Dynamics Homework 6 - ENGINEERING MECHANICS ~ Dynamics W.F....

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Unformatted text preview: ENGINEERING MECHANICS ~ Dynamics W.F. Riley & L.D. Sturges 16—5 A 450—lb crate 18 in. wide by 45 in. tall is sitting on the back of a truck as shown. The truck is traveling at 65 mi/h when the driver suddenly applies the brakes. Determine the minimum distance required to stop the truck (without causing the crate to tip over) and the minimum static coefficient of friction required between the crate and the truck to prevent the crate from slipping. Solution The crate is in translation with ay = a = 0. The equations of motion are +_) 217 _F __..4_§_0_ a 32.2 X N-450 o N(9-a) — 22.517 0 At the point of impending tip, the back corner of the crate begins to lift away from the bed of the truck and the normal force resultant acts at the front corner (a = 0). At this point N = 450 lb = constant F = 9N/22.5 = 180 1b = constant ax = -0.40g = -12.88 ft/s2 = constant and the minimum required coefficient of friction is u F/N = 9/22.5 = 0.4 Since a relationship between velocity and position is desired, rewrite the acceleration using the chain rule of differentiation and integrate I v dv I a dx = -12.88 I dx 2 V /2 = -12.88x + Cll = 4544.222 - 12.88x where the constant of integration is has been chosen such that V = 65 mi/h = 95.33 ft/s when x = 0. Then the truck will come to a stop (v = 0) when V = O ; 4544.222 — 12.88x when ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 16-27* The 20-1b slender bar AB shown is supported by a frictionless pin at A and a cord at B. If the cord at B breaks, determine the acceleration of the mass center of the bar and the reaction at support A at the instant that motion begins. Solution The bar rotates about a fixed axis through A. The equations of motion are 20 +9 2 = = —————— Fx Ax 32.2 aGX 20 +¢ = _ = ______ _ Egy 20 Ay 32.2 aGy = . = a C+ EMA _1 5(20) 1; where w = 0 (starts from rest) 2 2 a = —1.5w = O ft/s Gx — 1 5a ft/s2 aqy — . 20 2 I = —l—(—————— (3) 1 A 3 32.2 2 1.86335 slug-ft Therefore 2 a = 16.100 rad/s D 2 24.150 ft/s l $1) I! Gy — 2 ac = 24.2 ft/s l . . . . . . . . . . ... . . . . . . . . . . . ........... Ans. A = 0 lb X 3 II U1 5 _) ENGINEERING MECHANICS - Dynamics 16—35 A 300-lb force is applied to the end of a rope which is wrapped around the outside of a 6-ft diameter cable drum as shown in Fig. a. a. If the 600—lb cable drum can be treated as a uniform cylinder,determine the angular acceleration of the cable drum and the time required to rotate the cable drum 3 revolutions. b. Repeat part a if the 300—lb force is replaced with a 300—lb block as in Fig. b. Solution The wheel rotates about a fixed axis through its mass center G while the block moves in a straight vertical line. Therefore, separate free—body diagrams must be drawn of the two separate objects. The equations of motion of the drum and block are C+ ZMb 3T 16g 300 32.2 900 2 [ Jm +¢ SF 300 — T = a Y and l 2 32.2 2 125.7764 slug-ft a. If the to the drum, then T = 300 lb and Eq. a 2 a 7.16 rad/s D b. If the block is attached to the then solving Eqs. 245 lb 2 5.85 rad/s D T a W.F. Riley & L.D. Sturges 3UHb (M 300—lb force is applied directly gives drum, a and b together gives ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. sturges 16-44 A SO—kg rectangular plate of uniform thickness is supported by a smooth pin and a cable as shown. If the cable at B breaks, determine the angular velocity of the plate and the magnitude of the force exerted on the plate by the pin at support A a. When the mass center of the plate is directly below A. b. When side AB of the plate is vertical (corner B is directly below A). Solution The mass center of the plate travels a circular path about the fixed axis of rotation A. Therefore fl=raA+UA aG AG et en where rAG = 279.509 mm, ¢ = 26.5650, and the equations of motion are +F ZFt = 50(9.81) cos 6 — At = 50(o.27951 or) 50638 2 +5. 217'“ = An - 50(9.81) sin 9 = 50(o.27951 (a) +1) 214 = 50(9.81)(0.27951 cos 6) = I or A A 'QV‘ where r 2 2 2 2 IA = -I%-(50)[0.250 + 0.500 ] + 50[0.125 + 0.250 ] 2 = 5.20833 kg°m The moment equation can be solved for the A\ angular acceleration )6 a = 26.32313 cos 9 0L f‘ G .— Rewriting the angular acceleration using ‘R<\ the chain rule of differentiation Y;GPL Y. Us; Ac. a_ do _ dw d6 _w do) ' dt ' d6 dt ‘ ale and integrating gives (Problem 16—44 continues ...) ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 16-44 — cont.) I a d9 = I 26.32313 cos 6 d9 I... w2/2 = 26.32313 sin 9 — 11.77204 where the constant of integration has been 0 chosen so that w = 0 when 9 = ¢ = 26.565 ° 2 a. When 9 = 90 , a = O rad/s D m = 5.39 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. A = 897 N n A = t 0 N 2 2 A = A + A = 897 N . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. n t O 0 b. When 9 = 90 + ¢ = 116.565 , a = —11.77204 rad/S2 w 4.85 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. 767.759 N In ll At = -54.838 N A = 770 N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 16-54* Two 400-mm diameter disks and a 240-mm diameter disk are joined to form a spool as shown. The 125 kg spool has a radius of gyration of 125 mm with respect to the axis of the spool. If a force p of 500 N is applied to the spool through a cord that is wrapped around the cylinder, determine the acceleration 56 of the mass center of the spool and the angular acceleration a of the spool if a. The horizontal surface is smooth (u = O). b. The horizontal surface is rough (u = 0.25). Solution The equations of motion for the spool are +—) 2F = 500 — c = X f 125aG +T 2F = c - 125(9.81) = o y 12 (1+ ZMG = 0.120(500) + 0.2oocf = 160: K. where a = 0 since the mass center of ‘ Gn (if the spool has no motion normal to the surface, c 2 2 'n IG = 125(O.125) = 1.95313 kg°m a. If the surface is smooth (Cf = 0) then 2 a6 = 4 m/s -% .. . . . . .................. . . . . . . . . . ....... Ans. 2 a = 30.7 rad/s D ..... . . . . . ............... . . . . . . . . .. Ans. b. If the surface is rough (u = 0.25) then the spool may roll without f Guessing that the wheel will roll without slipping slipping (a6 = 0.2a; C S uCn) or it may slip (Cf = uch; aG ¢ 0.2a). C = 500 - 125(O.2a) = 9.76563a - 300 f 2 a = 23.0 rad/s D ................................... Ans. 2 a6 = 4.60 m/s -% ..... . . . . . . . . . . .................... Ans. an = 1226.25 N 1‘ cf = -75.28 N = 75.28 N —) Note that the amount of friction required to prevent slip (75.28 N) is less than the amount of friction available (0.25Cn = 306.6 N). Therefore, the assumption of rolling without slip is valid. ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 16-73 A 16-lb ladder 12 ft long slides in a smooth corner. The ladder is initially at rest 0 with 9 = 0 when the lower end is disturbed slightly. Determine the angle 9, the angular velocity 0, and the velocity 93 of end B when end A loses contact with the vertical wall. Solution The equations of motion of the ladder are +9 2F = A = ma x Gx +T 2F = B - m = ma Y 9 Gy (2+ ZIMG = B(L/2) sin 6 - A(L/2) cos 6 = I60: where 2 IG = mL /12 To get the relationship between the accelerations a , a , and a, must do Gx Gy the kinematics of the bar. First, using the known information about the Es motion of the ends of the bar I\ a = a + a as 3!! aB/A I“ A _ A 6 A _ e A x a3 1 — aA J + La(cos 1 + sm J) a 2 A A +Lw(—sin91+cosej) 8 get 2 aA = -La sin 9 — Lw cos 9 t:\ Then, a = - + a ac; aA aG/A = -(La sin 6 + sz cos 9) j + (L/2)a(cos 9 i + sin 9 j) + (L/2)w2(—sin 9 i + cos 6 3) (Problem 16—73 continues ...) ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 16—73 — cont.) Therefore, 2 . aGX - (L/2)(a cos 6 - w Sin 9) aGy = -(L/2)(a sin 9 + w2 cos 6) A = (mL/2)(a cos 6 — wz sin 6) B = mg - (mL/2)(a sin 6 + oz cos 6) and the moment equation can be simplified to 2Lu = 39 sin 6 Rewriting the angular acceleration using the Chain Rule of Differentiation do do d9 do a=—=——=w— dt d9 dt d9 and integrating to get the angular velocity gives I 2L0 dw I 2La d9 = I 39 sin 9 d9 2 L0 39(1 - cos 6) where the constant of integration has been chosen so that w = 0 when 6 = 0. Then, A will become zero when A = m(L/2)(a cos 9 —‘w2 sin 6) = O 2La cos 9 = 2sz sin 9 3g sin 9 cos 9 = 69(1 - cos 9) sin 6 II N 3 cos 9 9 = 48.190 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. (independent of m and L!) and at this angle a = 3.048 rad/s2 C m = 1.651 rad/s c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. B = 3.81 lb T and the relative velocity equation gives _. —. —. = + VB VA VB/A A = A + 6 A + ' 6 A VB 1 VA J Lw(cos 1 Sin J) v = 13.21 ft/s -9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ...
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Dynamics Homework 6 - ENGINEERING MECHANICS ~ Dynamics W.F....

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