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# hw3 - PROBLEM 12.59 I A turntable A is built into a stage...

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Unformatted text preview: PROBLEM 12.59 I A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable E 12 s after the turntable begins to rotate. Knowing that the trunk undergoes / H ‘ V i a constant tangential acceleration of 0.75 ft/sz, determine the coefﬁcient of static friction between the trunk and the turntable. : Uniformly accelerated motion on a circular path. ,0 = 8 ft v = v0 + a,t / F}, mat ’ ﬂ; = 0 + (0.75)(12) = 9 ft/s f 27\$ ‘ i \i» R, ma, F, = ma, = Ear: F, = Wﬂ : 93W = 0.0233 W 1 g g 32.2 «K 7 3 W 9 “ " 32,1161”:sz Fn=——(—)——=0.3144W gp (32.2)(8) F : JFf + Ff = 0.315 W This is the friction force available to cause the trunk to slide. The normal force N is calculated from equilibrium of forces in the vertical direction. ZF},=0:N—W=O N:W =0.315 as :0315 4 Since sliding is impending, y‘, = i; PROBLEM 12.73 The velocity of block A is 2 m/s to the right at the instant when r = 0.8 m and 6 = 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine, at this instant, (a) the tension in the cable, (/7) the acceleration of block A, (c) the acceleration of block B. SOLUTION AI” 9 J / g5 97 / ’ 1&9 /, «a m ‘ér L; ‘ mg, Let r and (9 be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block 3. Radial and transverse components of VA. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components. ' e, = —vl,,cos30° I” Z V 2 VA 1 r —2cos30° = 71.73205 m/s ll r9 = v6 2 VA -e0 = *vAsin30" : 2sin30° : 1.000 m/s2 9' = E 2 L000 : 1.25 rad/s r 0.8 Constraint of cable: r + y5 = constant, r'+vB :0, ﬂag :0 or 'r‘:—a,, (1) For block A, i. ZFX : mAa,1 : TcosH = mAaA or T = mAaA sec6 (2) For block B, +1 2F, 2 mBaB ; ,nt i T = mBaB (3) Adding Eq. (1) to Eq. (2) to eliminate T, mBg : mAaA secH + mBaB (4) Radial and transverse components of a4. Use a method similar to that used for the components of velocity. ii 7 r93 = (1,, 2 3/, -e,. : —a(,cos€ (5) Using Eq. (1) to eliminate F and changing signs gives a3 = aAcosB — r02 (6) PROBLEM 12.73 CONTINUED Substituting Eq. (6) into Eq. (4) and solving for a111, m3(g + r93) (25)[9.81+(0.8)(1.25)2] mA seca + mB c056 ’ 205ec30° + 25 cos30° (11/, z = 6.18 in/s2 From Eq. (6), a3 = 6.18cos30° — (0.8)(125)2 = 4.1mm2 (a) From Eq. (2), T : (20)(6.18)sec30° = 142.7 T : 142.7 N 4 (b) Acceleration ofblock A. 34 = 6.18 m/s2 —— ‘ (c) Acceleration ofblock B. a3 = 4.10 m/szl ‘ PROBLEM 12.91 A small ball swings in a horizontal circle at the end of a cord of length [1 , x ( "“X which forms an angle (91 with the vertical. The cord is then slowly drawn ' ‘ m,” ‘ “ through the support at 0 until the length ofthe free end is Z7 . a Derive a ’11 3;” a; a» relation among 1], 12, 01, and 62. (b) If the ball IS set in motion so that "' r-“Trif H" ‘ "*u f" initially Z = 2 ft and 0 = 40°, determine the angle :97 when 17 = 1.5 ft. {ff 1‘ )3" 9‘ :1 3,; 1 l - 2 KK3:;&~:: _ T _ : :n'. E: SOLUTION (a) For state 1 or 2, neglectin the vertical component of acceleration g ZFV =0: TcosQ—WzO T = W0036 ‘bl ,2- / w l o t . . mv‘ 11in 2 man: Tsm6l = WschosQ = p '- . m an But p 2 fsint? so that W v2 = ﬂsinzﬂcosa = l€gsin9tan6 m v1 2 twig sin61tant9] and v2 = ‘MnginHZ tanél2 ZMy = 0: Hy = constant rlmv1 : rzmvz or V151 sin6l1 = vzgzsinﬁ2 ring sin aplsinal tan .91 : £32” skier/sine2 tang2 3% sin3 {91 tantSll = sin3’6l2 tanﬂz 4 (b) With 191 = 40°, 51 = 2 ft, and £2 =1.5 ft (2)3 sin3 40°tan 40° = (1.5)3 sin3 93 tame2 sin362tan62 — 0.52824 = 0 62 = 49.8° 4 PROBLEM 13.5 mt An automobile weighing 4250 lb starts from rest at point A on a 60 \7'*’/ W incline and coasts through a distance of 500 ft to point B. The brakes are A ' “ » . .3. t?” ,6 then applied, causing the automobile to come to a stop at point C, 70 ft \ from B. Knowing that slipping is impending during the braking period and neglecting air resistance and rolling resistance, determine (a) the speed of the automobile at point B, (b) the coefﬁcient of static friction between the tires and the road. SOLUTION V» Given: Automobile weight, W 2 4250 lb Initial velocity at A, VA : 0 ft/s ' o ‘ . 0 ﬂ Incline angle, a — 6 Vehicle coasts 500 ft from A to B ’ Vehicle brakes at impending slip for 70 ft from B to C. Ki VF : F ‘6 Find: Speed of automobile at point B, VB “N Coefﬁcient ofstatic friction, y (a) UN]; 2 WM : (42501b)(500 ft)(sin6°) : 22.212><104 lb-ft l 7 UAW”; 2 TB *TA : Emv' —0 (4250 lb) 7 22.212 X1041b-ft = i 7 _ 2 (32.2 We) vB—O v3 2 58.0 ft/s 4 (b) UHF : A )L. — FdBﬁC : TC — TA : 0 613% : 70ft F = pN Where ,u = coefﬁcient of static friction UHC =(42501b)(sin 6°)(570 ft) — F(70 ft) F = ,u(42501b)cos6° (4250 Ib)(sin 6°)(570 a) = —#(42501b)(cos6°)(70 ft) = 0 ,u : ﬂtan6° : 0.856 70 ,u = 0.856 4 PROBLEM 13.21 The two blocks shown are released from rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine (a) the velocity of block A after it has moved 0.5 m, (b) the tension in the cable. W x Given: Blocks A, B released from rest and friction and masses of pulleys 8 Fl“ neglected. Find: (a) Velocity of Block A, VA, after moving down dA = 0.5 m. (b) The tension Tin the cable. (a) Constraint: VA + 3113 = 0 VB 2 £1214 1 Also, dB : —d4 3 . UH = W,1(sin30°)d‘4 — WB(sin30°)dB . . 0.5 : 10(9.81)(s1n30°)(0.5) — 8(9.81)(sm30°)(—3—] = 17.985 N~m T’OTWLmv'ZnLi ’2 1— a 2—2 4.4 ZmB‘B .y = 510M, + gang/1)- : 5.4444113, U| . = T2 — T1; 17.985 : 5.4444vﬁ VA 21.8175 v14 2 1.818 m/s A]. 4 (b) ForA alone UH = 124511130344 _ WA) 2%m14(v4) 2 10(9.8l)(0.5)(0.5) — T(O.5) = %(10)(1.818)2 : 16.517 T :16.016N T: 16.02N4 PROBLEM 13.64 A 6-lb collar can slide without friction on a vertical rod and is held so it just touches an undeformed spring. Determine the maximum deﬂection of the spring (a) if the collar is slowly released until it reaches an equilibrium position, (b) if the collar is suddenly released. SOLUTION “1" T \$ J. ! F=k\$=l5g , I0 = “Jr—o T“: gm (a) Collar is in equilibrium. +7 2F = (151b/in.)5 — 6 lb 5W = 0.4 in. 4 (6) Maximum compression occurs when velocity at ® is zero. 3:0 m:0 T2 = 0 V2 = ~W6max +ék6éax W — %k5m = ﬂ = 0.8in. m" (15 lb/in.) 6W = 0.8in.4 PROBLEM 13.67 The system shown is in equilibrium when 415 = 0. Knowing that initially Exam ¢ = 900 and that block C is given a slight nudge when the system is in T ‘ that position, determine the velocity of the block as it passes through the m mm r equilibrium position ¢ = 0. Neglect the mass of the rod. Find unstretched length of the spring 92 9 = tan“[ 0.1 19 = 71565" LBD = (0.3)2 + (0.1)2 = 0.3162m . m El “0 g = length at equilibrium 0.4m ~——-_l Equilibrium: (ML EMA = 0.1FS sin9 — 0.6(10 g) = 0 FS = 63.25g FS 2 kALBD: 63.25 g : (8000 N/m)(ALBD) 3 ALBD = 0.07756 m Unstretched length L0 = LED — ALBD = 0.3162 — 0.07756 2 0.23864 m Spring elongation, ALgD when 425 = 90° ALgD = (0.3 m + 0.1m) — L0 = 0.4 — 0.23864 = 0.16136m ¢=90° V120, T.=0 Vn=(Vx)e+(V1)g (VOL, = %k(ALgD)2 = @(ommf 2104.15N-m (mg = —10g(0.6) = —58.86N-m Vl = 104.15 — 58.86 = 45.29 N-m PROBLEM 13.67 CONTINUED At (3 ¢ = 0 (V2)e = %k(AL3D)2 = [\$ N/mj(0.07756 m)2 : 24.06N-m 1 « lOkY ') a T2 =§mv§ =[ Zgjvg = SVZ‘ T,+V1 =T3+ng 0+45.29=5v§+24.06 = 4.246 v2 : 2.06 m/s 4 [‘0 [g V PROBLEM 13.71 A 10—02 pellet is released from rest at A and slides Without friction along passed through E. the surface shown. Determine the force exerted on the pellet by the surface (a) just before the pellet reaches B, (b) immediately after it has SOLUTION (a) 4% we Vl = mg(4sin30°) = Q 20 B 300‘ __ C DATUM “zuz:0 = 1.251b-ft 710 16 v§ = (l009705v§ n+m=u+nzrﬁ=uwmmézsé=nmswﬁ 0é\$\$ \$7” 30° N 10 +/ 2F.:0: Ne#cos30 :0 16 N : 0.54] lb 4 PROBLEM 13.71 CONTINUED mv§ 0.625(1288 : N — 0.62500530° 2 ————-~ r 32.2 2.75 N — 0.54127 = 0.90909 J N = 1.450 lb { ...
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