 s In his ime, he ruc moves 543 m s 4
584 The ime i aes o reach he ground is 4 s The verical componen of he PAPS force is 450 N The horizonal componen is 400 N, 586 (a) he blocs will slide o he lef; (b) a = 0067g = 0658 m s (c) 44 N 588 390 g 590 05 N 59 (a) he blocs will have he same acceleraion, m s (b) 7 N (c) The sring will be slac The 400-g bloc will have a = 78 m s and he 800-g bloc will have a = 93 m s, unil he 400-g bloc overaes he 800-g bloc and collides wih i 594 (a) an θ = a / g (b) θ = 946 (c) 45 596 (a) m/s (b) 85 m/s 598 68 m s, abou 606 m h 500 ( ) v () = v + e m v 0 50 (a) v = v 0 m + 4m ; mv (b) = 0 mv (c) x = 3 3 0 x 3 v 0 0 = v m + m 5
504 (a) 30 N (b) 353 m/s The number of revoluions per second is v 353 m/s = = 0749 rev/s = 449 rev/min πr π(075 m) (c) 99 rev/min 506 (a) 84 m ; 06 s (b) 06 m 508 (a) 434 N (b) 98 m s The answer doesn depend on he car s mass, because he cenripeal force needed o hold i on he road is proporional o is mass and so o is weigh, which provides he cenripeal force in his siuaion 50 (a) θ = 773 above he horizonal The magniude of he ne force exered by he sea 854 N (b) The magniude of he force is he same, bu he horizonal componen is reversed 5 (a) 3 m s (b) The 5390 N 54 v = grm m 56 F = (58 N/s) and F = 440 N x y (b) The graph is given in Figure 56 (c)648 N a an angle of 3 Figure 56 58 (a) F = 68 N A 6
(b) F = 304 N, where he minus sign indicaes ha he rac pushes down on he B car 50 (a) gm A = ( M + m) an α+ ( M an α) a x gm = ( M + m) an α + ( M an α) gm ( + m) an α a = y ( M + m) an α+ ( M an α) (b) When M >> m, A 0, as expeced (he large bloc won move) Also, g an α a = g = gsin αcos αwhich is he acceleraion of he x an α+ ( an α) an α + bloc ( gsinα in his case), wih he facor of cos α giving he horizonal componen Similarly, a gsin α y (c) The rajecory is a spiral 5 IDENTIFY: Apply F = ma SET UP: Le + x be direced up he ramp EXECUTE: The normal force ha he ramp exers on he box will be n = wcosα Tsinα The rope provides a force of Tcosθ up he ramp, and he componen of he weigh down he ramp is wsinα Thus, he ne force up he ramp is F = Tcosθ wsin α μ ( wcos α Tsin θ) = T(cosθ + μ sin θ) w(sin α + μ cos α) The acceleraion will be he greaes when he firs erm in parenheses is greaes and his occurs when an θ = μ EVALUATE: Small θ means F is more nearly in he direcion of he moion Bu θ 90 means F is direced o reduce he normal force and hereby reduce fricion The opimum value of θ is somewhere in beween and depends on μ When μ = 0, he opimum value of θ is θ = 0 54 (a) The free-body diagram for he falling ball is seched in Figure 54 (b) Newon s Second Law is hen ma = mg Dv Iniially, when v = 0, he acceleraion is g, and he speed increases As he speed increases, he resisive force increases and hence he acceleraion decreases This coninues as he speed approaches he erminal speed mg (c) A erminal velociy, a = 0, so v = in agreemen wih Eq (53) D 7
dv g (d) The equaion of moion may be rewrien as = ( ) v v This is a separable d v equaion and may be expressed as ( ) v = v anh g v dv g v = d or arcanh g v v v v = v v 56 (a) F = 6 N, which is insufficien o raise eiher bloc; a = a = 0 (b) F = 47 N The larger bloc (of weigh 96 N) will no move, so a = 0, bu he smaller bloc, of weigh 98 N, has a ne upward force of 49 N applied o i, and so will 49 N accelerae upwards wih a = = 49 m s 00 g (c) F = N, so he ne upward force on bloc A is 6 N and ha on bloc B is 4 6 N 4 N N, so a = = 08 m s and a = = 4 m s 00 g 00 g 8