© 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. 11–1. Use the method of virtual work to determine the tensions in cable AC. The lamp weighs 10 lb. B C 45° SOLUTION A 30° Free Body Diagram: The tension in cable AC can be determined by releasing cable AC. The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only FAC and the weight of lamp (10 lb force) do work. Virtual Displacement: Force FAC and 10 lb force are located from the fixed point B using position coordinates yA and xA. xA = l cos u dxA = - l sin udu (1) yA = l sin u dyA = l cos udu (2) Virtual–Work Equation: When yA and xA undergo positive virtual displacements dyA and dxA, the 10 lb force and horizontal component of FAC, FAC cos 30°, do positive work while the vertical component of FAC, FAC sin 30°, does negative work. dU = 0; 10dyA - FAC sin 30°dyA + FAC cos 30°, dxA = 0 (3) Substituting Eqs. (1) and (2) into (3) yields (10 cos u - 0.5FAC cos u - 0.8660FAC sin u)ldu = 0 Since ldu Z 0, then FAC = 10 cos u 0.5 cos u + 0.8660 sin u At the equilibrium position u = 45°, FAC = 10 cos 45° = 7.32 lb 0.5 cos 45° + 0.8660 sin 45° Ans. Ans: FAC = 7.32 lb 1123 © 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. 11–2. The scissors jack supports a load P. Determine the axial force in the screw necessary for equilibrium when the jack is in the position u. Each of the four links has a length L and is pin-connected at its center. Points B and D can move horizontally. P SOLUTION x = L cos u, dx = -L sin u du y = 2L sin u, dy = 2L cos u du dU = 0; - Pdy - Fdx = 0 C D A B u -P(2L cos u du) - F(- L sin u du) = 0 Ans. F = 2P cot u Ans: F = 2P cot u 1124 © 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. 11–3. If a force of P = 5 lb is applied to the handle of the mechanism, determine the force the screw exerts on the cork of the bottle. The screw is attached to the pin at A and passes through the collar that is attached to the bottle neck at B. P 5 lb D SOLUTION A Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the force in the screw Fs and force P do work when the virtual displacements take place. u 30° 3 in. B Virtual Displacement: The position of the points of application for Fs and P are specified by the position coordinates yA and yD, measured from the fixed point B, respectively. yA = 2(3 sin u) dyA = 6 cos udu (1) yD = 6(3 sin u) dyD = 18 cos udu (2) Virtual Work Equation: Since P acts towards the positive sense of its corresponding virtual displacement, it does positive work. However, the work of Fs is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0; PdyD + A - FSdyA B = 0 (3) Substituting P = 5 lb, Eqs. (1) and (2) into Eq. (3), 5(18 cos udu)FS (6 cos udu) = 0 cos udu(90 - 6FS) = 0 Since cos udu Z 0, then 90 - 6FS = 0 Ans. FS = 15 lb Ans: FS = 15 lb 1125 © 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. *11–4. The disk has a weight of 10 lb and is subjected to a vertical force P = 8 lb and a couple moment M = 8 lb # ft. Determine the disk’s rotation u if the end of the spring wraps around the periphery of the disk as the disk turns. The spring is originally unstretched. M 8 lb ft 1.5 ft k 12 lb/ft SOLUTION P 8 lb dyF = dyP = 1.5du PdyP + Mdu - FdyF = 0 dU = 0; 8(1.5 du) + 8 du - F(1.5 du) = 0 u(20 - 1.5F) = 0 Since du Z 0 20 -1.5F = 0 F = 13.33 lb F = kx Where x = 1.5u 13.33 = 12(1.5u) Ans. u = 0.7407 rad = 42.4° Ans: u = 42.4° 1126 © 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. 11–5. The punch press consists of the ram R, connecting rod AB, and a flywheel. If a torque of M = 75 N # m is applied to the flywheel, determine the force F applied at the ram to hold the rod in the position u = 60°. B 200 mm 600 mm R u F M A Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only force F and couple moment M do work. Virtual Displacement. The position of force F is measured from fixed point O by position coordinate xA. Applying the law of cosines by referring to Fig. b, 0.62 = x2A + 0.22 - 2xA(0.2) cos u (1) Differentiating the above expression, 0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA dxA = 0.4xA sin u du 0.4 cos u - 2xA (2) Virtual–Work Equation. When point A undergoes a positive virtual displacement, and the flywheel undergoes positive virtual angular displacement du, both F and M do negative work. (3) - FdxA - Mdu = 0 dU = 0; Substituting Eq. (2) into (3) Since du ≠ 0, then ca 0.4 xA sin u bF + M d du = 0 0.4 cos u - 2xA 0.4 xA sin u F + M = 0 0.4 cos u - 2xA F = -a 0.4 cos u - 2xA bM 0.4 xA sin u (4) At the equilibrium position u = 60°, Eq. (1) gives 0.62 = xA2 + 0.22 - 2xA(0.2) cos 60° xA = 0.6745 m Substitute M = 75 N # m, u = 60° and this result into Eq. (4) F = -c 0.4 cos 60° - 2(0.6745) 0.4(0.6745) sin 60° = 368.81 N = 369 N d (75) Ans. Ans: F = 369 N 1127 © 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. 11–6. The flywheel is subjected to a torque of M = 75 N # m. Determine the horizontal compressive force F and plot the result of F (ordinate) versus the equilibrium position u (abscissa) for 0° … u … 180°. B 200 mm R u F M Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only force F and couple moment M do work. Virtual Displacement. The position of force F is measured from fixed point O by position coordinate xA. Applying the law of cosines by referring to Fig. b, 0.62 = xA2 + 0.22 - 2xA(0.2) cos u (1) Differentiating the above expression, 0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA dxA = 0.4xA sin u 0.4 cos u - 2xA (2) Virtual–Work Equation. When point A undergoes a positive virtual displacement and the flywheel undergoes positive virtual angular displacement du, both F and M do negative work. (3) - FdxA - Mdu = 0 dU = 0; Substituting Eq. (2) into (3) ca Since du ≠ 0, then 0.4 xA sin u bF + M d du = 0 0.4 cos u - 2xA a Here, M = 75 N # m. Then 0.4 xA sin u bF + M = 0 0.4 cos u - 2xA F = -M a 0.4 cos u - 2xA b 0.4 xA sin u F = - 75 a 0.4 cos u - 2xA b 0.4 xA sin u (4) Using Eq. (1) and (4), the following tabulation can be computed. Subsequently, the graph of F vs u shown in Fig. c can be plotted u(deg.) 0 xA(m) 0.80 F(N) ∞ 30 60 90 120 150 0.7648 0.6745 0.5657 0.4745 0.4184 580 369 375 524 1060 600 mm 180 0.4 15 73.0 0.7909 0.6272 ∞ 1128 1095 356 A © 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. 11–7. When u = 20°, the 50-lb uniform block compresses the two vertical springs 4 in. If the uniform links AB and CD each weigh 10 lb, determine the magnitude of the applied couple moments M needed to maintain equilibrium when u = 20°. k 2 lb/in. k B D 2 lb/ in. 1 ft 1 ft SOLUTION u Free Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the spring forces Fsp, the weight of the block (50 lb), the weights of the links (10 lb) and the couple moment M do work. u M A 1 ft 2 ft M C 4 ft Virtual Displacements: The spring forces Fsp, the weight of the block (50 lb) and the weight of the links (10 lb) are located from the fixed point C using position coordinates y3, y2 and y1 respectively. y3 = 1 + 4 cos u y2 = 0.5 + 4 cos u y1 = 2 cos u (1) dy3 = -4 sin udu (2) dy2 = -4 sin udu (3) dy1 = - 2 sin udu Virtual–Work Equation: When y1, y2 and y3 undergo positive virtual displacements dy1, dy2 and dy3, the spring forces Fsp, the weight of the block (50 lb) and the weights of the links (10 lb) do negative work. The couple moment M does negative work when the links undergo a positive virtual rotation du. dU = 0; - 2Fspdy3 - 50dy2 - 20dy1 - 2Mdu = 0 (4) Substituting Eqs. (1), (2) and (3) into (4) yields (8Fsp sin u + 240 sin u - 2M) du = 0 Since du Z 0, then 8Fsp sin u + 240 sin u - 2M = 0 M = sin u(4Fsp + 120) At the equilibrium position u = 20°, Fsp = kx = 2(4) = 8 lb. M = sin 20°[4(8) + 120] = 52.0 lb # ft Ans. Ans: M = 52.0 lb # ft 1129 © 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. *11–8. The bar is supported by the spring and smooth collar that allows the spring to be always perpendicular to the bar for any angle u. If the unstretched length of the spring is l0, determine the force P needed to hold the bar in the equilibrium position u. Neglect the weight of the bar. a A B u k C l SOLUTION s = a sin u, ds = a cos u du y = l sin u, dy = l cos u du Fs = k(a sin u - l0) dU = 0; P Pdy - Fsds = 0 Pl cos du - k(a sin u - l0) a cos u du = 0 P = ka (a sin u - l0) l Ans. Ans: P = 1130 ka (a sin u - l0) l © 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. 11–9. The 4-ft members of the mechanism are pin-connected at their centers. If vertical forces P1 = P2 = 30 lb act at C and E as shown, determine the angle u for equilibrium. The spring is unstretched when u = 45°. Neglect the weight of the members. P1 P2 E C k = 200 lb/ft 2 ft 2 ft SOLUTION y = 4 sin u, x = 4 cos u dy = 4 cos u du, dU = 0; B 2 ft dx = - 4 sin u du A - Fsdx - 30dy - 30dy = 0 θ 2 ft D 3 - Fs1 -4 sin u2 - 6014 cos u24du = 0 Fs = 60 a cos u b sin u Since Fs = k14 cos u - 4 cos 45°2 = 20014 cos u - 4 cos 45°2 60 cos u = 8001cos u - cos 45°2sin u sin u - 0.707 tan u - 0.075 = 0 Ans. u = 16.6° and Ans. u = 35.8° Ans: u = 16.6° u = 35.8° 1131 © 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. 11–10. The thin rod of weight W rest against the smooth wall and floor. Determine the magnitude of force P needed to hold it in equilibrium for a given angle u. B SOLUTION l Free-Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the weight of the rod W and force P do work. Virtual Displacements: The weight of the rod W and force P are located from the fixed points A and B using position coordinates yC and xA, respectively yC = 1 sin u 2 dyC = xA = l cos u 1 cos udu 2 dxA = - l sin udu P A θ (1) (2) Virtual-Work Equation: When points C and A undergo positive virtual displacements dyC and dxA, the weight of the rod W and force F do negative work. (3) dU = 0; - WdyC - PdyA = 0 Substituting Eqs. (1) and (2) into (3) yields a Pl sin u - Wl cos ub du = 0 2 Since du Z 0, then Pl sin u P = Wl cos u = 0 2 W cot u 2 Ans. Ans: P = 1132 W cot u 2 © 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. 11–11. If each of the three links of the mechanism have a mass of 4 kg, determine the angle u for equilibrium. The spring, which always remains vertical, is unstretched when u = 0°. A M 30 N m u 200 mm k 3 kN/ m D 200 mm Solution B Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the weights W1 = W2 = W3 = W, couple moment M, and spring force Fsp do work. Virtual Displacement. The positions of the weights W1, W2, W3 and spring force Fsp are measured from fixed point A using position coordinates y1, y2, y3 and y4 respectively y1 = 0.1 sin u dy1 = 0.1 cos u du (1) y2 = 0.2 sin u + 0.1 dy2 = 0.2 cos u du (2) y3 = 0.1 sin u + 0.2 dy3 = 0.1 cos u du (3) y4 = 0.5 sin u dy4 = 0.2 cos u du (4) Virtual Work Equation. When all the weights undergo positive virtual displacement, all of them do positive work. However, Fsp does negative work when its undergoes positive virtual displacement. Also, M does positive work when it undergoes positive virtual angular displacement. dU = 0; W1dy1 + W2dy2 + W3dy3 - Fspdy4 + Mdu = 0 (5) Substitute Eqs (1), (2) and (3) into (5), using W1 = W2 = W3 = W. W(0.1 cos u du) + W(0.2 cos u du) + W(0.1 cos u du) - Fsp(0.2 cos u du) + Mdu = 0 (0.4 W cos u - 0.2 Fsp cos u + M)du = 0 Since d ≠ 0, then 0.4 W cos u - 0.2 Fsp cos u + M = 0 Here M = 30 N # m, W = 4(9.81)N = 39.24 N and Fsp = kx = 3000(0.2 sin u) = 600 sin u. Substitute these results into this equation, 0.4(39.24) cos u - 0.2(600 sin u)cos u + 30 = 0 15.696 cos u - 120 sin u cos u + 30 = 0 1133 u 200 mm C © 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. 11–11. Continued Solve numerically: Ans. u = 23.8354° = 23.8° or Ans. u = 72.2895° = 72.3° Note: u = 23.8° is a stable equilibrium, while u = 72.3° is an unstable one. Ans: u = 23.8° u = 72.3° 1134 © 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. *11–12. The disk is subjected to a couple moment M. Determine the disk’s rotation u required for equilibrium. The end of the spring wraps around the periphery of the disk as the disk turns. The spring is originally unstretched. k 4 kN/ m 0.5 m M 300 N m Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the spring force Fsp and couple moment M do work. Virtual Work Equation. When the disk undergoes a positive angular displacement du, correspondingly point A undergoes a positive displacement of dxA. As a result couple moment M does positive work whereas spring force Fsp does negative work. dU = 0; (1) Mdu + ( - FspdxA) = 0 Here, Fsp = kxA = 4000(0.50) = 2000u and dxA = 0.5du. Substitute these results into Eq. (1) 300du - 2000u(0.5du) = 0 (300 - 1000u)du = 0 Since du ≠ 0, 300 - 1000u = 0 u = 0.3 rad = 17.19° = 17.2° Ans. Ans: u = 17.2° 1135 © 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. 11–13. A 5-kg uniform serving table is supported on each side by pairs of two identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg, determine the angle u where the table is in equilibrium. The springs each have a stiffness of k = 200 N>m and are unstretched when u = 90°. Neglect the mass of the links. 250 mm 150 mm A E k C 250 mm u u SOLUTION Free -Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work when the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = 200 A 0.25 cos u B = 50 cos u N. D B 150 mm Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb = 0.25 sin u + b dyGb = 0.25 cos udu (1) yGt = 0.25 sin u + a dyGt = 0.25 cos udu (2) xC = 0.25 cos u dxC = -0.25 sin udu (3) Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0; -WbdyGb + A - WtdyGt B + A - FspdxC B = 0 (4) 1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 50 cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have - 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 50 cos u( -0.25 sin udu) = 0 du A - 7.3575 cos u + 12.5 sin u cos u B = 0 Since du Z 0, then - 7.3575 cos u + 12.5 sin u cos u = 0 cos u( -7.3575 + 12.5 sin u) = 0 Solving the above equation, cos u = 0 Ans. u = 90° - 7.3575 + 12.5 sin u = 0 Ans. u = 36.1° Ans: u = 90° u = 36.1° 1136 © 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. 11–14. A 5-kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when u = 45°, determine the stiffness k of each spring. The springs are unstretched when u = 90°. Neglect the mass of the links. 250 mm 150 mm A E k C 250 mm u u SOLUTION Free -Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work when the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = k A 0.25 cos u B = 0.25 k cos u. D B 150 mm Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb = 0.25 sin u + b dyGb = 0.25 cos udu (1) yGt = 0.25 sin u + a dyGt = 0.25 cos udu (2) xC = 0.25 cos u dxC = -0.25 sin udu (3) Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0; -WbdyGb + A - WtdyGt B + A - FspdxC B = 0 (4) 1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 0.25k cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have - 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 0.25k cos u(-0.25 sin udu) = 0 du A -7.3575 cos u + 0.0625k sin u cos u B = 0 Since du Z 0, then - 7.3575 cos u + 0.0625k sin u cos u = 0 117.72 k = sin u When u = 45°, then 117.72 k = = 166 N>m sin 45° Ans. Ans: k = 166 N>m 1137 © 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. 11–15. The service window at a fast-food restaurant consists of glass doors that open and close automatically using a motor which supplies a torque M to each door. The far ends, A and B, move along the horizontal guides. If a food tray becomes stuck between the doors as shown, determine the horizontal force the doors exert on the tray at the position u. a a a u C M a u A M B D SOLUTION x = 2a cos u, dU = 0; dx = - 2a sin u du -M du - F dx = 0 -M du + F (2a sin u)du = 0 F = M 2a sin u Ans. Ans: F = 1138 M 2a sin u © 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. *11–16. The members of the mechanism are pin connected. If a vertical force of 800 N acts at A, determine the angle u for equilibrium. The spring is unstretched when u = 0°. Neglect the mass of the links. B k 6 kN/m 1m 1m 1m Solution D u A Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only spring force Fsp and force P do work. 800 N Virtual Displacement. The positions of Fsp and P are measured from fixed point B using position coordinates yc and yA respectively. yc = 1 sin u dyc = cos u du (1) yA = 3(1 sin u) dyA = 3 cos u du (2) Virtual Work Equation. When Fsp and P undergo their respective positive virtual displacement, P does positive work whereas Fsp does negative work. (3) - Fspdyc + PdyA = 0 dU = 0; Substitute Eqs. (1) and (2) into (3) - Fsp(cos udu) + P(3 cos udu) = 0 ( -Fsp cos u + 3P cos u)du = 0 Since du ≠ 0, and assuming u 6 90°, then - Fsp cos u + 3P cos u = 0 Fsp = 3P Here Fsp = kx = 6000(1 sin u) = 6000 sin u and P = 800 N, Then 6000 sin u = 3(800) sin u = 0.4 Ans. u = 23.58° = 23.6° Ans: u = 23.6° 1139 © 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. 11–17. When u = 30°, the 25-kg uniform block compresses the two horizontal springs 100 mm. Determine the magnitude of the applied couple moments M needed to maintain equilibrium. Take k = 3 kN>m and neglect the mass of the links. 50 mm k D 300 mm 100 mm 200 mm M C 100 mm Solution A u k u M B Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only spring force Fsp, the weight of the block W, and couple moment M do work. Virtual Displacement. The positions of Fsp and W are measured from fixed point B using position coordinates x and y respectively. x = 0.3 cos u + 0.05 dx = -0.3 sin u du (1) y = 0.3 sin u + 0.1 dy = 0.3 cos udu (2) Virtual–Work Equation. When Fsp, W and M undergo their respective positive virtual displacement, all of them do negative work. Thus dU = 0; - 2Fsp dx - Wdy - 2Mdu = 0 (3) Substitute Eqs. (1) and (2) into (3), - 2Fsp( -0.3 sin udu) - W(0.3 cos udu) - 2Mdu = 0 (0.6Fsp sin u - 0.3 W cos u - 2M)du = 0 Since du ≠ 0, then 0.6 Fsp sin u - 0.3 W cos u - 2M = 0 (4) M = 0.3 Fsp sin u - 0.15 W cos u When u = 30°, Fsp = kx = 3000(0.1) = 300 N. Also W = 25(9.81) = 245.25 N. Substitute these values into Eq. 4. M = 0.3(300) sin 30° - 0.15(245.25) cos 30° = 13.14 N # m = 13.1 N # m Ans. Ans: M = 13.1 N # m 1140 © 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. 11–18. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the angle u for equilibrium. The spring has a stiffness of k = 15 kN>m and is unstretched when u = 15°. E u 200 mm A k D SOLUTION B 200 mm C Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is a formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = 15(103) C 2(0.2 sin u) - 2(0.2 sin 15°) D = 6000(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u dyA = 0.2 cos udu (1) yB = 3(0.2 sin u) dyB = 0.6 cos udu (2) yC = 8(0.2 sin u) dyB = 1.6 cos udu (3) Virtual Work Equation: Since Fsp at point A and force P acts towards the positive sense of its corresponding virtual displacement, their work is positive. The work of Fsp at point B is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, Fsp dyA + A - FspdyB B + PdyC = 0 dU = 0; (4) Substituting Fsp = 6000(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), 6000(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C -2400(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 2400(sin u - 0.2588) + 960 = 0 Ans. u = 41.2° Ans: u = 41.2° 1141 P © 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. 11–19. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the stiffness k of the spring for equilibrium when u = 60°. The spring is unstretched when u = 15°. E u 200 mm A k D SOLUTION B 200 mm C P Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula. Fsp = kx = k C 2(0.2 sin u) - 2(0.2 sin 15°) D = (0.4)k(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u dyA = 0.2 cos udu (1) yB = 3(0.2 sin u) dyB = 0.6 cos udu (2) yC = 8(0.2 sin u) dyB = 0.6 cos udu (3) Virtual Work Equation: Since Fsp at point A and force P acts towards the positive sense of its corresponding virtual displacement, their work is positive. The work of Fsp at point B is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0; Fsp dyA + A - FspdyB B + PdyC = 0 (4) Substituting Fsp = k(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), (0.4)k(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C - 0.16k(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 0.16k(sin u - 0.2588) + 960 = 0 k = 6000 sin u - 0.2588 When u = 60°, k = 6000 = 9881N>m = 9.88 kN>m sin 60° - 0.2588 Ans. Ans: k = 9.88 kN>m 1142 © 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. *11–20. The crankshaft is subjected to a torque of M = 50 N # m. Determine the horizontal compressive force F applied to the piston for equilibrium when u = 60°. 400 mm 100 mm u F M SOLUTION (0.4)2 = (0.1)2 + x2 - 2(0.1)(x)(cos u) 0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx dU = 0; For u = 60°, - 50du - Fdx = 0 x = 0.4405 m Ans. dx = - 0.09769 du ( -50 + 0.09769F) du = 0 Ans. F = 512 N Ans: dx = - 0.09769 du F = 512 N 1143 © 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. 11–21. The crankshaft is subjected to a torque of M = 50 N # m. Determine the horizontal compressive force F and plot the result of F (ordinate) versus u (abscissa) for 0° … u … 90°. 400 mm 100 mm u F M SOLUTION (0.4)2 = (0.1)2 + x 2 - 2(0.1)(x)(cos u) (1) 0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx dx = a dU = 0; - 50du - F a F = 0.2x sin u b du 0.2 cos u - 2x -50du - Fdx = 0 0.2x sin u b du = 0, 0.2 cos u - 2x du Z 0 50(2x - 0.2 cos u) 0.2x sin u From Eq. (1) x2 - 0.2x cos u - 0.15 = 0 x = 0.2 cos u ; 20.04 cos2 u + 0.6 , 2 x = 0.2 cos u + 20.04 cos2 u + 0.6 2 F = since20.04 cos2 u + 0.6 7 0.2 cos u 500 20.04 cos2 u + 0.6 Ans. (0.2 cos u + 20.04 cos2 u + 0.6) sin u Ans: F = 1144 50020.04 cos2 u + 0.6 (0.2 cos u + 20.04 cos2 u + 0.6) sin u © 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. 11–22. The spring is unstretched when u = 0°. If P = 8 lb, determine the angle u for equilibrium. Due to the roller guide, the spring always remains vertical. Neglect the weight of the links. k 50 lb/ft 2 ft 2 ft SOLUTION y1 = 2 sin u, dy1 = 2 cos u du y2 = 4 sin u + 4, dy2 = 4 cos u du Fs = 50(2 sin u) = 100 sin u dU = 0; 4 ft u - Fs dy1 + Pdy2 = 0 4 ft - 100 sin u(2 cos u du) + 8(4 cos u du) = 0 Assume ® 6 90°, so cos ® Z 0. P 200 sin u = 32 Ans. u = 9.21° Ans: u = 9.21° 1145 © 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. 11–23. Determine the weight of block G required to balance the differential lever when the 20-lb load F is placed on the pan. The lever is in balance when the load and block are not on the lever. Take x = 12 in. 4 in. x 4 in. C A G B E D 2 in. SOLUTION Free-Body Diagram: When the lever undergoes a virtual angular displacement of du about point B, the dash line configuration shown in Fig. a is formed. We observe that only the weight WG of block G and the weight WF of load F do work when the virtual displacements take place. F Virtual Displacement: Since dyG is very small, the vertical virtual displacement of block G and load F can be approximated as dyG = (12 + 4)du = 16du (1) dyF = 2du (2) Virtual Work Equation: Since WG acts towards the positive sense of its corresponding virtual displacement, its work is positive. However, force WF does negative work since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0; WGdyG + A - WFdyF B = 0 (3) Substituting WG = 20 lb and Eqs. (1) and (2) into Eq. (3), WG A 16du B - 20(2du) = 0 du(16WG - 40) = 0 Since du Z 0, then 16WG - 40 = 0 Ans. WG = 2.5 lb Ans: WG = 2.5 lb 1146 © 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. *11–24. If the load F weighs 20 lb and the block G weighs 2 lb, determine its position x for equilibrium of the differential lever. The lever is in balance when the load and block are not on the lever. 4 in. x 4 in. C A G B E D 2 in. SOLUTION Free -Body Diagram: When the lever undergoes a virtual angular displacement of du about point B, the dash line configuration shown in Fig. a is formed. We observe that only the weight WG of block G and the weight WF of load F do work when the virtual displacements take place. F Virtual Displacement: Since dyG is very small, the vertical virtual displacement of block G and load F can be approximated as dyG = (4 + x)du (1) dyF = 2du (2) Virtual Work Equation: Since WG acts towards the positive sense of its corresponding virtual displacement, its work is positive. However, force WF does negative work since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0; WGdyG + A -WFdyF B = 0 (3) Substituting WF = 20 lb, WG = 2 lb, Eqs.(1) and (2) into Eq. (3), 2(4 + x)du - 20(2du) = 0 du C 2(4 + x) - 40 D = 0 Since du Z 0, then 2(4 + x) - 40 = 0 Ans. x = 16 in. Ans: x = 16 in. 1147 © 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. 11–25. The dumpster has a weight W and a center of gravity at G. Determine the force in the hydraulic cylinder needed to hold it in the general position u. b a SOLUTION G d θ s = 2a2 + c2 - 2a c cos (u + 90°) = 2a2 + c2 + 2a c sin u c 1 ds = (a2 + c2 + 2a c sin u)- 2 ac cos u du y = (a + b) sin u + d cos u dy = (a + b) cos u du - d sin u du dU = 0; Fds - Wdy = 0 1 F(a2 + c2 + 2a c sin u)- 2 ac cos u du - W(a + b) cos u du + Wd sin u du = 0 F = a W(a + b - d tan u) b 2a2 + c2 + 2a c sin u ac Ans. Ans: F = 1148 W(a + b - d tan u) ac 2a2 + c 2 + 2ac sin u © 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. 11–26. The potential energy of a one-degree-of-freedom system is defined by V = (20x3 - 10x2 - 25x - 10) ft # lb, where x is in ft. Determine the equilibrium positions and investigate the stability for each position. SOLUTION Equilibrium Configuration: Taking the first derivative of V, we have dV = 60x2 - 20x - 25 dx Equilibrium requires dV = 0. Thus, dx 60x 2 - 20x - 25 = 0 x = 20 ; 3( -20)2 - 4(60)(- 25) 2(60) Ans. x = 0.833 ft and - 0.5 ft Stability: The second derivative of V is d2V = 120x - 20 dx2 At x = 0.8333 ft, d2V 2 = 120(0.8333) - 20 = 80 7 0 dx2 x = 0.8333 ft Ans. Thus, the equilibrium configuration at x = 0.8333 ft is stable. At x = -0.5ft, d 2V 2 = 120( - 0.5) - 20 = - 80 6 0 dx2 x = -0.5 ft Ans. Thus, the equilibrium configuration at x = -0.5 ft is unstable. Ans: x = - 0.5 ft unstable x = 0.833 ft stable 1149 © 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. 11–27. If the potential function for a conservative one-degree-offreedom system is V = (12 sin 2u + 15 cos u) J,where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions. SOLUTION V = 12 sin 2u + 15 cos u dV = 0; du 24 cos 2u - 15 sin u = 0 2411 - 2 sin2 u2 - 15 sin u = 0 48 sin2 u + 15 sin u - 24 = 0 Choosing the angle 0° 6 u 6 180° u = 34.6° Ans. u = 145° Ans. and d2V = - 48 sin 2u - 15 cos u du2 u = 34.6°, d 2V = - 57.2 6 0 du2 Unstable Ans. u = 145°, d2V = 57.2 7 0 du2 Stable Ans. Ans: Unstable at u = 34.6° stable at u = 145° 1150 © 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. *11–28. If the potential function for a conservative one-degree-offreedom system is V = 18x3 - 2x 2 - 102 J, where x is given in meters, determine the positions for equilibrium and investigate the stability at each of these positions. SOLUTION V = 8x3 - 2x2 - 10 dV = 24x2 - 4x = 0 dx 124x - 42x = 0 x = 0 Ans. and x = 0.167 m d2V = 48x - 4 dx2 x = 0, d 2V = -4 6 0 dx2 x = 0.167 m, Ans. Unstable d2V = 4 7 0 dx2 Ans. Stable Ans: x = 0.167 m d 2V = -4 6 0 Unstable dx2 d 2V = 4 7 0 Stable dx2 1151 © 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. 11–29. If the potential function for a conservative one-degree-offreedom system is V = 110 cos 2u + 25 sin u2 J, where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions. SOLUTION V = 10 cos 2u + 25 sin u For equilibrium: dV = - 20 sin 2u + 25 cos u = 0 du 1- 40 sin u + 252 cos u = 0 u = sin-1 a 25 b = 38.7° and 141° 40 Ans. and u = cos-1 0 = 90° Ans. Stability: d2V = - 40 cos 2u - 25 sin u du2 u = 38.7°, d2V = - 24.4 6 0, du2 Unstable Ans. u = 141°, d2V = - 24.4 6 0, du2 Unstable Ans. Stable Ans. u = 90°, d 2V = 15 7 0, du2 Ans: u = 38.7° unstable u = 90° stable u = 141° unstable 1152 © 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. 11–30. If the potential energy for a conservative one-degree-offreedom system is expressed by the relation V = (4x3 - x 2 - 3x + 10) ft # lb, where x is given in feet, determine the equilibrium positions and investigate the stability at each position. SOLUTION V = 4x3 - x2 - 3x + 10 Equilibrium Position: dV = 12x 2 - 2x - 3 = 0 dx x = 2 ; 2(- 2)2 - 4(12)(- 3) 24 x = 0.590 ft and Ans. - 0.424 ft Stability: d 2V = 24x - 2 dx 2 At x = 0.590 ft d2V = 24(0.590) - 2 = 12.2 7 0 dx 2 Stable Ans. At x = -0.424 ft d 2V = 24( - 0.424) - 2 = -12.2 6 0 dx 2 Unstable Ans. Ans: x = - 0.424 ft unstable x = 0.590 ft stable 1153 © 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. 11–31. The uniform link AB has a mass of 3 kg and is pin connected at both of its ends. The rod BD, having negligible weight, passes through a swivel block at C. If the spring has a stiffness of k = 100 N>m and is unstretched when u = 0°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. Neglect the size of the swivel block. 400 mm D k 100 N/m A C u SOLUTION 400 mm s = 2(0.4)2 + (0.4)2 - 2(0.4)2 cos u B = (0.4)22(1 - cos u) V = Vg + Ve = -(0.2)(sin u)3(9.81) + 1 (100) C (0.4)2(2)(1 - cos u) D 2 dV = - (5.886) cos u + 16(sin u) = 0 du (1) Ans. u = 20.2° d2V = 5.886 sin u + (16) cos u = 17.0 7 0 du2 stable Ans. Ans: u = 20.2° stable 1154 © 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. *11–32. The spring of the scale has an unstretched length of a. Determine the angle u for equilibrium when a weight W is supported on the platform. Neglect the weight of the members. What value W would be required to keep the scale in neutral equilibrium when u = 0°? W L L k SOLUTION Potential Function: The datum is established at point A. Since the weight W is above the datum, its potential energy is positive. From the geometry, the spring stretches x = 2L sin u and y = 2L cos u. V = Ve + Vg L L u u a = 1 2 kx + Wy 2 = 1 (k)(2 L sin u)2 + W(2L cos u) 2 = 2kL2 sin 2 u + 2WL cos u Equilibrium Position: The system is in equilibrium if dV = 0. du dV = 4kL2 sin u cos u - 2WL sin u = 0 du dV = 2kL2 sin 2u - 2WLsin u = 0 du Solving, u = 0° or u = cos-1 a Stability: To have neutral equilibrium at u = 0°, W b 2kL Ans. d2V 2 = 0. du2 u - 0° d 2V = 4kL2 cos 2 u - 2WL cos u du2 d 2V 2 = 4kL2 cos 0° - 2WLcos 0° = 0 du2 u - 0° Ans. W = 2kL Ans: u = cos-1a W = 2kL 1155 W b 2kL © 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. 11–33. The uniform bar has a mass of 80 kg. Determine the angle u for equilibrium and investigate the stability of the bar when it is in this position. The spring has an unstretched length when u = 90°. B 4m k 400 N/ m Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is above the datum, its potential energy is positive. Here, y = 2 sin u and the spring stretches x = 4(1 - sin u) m. Thus, A u V = Ve + Vg 1 2 kx + Wy 2 1 = (400) 3 4(1 - sin u) 4 2 + 80(9.81)(2 sin u) 2 = = 3200 sin2 u - 4830.4 sin u + 3200 Equilibrium Position. The bar is in equilibrium of dV = 0 du dV = 6400 sin u cos u - 4830.4 cos u = 0 du cos u(6400 sin u - 4830.4) = 0 Solving, Ans. u = 90° or u = 49.00° = 49.0° Using the trigonometry identify sin 2u = 2 sin u cos u, dV = 3200 sin 2u - 4830.4 cos u du d 2V = 6400 cos 2u + 4830.4 sin u du 2 d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0 du 2 du 2 d 2V and neutral if = 0. du 2 At u = 90°, d 2V 2 = 6400 cos [2(90°)] + 4830.4 sin 90° = -1569.6 6 0 du 2 u = 90° Thus, the bar is in unstable equilibrium at u = 90° At u = 49.00°, d 2V = 6400 cos [2(49.00°)] + 4830.4 sin 49.00° = 2754.26 7 0 du2 Thus, the bar is in stable equilibrium at u = 49.0°. Ans: Unstable equilibrium at u = 90° Stable equilibrium at u = 49.0° 1156 © 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. 11–34. The uniform bar AD has a mass of 20 kg. If the attached spring is unstretched when u = 90°, determine the angle u for equilibrium. Note that the spring always remains in the vertical position due to the roller guide. Investigate the stability of the bar when it is in the equilibrium position. C k 2 kN/m A u B 0.5 m Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is below the datum, its potential energy is negative. Here, y = 0.75 cos u and the spring stretches x = 0.5 cos u. Thus, 1m D V = Ve + Vg 1 2 kx + ΣWy 2 1 = (2000)(0.5 cos u)2 + 2 = 3 - 20(9.81)(0.75 cos u) 4 = 250 cos2 u - 147.15 cos u Equilibrium Position. The bar is in equilibrium if dV = 0. du dV = - 500 sin u cos u + 147.15 sin u = 0 du sin u(147.15 - 500 cos u) = 0 Solving, u = 0° Ans. u = 72.88° = 72.9° Using the trigonometry identity sin 2u = 2 sin u cos u, dV = - 250 sin 2u + 147.15 sin u du d 2V = - 500 cos 2u + 147.15 cos u du 2 d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0 2 du du 2 d 2V and neutral if = 0. du 2 d 2V 2 = - 500 cos 0° + 147.15 cos 0° = -352.85 6 0 du 2 u = 0° Thus, the bar is in unstable equilibrium at u = 0°. At u = 72.88°, d 2V 2 = - 500 cos [2(72.88°)] + 147.15 cos 72.88° = 456.69 7 0 du2 u = 72.88° Thus, the bar is in stable equilibrium at u = 72.9°. Ans: Unstable equilibrium at u = 0° Stable equilibrium at u = 72.9° 1157 © 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. 11–35. The two bars each have a weight of 8 lb. Determine the required stiffness k of the spring so that the two bars are in equilibrium when u = 30°. The spring has an unstretched length of 1 ft. B 2 ft A SOLUTION V = 2(8)(1 sin u) + θ 2 ft k C 1 k(4 cos u - 1)2 2 dV = 16 cos u + k(4 cos u - 1)( - 4 sin u) du dV = 16 cos u - 4k(4 cos u - 1) sin u du u = 30°, dV = 0 du 16 cos 30° - 4k(4 cos 30° - 1) sin 30° = 0 Ans. k = 2.81 lb>ft Ans: k = 2.81 lb>ft 1158 © 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. *11–36. Determine the angle u for equilibrium and investigate the stability at this position. The bars each have a mass of 3 kg and the suspended block D has a mass of 7 kg. Cord DC has a total length of 1 m. A SOLUTION 500 mm 500 mm u u C 500 mm l = 500 mm y1 = l sin u 2 D y2 = l + 2l(1 - cos u) = l(3 - 2 cos u) V = 2Wy1 - WDy 2 = Wl sin u - WD l(3 - 2 cos u) dV = l(W cos u - 2WD sin u) = 0 du tan u = 3(9.81) W = 0.2143 = 2WD 14(9.81) Ans. u = 12.1° d2V = l( -W sin u - 2WD cos u) du2 u = 12.1°, d 2V = 0.5[- 3(9.81) sin 12.1° - 14(9.81) cos 12.1°] du2 = -70.2 6 0 Ans. Unstable Ans: u = 12.1° Unstable 1159 © 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. 11–37. Determine the angle u for equilibrium and investigate the stability at this position. The bars each have a mass of 10 kg and the spring has an unstretched length of 100 mm. 500 mm k 1.5 kN/ m u u A 500 mm Solution Potential Function. The Datum the centers of gravity of the energies are positive. Here x = 3 2(0.5 cos u) - 0.5 4 - 0.1 = is established through point A, Fig. a. Since bars are above the datum, their potential y = 0.25 sin u and the spring stretches (cos u - 0.6) m. Thus V = Ve + Vg 1 2 kx + ΣWy 2 1 = (1500)(cos u - 0.6)2 + 2[10(9.81)](0.25 sin u) 2 = = 750 cos2 u - 900 cos u + 49.05 sin u + 270 Equilibrium Position. The system is in equilibrium if dV = 0. du dV = - 1500 sin u cos u + 900 sin u + 49.05 cos u = 0 du Solving numerically, u = 4.713° = 4.71° or u = 51.22° = 51.2° Ans. Using the trigonometry identity sin 2u = 2 sin u cos u, dV = - 750 sin 2u + 900 sin u + 49.05 cos u du d 2V = 900 cos u - 1500 cos 2u - 49.05 sin u du 2 d 2V d 2V Stability. The equilibrium configurate is stable if 7 0, unstable if 6 0 and du 2 du 2 d 2V neutral if = 0. du 2 1160 500 mm C © 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. 11–37. Continued At u = 51.22°, d 2V 2 = 900 cos 51.22° - 49.05 sin 51.22° - 1500 cos [2(51.22°)] = 848.77 7 0 du 2 u = 51.22° Thus, the system is in stable equilibrium at u = 51.2°. At u = 4.713°, d 2V 2 = 900 cos 4.713° - 49.05 sin 4.713° - 1500 cos [2(4.713°)] = - 586.82 6 0 du2 u = 4.713° Thus, the system is in unstable equilibrium at u = 4.71°. Ans: Stable equilibrium at u = 51.2° Unstable equilibrium at u = 4.71° 1161 © 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. 11–38. The two bars each have a mass of 8 kg. Determine the required stiffness k of the spring so that the two bars are in equilibrium when u = 60°. The spring has an unstretched length of 1 m. Investigate the stability of the system at the equilibrium position. A u 1.5 m B k Solution 1.5 m Potential Function. The Datum is established through point A, Fig. a. Since the centers of gravity of the bars are below the datum, their potential energies are negative. Here, y1 = 0.75 cos u, y2 = 1.5 cos u + 0.75 cos u = 2.25 cos u and the spring stretches x = 2(1.5 cos u) - 1 = (3 cos u - 1) m. Thus, V = Ve + Vg = = 1 2 kx + ΣWy 2 1 k(3 cos u - 1)2 + [ - 8(9.81)(0.75 cos u)] + [ - 8(9.81)(2.25 cos u)] 2 = 4.5 k cos2 u - 3 k cos u + 0.5 k - 235.44 cos u = 4.5 k cos2 u - (3 k + 235.44) cos u + 0.5 k Equilibrium Position. The system is in equilibrium if dV = 0 du dV = - 9 k sin u cos u + (3 k + 235.44) sin u = 0 du sin u( -9 k cos u + 3 k + 235.44) = 0 Since sin u ≠ 0, then - 9 k cos u + 3 k + 235.44 = 0 k = When u = 60°, k = 235.44 9 cos u - 3 235.44 = 156.96 N>m = 157 N>m 9 cos 60° - 3 Using this result, dV = -9(156.96) sin u cos u + [3(156.96) + 235.44] sin u du = - 1412.64 sin u cos u + 706.32 sin u Using the trigonometry identity sin 2u = 2 sin u cos u, dV = 706.32 sin u - 706.32 sin 2u du d 2V = 706.32 cos u - 1412.64 cos 2u du 2 1162 Ans. C © 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. 11–38 Continued d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0, 2 du du 2 d 2V and neutral if = 0. du 2 At u = 60°, d 2V 2 = 706.32 cos 60° - 1412.64 cos [2(60°)] = 1059.16 7 0 du 2 u = 60° Thus, the system is in stable equilibrium at u = 60°. Ans: k = 157 N>m Stable equilibrium at u = 60° 1163 © 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. 11–39. A spring with a torsional stiffness k is attached to the pin at B. It is unstretched when the rod assembly is in the vertical position. Determine the weight W of the block that results in neutral equilibrium. Hint: Establish the potential energy function for a small angle u. i.e., approximate sin u L 0, and cos u L 1 - u2>2. L 2 A L 2 SOLUTION Potential Function: With reference to the datum, Fig. a, the gravitational potential B k energy of the block is positive since its center of gravity is located above the datum. Here, the rods are tilted with a small angle u. Thus, y = 3 L cos u + L cos u = L cos u. 2 2 u2 .Thus, 2 u2 3WL u2 3 b = a1 - b Vg = Wy = W a L b a1 2 2 2 2 L 2 C 1 - However, for a small angle u, cos u The elastic potential energy of the torsional spring can be computed using 1 Ve = kb 2, where b = 2u. Thus, 2 Vg = 1 k(2u)2 = 2ku2 2 The total potential energy of the system is V = Vg + Ve = u2 3WL a1 - b + 2ku2 2 2 Equilibrium Configuration: Taking the first derivative of V, we have 3WL 3WL dV = u + 4ku = ua + 4k b du 2 2 dV = 0. Thus, du 3WL + 4k b = 0 ua 2 Equilibrium requires u = 0° Stability: The second derivative of V is 3WL d2V + 4k = 2 du2 To have neutral equilibrium at u = 0°, 3WL + 4k = 0 2 8k W = 3L d2V 2 = 0. Thus, du2 u = 0° - Ans. Note: The equilibrium configuration of the system at u = 0° is stable if 8k d2V 8k d 2V and is unstable if W > W< > 0 < 0 . 3L du2 3L du2 1164 Ans: W = 8k 3L © 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. *11–40. A conical hole is drilled into the bottom of the cylinder, and it is then supported on the fulcrum at A. Determine the minimum distance d in order for it to remain in stable equilibrium. SOLUTION Potential Function: First, we must determine the center of gravity of the cylinder. By referring to Fig. a, d 1 h (rpr2h) - a rpr2d b ©yCm 2 4 3 6h2 - d 2 = y = = 1 ©m 4(3h - d) rpr 2h - rpr2d 3 (1) With reference to the datum, Fig. a, the gravitational potential energy of the cylinder is positive since its center of gravity is located above the datum. Here, y = (y - d)cos u = B 6h2 - d2 6h2 - 12hd + 3d2 - d R cos u = B R cos u 4(3h - d) 4(3h - d) Thus, V = Vg = Wy = W B 6h2 - 12hd + 3d2 R cos u 4(3h - d) Equilibrium Configuration: Taking the first derivative of V, dV 6h2 - 12hd + 3d2 = -W B R sin u du 4(3h - d) Equilibrium requires dV = 0. Thus, du -WB 6h2 - 12hd + 3d2 R sin u = 0 4(3h - d) sin u = 0 u = 0° Stability: The second derivative of V is 6h2 - 12hd + 3d2 d 2V = -W B R cos u 2 4(3h - d) du To have neutral equilibrium at u = 0°, -WB d 2V 2 = 0. Thus, d2u u = 0° 6h2 - 12hd + 3d2 R cos 0° = 0 4(3h - d) 6h2 - 12hd + 3d2 = 0 d = 12h ; 2( - 12h)2 - 4(3)(6h2) = 0.5858h = 0.586h 2(3) Ans. Note: If we substitute d = 0.5858h into Eq. (1), we notice that the fulcrum must be at the center of gravity for neutral equilibrium. 1165 Ans: d = 0.586h © 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. 11–41. The uniform rod has a mass of 100 kg. If the spring is unstretched when u = 60°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. The spring is always in the horizontal position due to the roller guide at B A u 2m k 500 N/m B 2m Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is below the datum, its potential energy is negative. Here y = 2 cos u, and the spring stretches x = 2 sin 60° - 2 sin u = 2(sin 60° - sin u). Thus V = Ve + Vg = 1 2 kx + Wy 2 = 1 (500)[2(sin 60° - sin u)]2 + [ - 100(9.81)(2 cos u)] 2 = 1000 sin2 u - 100013 sin u - 1962 cos u + 750 dV = 0. Equilibrium Position. The bar is in equilibrium if du dV = 2000 sin u cos u - 100013 cos u + 1962 sin u du Using the trigonometry identity sin 2u = 2 sin u cos u, dV = 1000 sin 2u - 100013 cos u + 1962 sin u = 0 du Solved numerically, Ans. u = 24.62° = 24.6° d 2V = 2000 cos 2u + 100013 sin u + 1962 cos u du 2 d 2V d 2V 7 0, unstable if 6 0 Stability. The equilibrium configuration is stable if 2 du du 2 d 2V and neutral if = 0. du 2 At u = 24.62°, d 2V = 2000 cos[2(24.62°)] + 100013 sin 24.62° + 1962 cos 24.62° = 3811.12 7 0 du 2 Thus, the bar is in stable equilibrium at u = 24.6°. Ans: Stable equilibrium at u = 24.6° 1166 © 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. 11–42. Each bar has a mass per length of m0. Determine the angles u and f at which they are suspended in equilibrium. The contact at A is smooth, and both are pin connected at B. B l u 3l 2 Solution A l 2 Require G for system to be at its lowest point. 0 x = f l l 3 a b - (0.75 l sin 26.565°)a lb 4 2 2 = -0.20937 l l 3 l + + l 2 2 lx 1 3 - a b(l) - l a b - (0.75 l cos 26.565°)a lb 2 2 2 y = = -0.66874 l l 3 l + + l 2 2 f = tan-1a 0.20937 l b = 17.38° = 17.4° 0.66874 l Ans. Ans. u = 26.565° - 17.38° = 9.18° Ans: f = 17.4° u = 9.18° 1167 © 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. 11–43. The truck has a mass of 20 Mg and a mass center at G. Determine the steepest grade u along which it can park without overturning and investigate the stability in this position. G 3.5 m SOLUTION Potential Function: The datum is established at point A. Since the center of gravity for the truck is above the datum, its potential energy is positive. Here, y = (1.5 sin u + 3.5 cos u) m. 1.5 m u 1.5 m V = Vg = Wy = W(1.5 sin u + 3.5 cos u) Equilibrium Position: The system is in equilibrium if dV = 0 du dV = W(1.5 cos u - 3.5 sin u) = 0 du Since W Z 0, 1.5 cos u - 3.5 sin u = 0 Ans. u = 23.20° = 23.2° Stability: d2V = W(-1.5 sin u - 3.5 cos u) du2 d2V du2 u = 23.20° = W( - 1.5 sin 23.20° - 3.5 cos 23.20°) = -3.81W 6 0 Thus, the truck is in unstable equilibrium at u = 23.2° Ans. Ans: Unstable equilibrium at u = 23.2° 1168 © 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. *11–44. The small postal scale consists of a counterweight W1, connected to the members having negligible weight. Determine the weight W2 that is on the pan in terms of the angles u and f and the dimensions shown. All members are pin connected. W2 a u SOLUTION b y1 = b cos u a f f W1 y2 = a sin f = a sin (90° - u - g) where g is a constant and f = (90° - u - g) V = -W1y1 + W2y2 = -W1 b cos u + W2 a sin (90° - u - g) dV = W1 b sin u - W2 a cos (90° - u - g) du b sin u W2 = W1 a b a cos f Ans. Ans: b sin u W2 = W1a b a cos f 1169 © 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. 11–45. A 3-lb weight is attached to the end of rod ABC. If the rod is supported by a smooth slider block at C and rod BD, determine the angle u for equilibrium. Neglect the weight of the rods and the slider. C 6 in. B θ D 4 in. 10 in. SOLUTION x = 2(6)2 - (4 sin u)2 = 236 - 16 sin2 u A 16 6 = x (x + 4 cos u) + y x + 4 cos u + y = 2.667x y = - 4 cos u + 1.667 236 - 16 sin2 u 1 1 dy = 4 sin u du + 1.667a b (36 - 16 sin2 u)- 2 ( -32 sin u cos u)du 2 dU = 0; W dy = 0 W c 4 - 0.8333(36 - 16 sin2 u)- 2 (32 cos u) d sin u du = 0 1 Thus, sin u = 0 Ans. u = 0° or, A 36 - 16 sin2u B 2 = 6.667 cos u 1 Ans. u = 33.0° Ans: u = 0° u = 33.0° 1170 © 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. 11–46. If the uniform rod OA has a mass of 12 kg, determine the mass m that will hold the rod in equilibrium when u = 30°. Point C is coincident with B when OA is horizontal. Neglect the size of the pulley at B. B C m 3m SOLUTION Geometry: Using the law of cosines, 2 A 1m 2 lA¿B = 21 + 3 - 2112132 cos190° - u2 = 210 - 6 sin u e O lAB = 212 + 32 = 210 m l = lAB - lA¿B = 210 - 210 - 6 sin u Potential Function: The datum is established at point O. Since the center of gravity of the rod and the block are above the datum, their potential energy is positive. Here, y1 = 3 - l = 33 - 1210 - 210 - 6 sin u24 m and y2 = 0.5 sin u m. V = Vg = W1y1 + W2y2 = 9.81m33 - 1210 - 210 - 6 sin u24 + 117.7210.5 sin u2 = 29.43m - 9.81m1210 - 210 - 6 sin u2 + 58.86 sin u Equilibrium Position: The system is in equilibrium if dV = 0. ` du u = 30° 1 1 dV = - 9.81m c - 110 - 6 sin u2- 21- 6 cos u2 d + 58.86 cos u du 2 = - 29.43m cos u 210 - 6 sin u + 58.86 cos u At u = 30°, dV 29.43m cos 30° = + 58.86 cos 30° = 0 ` du u = 30° 210 - 6 sin 30° Ans. m = 5.29 kg Ans: m = 5.29 kg 1171 © 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. 11–47. The cylinder is made of two materials such that it has a mass of m and a center of gravity at point G. Show that when G lies above the centroid C of the cylinder, the equilibrium is unstable. G C r SOLUTION Potential Function: The datum is established at point A. Since the center of gravity of the cylinder is above the datum, its potential energy is positive. Here, y = r + d cos u. V = Vg = Wy = mg(r + d cos u) Equilibrium Position: The system is in equilibrium if dV = 0. du dV = - mgd sin u = 0 du sin u = 0 u = 0° Stability: d 2V = - mgd cos u du2 d 2V 2 = - mgd cos 0° = - mgd 6 0 du2 u = 0° Thus, the cylinder is in unstable equilibrium at u = 0° (Q.E.D.) 1172 a © 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. *11–48. The bent rod has a weight of 5 lb ft. A pivot of negligible size is attached at its center A and the rod is balanced as shown. Determine the length L of its vertical segments so that it remains in neutral equilibrium. Neglect the thickness of the rod. 8 in. 2 in. L 8 in. A L SOLUTION To remain in neutral equilibrium, the center of gravity must be located at A. y = 0 = ©yW ©W 2a L 2L 16 b (5) - a - 2 b a b (5) 12 2 12 a 2L 16 b (5) + a b (5) 12 12 0 = 5L2 - 20L - 160 L = 20 ; 2( - 20)2 - 4(5)(- 160) 10 Take the positive root Ans. L = 8 in. Ans: L = 8 in. 1173 © 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. 11–49. The triangular block of weight W rests on the smooth corners which are a distance a apart. If the block has three equal sides of length d, determine the angle u for equilibrium. d 60 G 60 u SOLUTION a AF = AD sin f = AD sin (60° - u) a AD = sin a sin 60° AD = a (sin (60° + u)) sin 60° AF = a (sin (60° + u)) sin (60° - u) sin 60° = y= a (0.75 cos2 u - 0.25 sin2 u) sin 60° d 23 cos u - AF V = Wy dV a ( -1.5 sin u cos u - 0.5 sin u cos u)d = 0 = Wc( -0.5774 d) sin u du sin 60° Require, sin u = 0 or - 0.5774 d - Ans. u = 0° a ( -2 cos u) = 0 sin 60° u = cos-1 a d b 4a Ans. Ans: u = 0° u = cos-1a 1174 d b 4a