© 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) 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°, 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 = (3) Substituting Eqs (1) and (2) into (3) yields (10 cos u - 0.5FAC cos u - 0.8660FAC sin u)ldu = 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 = C D A B u -P(2L cos u du) - F(- L sin u du) = 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 = 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 ϭ 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 work when the virtual displacements take place u ϭ 30° 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 = 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 = (3) Substituting P = lb, Eqs (1) and (2) into Eq (3), 5(18 cos udu)FS (6 cos udu) = cos udu(90 - 6FS) = Since cos udu Z 0, then 90 - 6FS = 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 = lb and a couple moment M = 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 ϭ lb и ft 1.5 ft k ϭ 12 lb/ft SOLUTION P ϭ lb dyF = dyP = 1.5du PdyP + Mdu - FdyF = dU = 0; 8(1.5 du) + du - F(1.5 du) = u(20 - 1.5F) = Since du Z 20 -1.5F = 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 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, = 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 negative work (3) - FdxA - Mdu = dU = 0; Substituting Eq (2) into (3) Since du ≠ 0, then ca 0.4 xA sin u bF + M d du = 0.4 cos u - 2xA 0.4 xA sin u F + M = 0.4 cos u - 2xA F = -a 0.4 cos u - 2xA bM (4) 0.4 xA sin u 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 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, = 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 negative work (3) - FdxA - Mdu = dU = 0; Substituting Eq (2) into (3) ca Since du ≠ 0, then 0.4 xA sin u bF + M d du = 0.4 cos u - 2xA a Here, M = 75 N # m Then 0.4 xA sin u bF + M = 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 (4) 0.4 xA sin u 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.) 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 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 lb/in k B D lb/ in ft 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 work u M A ft ft M C 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 = + cos u y2 = 0.5 + cos u y1 = cos u (1) dy3 = -4 sin udu (2) dy2 = -4 sin udu (3) dy1 = - 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) negative work The couple moment M does negative work when the links undergo a positive virtual rotation du dU = 0; - 2Fspdy3 - 50dy2 - 20dy1 - 2Mdu = (4) Substituting Eqs (1), (2) and (3) into (4) yields (8Fsp sin u + 240 sin u - 2M) du = Since du Z 0, then 8Fsp sin u + 240 sin u - 2M = M = sin u(4Fsp + 120) At the equilibrium position u = 20°, Fsp = kx = 2(4) = 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 = Pl cos du - k(a sin u - l0) a cos u du = 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 ft ft SOLUTION y = sin u, x = cos u dy = cos u du, dU = 0; B ft dx = - sin u du A - Fsdx - 30dy - 30dy = θ ft D - Fs1 -4 sin u2 - 6014 cos u24du = Fs = 60 a cos u b sin u Since Fs = k14 cos u - cos 45°2 = 20014 cos u - cos 45°2 60 cos u = 8001cos u - cos 45°2sin u sin u - 0.707 tan u - 0.075 = 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 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 = sin u dyC = xA = l cos u cos udu 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 negative work (3) dU = 0; - WdyC - PdyA = Substituting Eqs (1) and (2) into (3) yields a Pl sin u - Wl cos ub du = Since du Z 0, then Pl sin u P = Wl cos u = W cot u Ans Ans: P = 1132 W cot 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 11–37 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 ϭ1.5 kN/ m u u A 500 mm Solution Potential Function The Datum the centers of gravity of the energies are positive Here x = 2(0.5 cos u) - 0.5 - 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 kx + ΣWy = (1500)(cos u - 0.6)2 + 2[10(9.81)](0.25 sin u) = = 750 cos2 u - 900 cos u + 49.05 sin u + 270 Equilibrium Position The system is in equilibrium if dV = du dV = - 1500 sin u cos u + 900 sin u + 49.05 cos u = du Solving numerically, u = 4.713° = 4.71° or u = 51.22° = 51.2° Ans Using the trigonometry identity sin 2u = 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 d 2V d 2V Stability The equilibrium configurate is stable if 0, unstable if and du du d 2V neutral if = du 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 = 900 cos 51.22° - 49.05 sin 51.22° - 1500 cos [2(51.22°)] = 848.77 du u = 51.22° Thus, the system is in stable equilibrium at u = 51.2° At u = 4.713°, d 2V = 900 cos 4.713° - 49.05 sin 4.713° - 1500 cos [2(4.713°)] = - 586.82 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 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 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) - = (3 cos u - 1) m Thus, V = Ve + Vg = = kx + ΣWy k(3 cos u - 1)2 + [ - 8(9.81)(0.75 cos u)] + [ - 8(9.81)(2.25 cos u)] = 4.5 k cos2 u - 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 = du dV = - k sin u cos u + (3 k + 235.44) sin u = du sin u( -9 k cos u + k + 235.44) = Since sin u ≠ 0, then - k cos u + k + 235.44 = k = When u = 60°, k = 235.44 cos u - 235.44 = 156.96 N>m = 157 N>m cos 60° - 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 = 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 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 0, unstable if 0, du du d 2V and neutral if = du At u = 60°, d 2V = 706.32 cos 60° - 1412.64 cos [2(60°)] = 1059.16 du 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 - u2>2 L A L 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 = L cos u + L cos u = L cos u 2 u2 Thus, u2 3WL u2 b = a1 - b Vg = Wy = W a L b a1 2 2 L C - However, for a small angle u, cos u The elastic potential energy of the torsional spring can be computed using Ve = kb 2, where b = 2u Thus, Vg = k(2u)2 = 2ku2 The total potential energy of the system is V = Vg + Ve = u2 3WL a1 - b + 2ku2 2 Equilibrium Configuration: Taking the first derivative of V, we have 3WL 3WL dV = u + 4ku = ua + 4k b du 2 dV = Thus, du 3WL + 4k b = ua Equilibrium requires u = 0° Stability: The second derivative of V is 3WL d2V + 4k = du2 To have neutral equilibrium at u = 0°, 3WL + 4k = 8k W = 3L d2V = 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< > < 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 h (rpr2h) - a rpr2d b ©yCm 6h2 - d = y = = ©m 4(3h - d) rpr 2h - rpr2d (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 = Thus, du -WB 6h2 - 12hd + 3d2 R sin u = 4(3h - d) sin u = u = 0° Stability: The second derivative of V is 6h2 - 12hd + 3d2 d 2V = -W B R cos u 4(3h - d) du To have neutral equilibrium at u = 0°, -WB d 2V = Thus, d2u u = 0° 6h2 - 12hd + 3d2 R cos 0° = 4(3h - d) 6h2 - 12hd + 3d2 = 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 ϭ 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 = cos u, and the spring stretches x = sin 60° - sin u = 2(sin 60° - sin u) Thus V = Ve + Vg = kx + Wy = (500)[2(sin 60° - sin u)]2 + [ - 100(9.81)(2 cos u)] = 1000 sin2 u - 100013 sin u - 1962 cos u + 750 dV = 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 = sin u cos u, dV = 1000 sin 2u - 100013 cos u + 1962 sin u = du Solved numerically, Ans u = 24.62° = 24.6° d 2V = 2000 cos 2u + 100013 sin u + 1962 cos u du d 2V d 2V 0, unstable if Stability The equilibrium configuration is stable if du du d 2V and neutral if = du At u = 24.62°, d 2V = 2000 cos[2(24.62°)] + 100013 sin 24.62° + 1962 cos 24.62° = 3811.12 du 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 Solution A l Require G for system to be at its lowest point x = f l l a b - (0.75 l sin 26.565°)a lb 2 = -0.20937 l l l + + l 2 lx - a b(l) - l a b - (0.75 l cos 26.565°)a lb 2 y = = -0.66874 l l l + + l 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 = du dV = W(1.5 cos u - 3.5 sin u) = du Since W Z 0, 1.5 cos u - 3.5 sin u = 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 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 in B θ D in 10 in SOLUTION x = 2(6)2 - (4 sin u)2 = 236 - 16 sin2 u A 16 = x (x + cos u) + y x + cos u + y = 2.667x y = - cos u + 1.667 236 - 16 sin2 u 1 dy = sin u du + 1.667a b (36 - 16 sin2 u)- ( -32 sin u cos u)du dU = 0; W dy = W c - 0.8333(36 - 16 sin2 u)- (32 cos u) d sin u du = Thus, sin u = Ans u = 0° or, A 36 - 16 sin2u B = 6.667 cos u 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, A 1m lA¿B = 21 + - 2112132 cos190° - u2 = 210 - sin u e O lAB = 212 + 32 = 210 m l = lAB - lA¿B = 210 - 210 - 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 = - l = 33 - 1210 - 210 - sin u24 m and y2 = 0.5 sin u m V = Vg = W1y1 + W2y2 = 9.81m33 - 1210 - 210 - sin u24 + 117.7210.5 sin u2 = 29.43m - 9.81m1210 - 210 - sin u2 + 58.86 sin u Equilibrium Position: The system is in equilibrium if dV = ` du u = 30° 1 dV = - 9.81m c - 110 - sin u2- 21- cos u2 d + 58.86 cos u du = - 29.43m cos u 210 - sin u + 58.86 cos u At u = 30°, dV 29.43m cos 30° = + 58.86 cos 30° = ` du u = 30° 210 - 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 = du dV = - mgd sin u = du sin u = u = 0° Stability: d 2V = - mgd cos u du2 d 2V = - mgd cos 0° = - mgd 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 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 in in L in A L SOLUTION To remain in neutral equilibrium, the center of gravity must be located at A y = = ©yW ©W 2a L 2L 16 b (5) - a - b a b (5) 12 12 a 2L 16 b (5) + a b (5) 12 12 = 5L2 - 20L - 160 L = 20 ; 2( - 20)2 - 4(5)(- 160) 10 Take the positive root Ans L = in Ans: L = 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 = = Wc( -0.5774 d) sin u du sin 60° Require, sin u = or - 0.5774 d - Ans u = 0° a ( -2 cos u) = sin 60° u = cos-1 a d b 4a Ans Ans: u = 0° u = cos-1a 1174 d b 4a ... 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. .. 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... 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,