Dynamics 14th edition by r c hibbeler chapter 14

© 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–1 The 20-kg crate is subjected to a force having a constant direction and a magnitude F = 100 N When s = 15 m, the crate is moving to the right with a speed of m/s Determine its speed when s = 25 m The coefficient of kinetic friction between the crate and the ground is mk = 0.25 F 30° SOLUTION Equation of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = mkN = 0.25N Applying Eq 13–7, we have + c a Fy = may ; N + 100 sin 30° - 20(9.81) = 20(0) N = 146.2 N Principle of Work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force Ff = 0.25(146.2) = 36.55 N does negative work since it acts in the opposite direction to that of displacement The normal reaction N, the vertical component of force F and the weight of the crate not displace hence no work Applying Eq.14–7, we have T1 + a U1 - = T2 25 m (20)(8 2) + L15 m 100 cos 30° ds 25 m - L15 m 36.55 ds = (20)v2 Ans v = 10.7 m s Ans: v = 10.7 m>s 377 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–2 F (lb) For protection, the barrel barrier is placed in front of the bridge pier If the relation between the force and deflection of the barrier is F = (90(103)x1>2) lb, where x is in ft, determine the car’s maximum penetration in the barrier The car has a weight of 4000 lb and it is traveling with a speed of 75 ft>s just before it hits the barrier F ϭ 90(10)3 x1/2 x (ft) SOLUTION Principle of Work and Energy: The speed of the car just before it crashes into the barrier is v1 = 75 ft>s The maximum penetration occurs when the car is brought to a stop, i.e., v2 = Referring to the free-body diagram of the car, Fig a, W and N no work; however, Fb does negative work T1 + ©U1 - = T2 4000 a b (752) + c 32.2 L0 xmax 90(103)x1>2dx d = Ans xmax = 3.24 ft Ans: x max = 3.24 ft 378 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–3 The crate, which has a mass of 100 kg, is subjected to the action of the two forces If it is originally at rest, determine the distance it slides in order to attain a speed of m>s The coefficient of kinetic friction between the crate and the surface is mk = 0.2 1000 N 800 N 30 SOLUTION Equations of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = mk N = 0.2N Applying Eq 13–7, we have + c ©Fy = may; N + 1000 a b - 800 sin 30° - 100(9.81) = 100(0) N = 781 N Principle of Work and Energy: The horizontal components of force 800 N and 1000 N which act in the direction of displacement positive work, whereas the friction force Ff = 0.2(781) = 156.2 N does negative work since it acts in the opposite direction to that of displacement The normal reaction N, the vertical component of 800 N and 1000 N force and the weight of the crate not displace, hence they no work Since the crate is originally at rest, T1 = Applying Eq 14–7, we have T1 + a U1-2 = T2 + 800 cos 30°(s) + 1000 a bs - 156.2s = (100) A 62 B Ans s = 1.35m Ans: s = 1.35 m 379 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *14–4 The 100-kg crate is subjected to the forces shown If it is originally at rest, determine the distance it slides in order to attain a speed of v = m>s The coefficient of kinetic friction between the crate and the surface is mk = 0.2 500 N 400 N 30Њ 45Њ Solution Work Consider the force equilibrium along the y axis by referring to the FBD of the crate, Fig a, + c ΣFy = 0; N + 500 sin 45° - 100(9.81) - 400 sin 30° = N = 827.45 N Thus, the friction is Ff = mkN = 0.2(827.45) = 165.49 N Here, F1 and F2 positive work whereas Ff   does negative work W and N no work UF1 = 400 cos 30° s = 346.41 s UF2 = 500 cos 45° s = 353.55 s UFf = - 165.49 s Principle of Work And Energy Applying Eq 14–7, T1 + ΣU1 - = T2 + 346.41 s + 353.55 s + ( -165.49 s) = (100) ( 82 ) s = 5.987 m = 5.99 m Ans Ans: s = 5.99 m 380 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–5 Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill A it will reach a speed of 100 km>h when it comes to the bottom B Also, what should be the minimum radius of curvature r for the track at B so that the passengers not experience a normal force greater than 4mg = (39.24m) N? Neglect the size of the car and passenger A h r B Solution 100 km>h = 100 ( 103 ) 3600 = 27.778 m>s T1 + ΣU1 - = T2 + m(9.81)h = m(27.778)2 Ans h = 39.3 m + c ΣFn = man; (27.778)2 39.24 m - mg = ma b r Ans r = 26.2 m Ans: h = 39.3 m r = 26.2 m 381 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–6 When the driver applies the brakes of a light truck traveling 40 km>h, it skids m before stopping How far will the truck skid if it is traveling 80 km>h when the brakes are applied? Solution 40 km>h = 40 ( 103 ) 3600 = 11.11 m>s 80 km>h = 22.22 m>s T1 + ΣU1 - = T2 m(11.11)2 - mk mg(3) = mk g = 20.576 T1 + ΣU1 - = T2 m(22.22)2 - (20.576)m(d) = Ans d = 12 m Ans: d = 12 m 382 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–7 As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame Show that this is so, by considering the 10-kg block which rests on the smooth surface and is subjected to a horizontal force of N If observer A is in a fixed frame x, determine the final speed of the block if it has an initial speed of m>s and travels 10 m, both directed to the right and measured from the fixed frame Compare the result with that obtained by an observer B, attached to the x¿ axis and moving at a constant velocity of m>s relative to A Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy A B x x¿ m/s m/s 6N 10 m SOLUTION Observer A: T1 + ©U1 - = T2 1 (10)(5)2 + 6(10) = (10)v22 2 Ans v2 = 6.08 m>s Observer B: F = ma = 10a + B A: a = 0.6 m>s2 s = s0 + v0t + at c 10 = + 5t + (0.6)t2 t2 + 16.67t - 33.33 = t = 1.805 s At v = m>s, s¿ = 2(1.805) = 3.609 m Block moves 10 - 3.609 = 6.391 m Thus T1 + ©U1 - = T2 1 (10)(3)2 + 6(6.391) = (10)v22 2 Ans v2 = 4.08 m>s Note that this result is m>s less than that observed by A Ans: Observer A: v2 = 6.08 m>s Observer B: v2 = 4.08 m>s 383 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *14–8 A force of F = 250 N is applied to the end at B Determine the speed of the 10-kg block when it has moved 1.5 m, starting from rest Solution Work with reference to the datum set in Fig a, B SW + 2sF = l (1) dSW + 2dsF = 0 A F Assuming that the block moves upward 1.5 m, then dSW = -1.5 m since it is directed in the negative sense of SW Substituted this value into Eq (1), - 1.5 + 2dsF = 0  dsF = 0.75 m Thus, UF = FdSF = 250(0.75) = 187.5 J UW = - WdSW = - 10(9.81)(1.5) = -147.15 J Principle of Work And Energy Applying Eq 14–7, T1 + U1 - = T2 + 187.5 + ( -147.15) = (10)v2 Ans v = 2.841 m>s = 2.84 m>s Ans: v = 2.84 m>s 384 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–9 The “air spring” A is used to protect the support B and prevent damage to the conveyor-belt tensioning weight C in the event of a belt failure D The force developed by the air spring as a function of its deflection is shown by the graph If the block has a mass of 20 kg and is suspended a height d = 0.4 m above the top of the spring, determine the maximum deformation of the spring in the event the conveyor belt fails Neglect the mass of the pulley and belt F (N) D 1500 C d 0.2 s (m) A B Solution Work Referring to the FBD of the tensioning weight, Fig a, W does positive work whereas force F does negative work Here the weight displaces downward SW = 0.4 + xmax where xmax is the maximum compression of the air spring Thus UW = 20(9.81) ( 0.4 + xmax ) = 196.2 ( 0.4 + xmax ) The work of F is equal to the area under the F-S graph shown shaded in Fig b, Here F 1500    ; F = 7500xmax Thus = xmax 0.2 UF = - (7500 xmax)(xmax) = - 3750x2max Principle of Work And Energy Since the block is at rest initially and is required to stop momentarily when the spring is compressed to the maximum, T1 = T2 = Applying Eq 14–7, T1 + ΣU1 - = T2 + 196.2(0.4 + xmax) + ( - 3750x2max ) = 3750x2max - 196.2xmax - 78.48 = xmax = 0.1732 m = 0.173 m 0.2 m      (O.K!) Ans Ans: xmax = 0.173 m 385 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–10 v The force F, acting in a constant direction on the 20-kg block, has a magnitude which varies with the position s of the block Determine how far the block must slide before its velocity becomes 15 m>s When s = the block is moving to the right at v = m>s The coefficient of kinetic friction between the block and surface is mk = 0.3 F (N) F F ϭ 50s 1/2 Solution Work Consider the force equilibrium along y axis, by referring to the FBD of the block, Fig a, s (m) + c ΣFy = 0  ;   N - 20(9.81) = 0   N = 196.2 N Thus, the friction is Ff = mkN = 0.3(196.2) = 58.86 N Here, force F does positive work whereas friction Ff does negative work The weight W and normal reaction N no work UF = L Fds = L0 s 50s2 ds = 100 s2 UFf = - 58.86 s Principle of Work And Energy Applying Eq 14–7, T1 + ΣU1 - = T2 100 (20)(62) + s2 + ( - 58.86s) = (20)(152) 100 s2 - 58.86s - 1890 = Solving numerically, Ans s = 20.52 m = 20.5 m Ans: s = 20.5 m 386 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–83 A rocket of mass m is fired vertically from the surface of the earth, i.e., at r = r1 Assuming no mass is lost as it travels upward, determine the work it must against gravity to reach a distance r2 The force of gravity is F = GMem>r (Eq 13–1), where Me is the mass of the earth and r the distance between the rocket and the center of the earth r2 r SOLUTION F = G F1-2 = r1 Mem r2 L r2 F dr = GMem = GMema dr Lr1 r 1 - b r1 r2 Ans Ans: F = GMem a 459 1 - b r1 r2 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *14–84 The 4-kg smooth collar has a speed of m>s when it is at s = Determine the maximum distance s it travels before it stops momentarily The spring has an unstretched length of m 1.5 m A m/s s k ϭ 100 N/m Solution Potential Energy With reference to the datum set through A the gravitational potential energies of the collar at A and B are B (Vg)A = 0    (Vg)B = - mghB = - 4(9.81) Smax = -39.24 Smax At A and B, the spring stretches xA = 1.5 - = 0.5 m and xB = 2S2max + 1.52 - Thus, the elastic potential Energies in the spring when the collar is at A and B are (Ve)A = (Ve)B = kx = (100) ( 0.52 ) = 12.5 J A 2 kxB = (100) ( 2S2max + 1.52 - ) = 50 ( S2max - 22S2max + 1.52 + 3.25 ) 2 Conservation of Energy Since the collar is required to stop momentarily at B, TB = TA + VA = TB + VB (4) ( 32 ) + + 12.5 = + ( - 39.24 Smax) + 50 ( S2max - 22S2max + 1.52 + 3.25 ) 50 S2max - 1002S2max + 1.52 - 39.24 Smax + 132 = Solving numerically, Ans Smax = 1.9554 m = 1.96 m Ans: Smax = 1.96 m 460 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–85 A 60-kg satellite travels in free flight along an elliptical orbit such that at A, where rA = 20 Mm, it has a speed vA = 40 Mm>h What is the speed of the satellite when it reaches point B, where rB = 80 Mm? Hint: See Prob 14–82, where Me = 5.976(1024) kg and G = 66.73(10-12) m3>(kg # s2) B vB 80 Mm rB rA 20 Mm vA A SOLUTION yA = 40 Mm>h = 11 111.1 m>s Since V = - GMe m r T1 + V1 = T2 + V2 66.73(10) - 12(5.976)(10)23(60) 66.73(10) - 12(5.976)(10)24(60) 1 (60)(11 111.1)2 = (60)v2B 2 20(10) 80(10)6 Ans vB = 9672 m>s = 34.8 Mm>h Ans: vB = 34.8 Mm>h 461 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–86 The skier starts from rest at A and travels down the ramp If friction and air resistance can be neglected, determine his speed vB when he reaches B Also, compute the distance s to where he strikes the ground at C, if he makes the jump traveling horizontally at B Neglect the skier’s size He has a mass of 70 kg A vB 50 m B 4m Solution s TA + VA = TB + VB C 30Њ + 70(9.81) (46) = (70)v2 + Ans v = 30.04 m>s = 30.0 m>s ( + T) sy = (sy)0 + (v0)yt + at c + s sin 30° = + + (9.81)t 2 (1) + )s = v t (d x x (2) s cos 30° = 30.04t Ans s = 130 m t = 3.75 s Ans: s = 130 m 462 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–87 The block has a mass of 20 kg and is released from rest when s 0.5 m If the mass of the bumpers A and B can be neglected, determine the maximum deformation of each spring due to the collision kA = 500 N/m s = 0.5 m SOLUTION Datum at initial position: A B kB = 800 N/m T1 + V1 = T2 + V2 + = + 1 (500)s2A + (800)s2B + 20(9.81) C - (sA + s B) - 0.5 D 2 Also, Fs = 500sA = 800sB (1) (2) sA = 1.6sB Solving Eqs (1) and (2) yields: sB = 0.638 m Ans sA = 1.02 m Ans Ans: sB = 0.638 m sA = 1.02 m 463 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *14–88 The 2-lb collar has a speed of ft>s at A The attached spring has an unstretched length of ft and a stiffness of k = 10 lb>ft If the collar moves over the smooth rod, determine its speed when it reaches point B, the normal force of the rod on the collar, and the rate of decrease in its speed y A y ϭ 4.5 Ϫ 12 x2 4.5 ft SOLUTION k ϭ 10 lb/ft Datum at B: TA + VA = TB + VB B 2 1 1 a b (5)2 + (10)(4.5 - 2)2 + 2(4.5) = a b (nB)2 + (10)(3 - 2)2 + 32.2 2 32.2 Ans nB = 34.060 ft>s = 34.1 ft>s y = 4.5 - ft x dy = tanu = - x ` = -3 dx x=3 d2y u = -71.57° r = c1 + a ` dy 2 b d dx d2y dx2 = -1 dx2 ` +b©Fn = man ; = [1 + ( -3)2]2 = 31.623 ft | - 1| - N + 10 cos 18.43° + cos 71.57° = a (34.060)2 ba b 32.2 31.623 Ans N = 7.84 lb + R©Ft = mat ; sin 71.57° - 10 sin 18.43° = a ba 32.2 t at = -20.4 ft>s2 Ans Ans: vB = 34.1 ft>s N = 7.84 lb at = - 20.4 ft>s2 464 x © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–89 When the 6-kg box reaches point A it has a speed of vA = m>s Determine the angle u at which it leaves the smooth circular ramp and the distance s to where it falls into the cart Neglect friction vA = m/s 20° A θ B 1.2 m SOLUTION s At point B: +b©Fn = man; 6(9.81) cos f = 6a n2B b 1.2 (1) Datum at bottom of curve: TA + VA = TB + VB 1 (6)(2)2 + 6(9.81)(1.2 cos 20°) = (6)(vB)2 + 6(9.81)(1.2 cos f) 2 13.062 = 0.5v2B + 11.772 cos f (2) Substitute Eq (1) into Eq (2), and solving for vB, vB = 2.951 m>s Thus, f = cos - a (2.951)2 b = 42.29° 1.2(9.81) Ans u = f - 20° = 22.3° A+cB s = s0 + v0 t + 12 ac t2 - 1.2 cos 42.29° = - 2.951(sin 42.29°)t + ( -9.81)t2 4.905t2 + 1.9857t - 0.8877 = Solving for the positive root: t = 0.2687 s + b a: s = s0 + v0 t s = + (2.951 cos 42.29°)(0.2687) Ans s = 0.587 m Ans: u = 22.3° s = 0.587 m 465 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–90 y When the 5-kg box reaches point A it has a speed vA = 10 m>s Determine the normal force the box exerts on the surface when it reaches point B Neglect friction and the size of the box yϭx 9m x 1/ ϩ y1/ ϭ B Solution A Conservation of Energy At point B, y = x x 9m x2 + x2 = x = m m With reference to the datum set to coincide with the x axis, the gravitational potential energies of the box at points A and B are Then y = (Vg)A = 0    (Vg)B = mghB = 5(9.81)a b = 110.3625 J Applying the energy equation, TA + VA = TB + VB 1 (5) ( 102 ) + = (5)v2B + 110.3625 2 v2B = 55.855 m2 >s2 dy 1 = ( - x2 ) a - x - b dx = At point B, x = m Thus, 2x2 Equation of Motion Here, y = ( - x2 ) Then, = x2 - x = - x and tan uB = d 2y dx ` x=4 m d 2y dx2 = -3 x 2 dy ` = dx x = 94 m = 3 ( 94 ) ( )2 = - 1  uB = -45° = 45° = 0.4444 The radius of curvature at B is PB = [1 + (dy>dx)2]2 d 2y>dx2 = [1 + ( - 1)2]2 0.4444 = 6.3640 m Referring to the FBD of the box, Fig a ΣFn = man;   N - 5(9.81) cos 45° = 5a 55.855 b 6.3640 Ans N = 78.57 N = 78.6 N Ans: N = 78.6 N 466 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–91 y When the 5-kg box reaches point A it has a speed vA = 10 m>s Determine how high the box reaches up the surface before it comes to a stop Also, what is the resultant normal force on the surface at this point and the acceleration? Neglect friction and the size of the box yϭx 9m x 1/ ϩ y1/ ϭ B Solution A Conservation of Energy With reference to the datum set coincide with x axis, the gravitational potential energy of the box at A and C (at maximum height) are x 9m (Vg)A = 0    (Vg)C = mghc = 5(9.81)(y) = 49.05y It is required that the box stop at C Thus, Tc = TA + VA = TC + VC (5) ( 102 ) + = + 49.05y y = 5.0968 m = 5.10 m Ans Then, 1 x + 5.0968 = x = 0.5511 m Equation of Motion Here, y = ( - x2 ) Then, = x2 - x = - x and d 2y dx2 = dy 1 = ( - x2 ) a - x2 b dx -3 x2 = At point C, x = 0.5511 m 2x Thus tan uc = dy ` = = - 3.0410   uC = -71.80° = 71.80° dx x = 0.5511 m 0.55112 Referring to the FBD of the box, Fig a, ΣFn = man ; N - 5(9.81) cos 71.80° = a 02 b rC Ans N = 15.32 N = 15.3 N ΣFt = mat ; - 5(9.81) sin 71.80° = 5at at = - 9.3191 m>s2 = 9.32 m>s2 R Since an = 0, Then a = at = 9.32 m>s2 R Ans Ans: y = 5.10 m N = 15.3 N a = 9.32 m>s2 R 467 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *14–92 y The roller-coaster car has a speed of 15 ft>s when it is at the crest of a vertical parabolic track Determine the car’s velocity and the normal force it exerts on the track when it reaches point B Neglect friction and the mass of the wheels The total weight of the car and the passengers is 350 lb 15 ft/s vA A y 200 (40 000 x2) 200 ft SOLUTION y = (40 000 - x2) 200 dy = x` = - 2, dx 100 x = 200 d2y dx2 = - B O 200 ft u = tan - 1( - 2) = - 63.43° 100 Datum at A: TA + VA = TB + VB 350 350 a b (15)2 + = a b (nB)2 - 350(200) 32.2 32.2 Ans nB = 114.48 = 114 ft>s r = c1 + a ` dy b d dx d 2y dx ` +b©Fn = man; 3 = [1 + (- 2)2]2 `- ` 100 = 1118.0 ft 350 cos 63.43° - NB = a 350 (114.48)2 b 32.2 1118.0 Ans NB = 29.1 lb Ans: vB = 114 ft>s NB = 29.1 lb 468 x © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–93 The 10-kg sphere C is released from rest when u = 0° and the tension in the spring is 100 N Determine the speed of the sphere at the instant u = 90° Neglect the mass of rod AB and the size of the sphere E 0.4 m SOLUTION k 500 N/m A Potential Energy: With reference to the datum set in Fig a, the gravitational potential u 44.145 J and A Vg B = mgh2 = 10(9.81)(0) = When the sphere is at position (1), 100 = 0.2 m Thus, the unstretched length of the spring is the spring stretches s1 = 500 0.3 m energy of the sphere at positions (1) and (2) are A Vg B = mgh1 = 10(9.81)(0.45) = D B C 0.15 m l0 = 30.32 + 0.42 - 0.2 = 0.3 m, and the elastic potential energy of the spring is 1 A Ve B = ks12 = (500)(0.22) = 10 J When the sphere is at position (2), the spring 2 stretches s2 = 0.7 - 0.3 = 0.4 m, and the elastic potential energy of the spring is 1 A Ve B = ks22 = (500)(0.42) = 40 J 2 Conservation of Energy: T1 + V1 = T2 + V2 1 m A v B + c A Vg B + A Ve B d = ms A vs B 2 + c A Vg B + A Ve B d s s + (44.145 + 10) = (10)(vs)22 + (0 + 40) Ans (vs)2 = 1.68 m>s Ans: v = 1.68 m>s 469 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–94 A quarter-circular tube AB of mean radius r contains a smooth chain that has a mass per unit length of m0 If the chain is released from rest from the position shown, determine its speed when it emerges completely from the tube A O r B SOLUTION Potential Energy: The location of the center of gravity G of the chain at positions (1) and (2) are shown in Fig a The mass of the chain is p p 2r p - m = m a r b = m 0r Thus, the center of mass is at h1 = r = a br p p 2 With reference to the datum set in Fig a the gravitational potential energy of the chain at positions (1) and (2) are A Vg B = mgh1 = a m 0rg b a p p - p - br = a b m 0r2g p and A Vg B = mgh = Conservation of Energy: T1 + V1 = T2 + V2 1 mv1 + A Vg B = mv2 + A Vg B 2 0+ a p - p b m 0r2g = a m 0r b v 2 + 2 v2 = (p - 2)gr Ap Ans Ans: v2 = 470 (p - 2)gr Ap © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–95 The cylinder has a mass of 20 kg and is released from rest when h = Determine its speed when h = m Each spring has a stiffness k = 40 N>m and an unstretched length of m 2m k 2m k h Solution T1 + V1 = T2 + V2 1 + = + 2c (40) ( 232 + 22 - 22 ) d - 20(9.81)(3) + (20)v2 2 v = 6.97 m>s Ans Ans: v = 6.97 m>s 471 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *14–96 If the 20-kg cylinder is released from rest at h = 0, determine the required stiffness k of each spring so that its motion is arrested or stops when h = 0.5 m Each spring has an unstretched length of m 2m k 2m k h Solution T1 + V1 = T2 + V2 1 + 2c k(2 - 1)2 d = - 20(9.81)(0.5) + 2c k ( 2(2)2 + (0.5)2 - ) d 2 k = -98.1 + 1.12689 k Ans k = 773 N>m Ans: k = 773 N>m 472 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 14–97 A pan of negligible mass is attached to two identical springs of stiffness k = 250 N>m If a 10-kg box is dropped from a height of 0.5 m above the pan, determine the maximum vertical displacement d Initially each spring has a tension of 50 N 1m k 250 N/m 1m k 250 N/m 0.5 m d SOLUTION Potential Energy: With reference to the datum set in Fig a, the gravitational potential energy of the box at positions (1) and (2) are A Vg B = mgh1 = 10(9.81)(0) = and A Vg B = mgh2 = 10(9.81) C - A 0.5 + d B D = - 98.1 A 0.5 + d B Initially, the spring 50 = 0.2 m Thus, the unstretched length of the spring 250 is l0 = - 0.2 = 0.8 m and the initial elastic potential of each spring is A Ve B = (2) ks1 = 2(250 > 2)(0.22) = 10 J When the box is at position (2), the stretches s1 = spring stretches s2 = a 2d2 + 12 - 0.8 b m The elastic potential energy of the springs when the box is at this position is A Ve B = (2) ks2 = 2(250 > 2) c 2d2 + - 0.8 d = 250a d2 - 1.62d2 + + 1.64b 2 Conservation of Energy: T1 + V1 + T2 + V2 1 mv1 + B a Vg b + A Ve B R = mv2 + B aVg b + A Ve B R 2 + A + 10 B = + B - 98.1 A 0.5 + d B + 250 ¢ d2 - 1.6 2d2 + + 1.64 ≤ R 250d2 - 98.1d - 400 2d2 + + 350.95 = Solving the above equation by trial and error, Ans d = 1.34 m Ans: d = 1.34 m 473 ... publisher *14 20 Barrier stopping force (kip) The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-absorbing material The barrier stopping force is measured versus... any form or by any means, without permission in writing from the publisher 14 2 F (lb) For protection, the barrel barrier is placed in front of the bridge pier If the relation between the force... 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced,