Engineering Mechanics - Statics Chapter 11 Problem 11-9 A force P is applied to the end of the lever. Determine the horizontal force F on the piston for equilibrium. Solution: δ s 2 l δθ = x 2 l cos θ () = δ x 2− l sin θ () δθ = δ UP− δ sF δ x−= 0= P− 2l δθ F2lsin θ () δθ + 0= FPcsc θ () = Problem 11-10 The mechanism consists of the four pin-connected bars and three springs, each having a stiffness k and an unstretched length l 0 Determine the horizontal forces P that must be applied to the pins in order to hold the mechanism in the horizontal position for equilibrium. 1081 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 Solution: xlcos θ () = δ xl− sin θ () δθ = y 2 l sin θ () = δ y 2 l cos θ () δθ = δ U = 0; 2− P δ x 3 F s − 0= 2 Plsin θ () δθ 3 F s 2lcos θ () δθ − 0= P sin θ () 3 F s cos θ () = Since F s k 2 lsin θ () l 0 − () = , then P 3 kcot θ () 2 lsin θ () l 0 − () = Problem 11-11 When θ = θ 0 , the uniform block of weight W b compresses the two vertical springs a distance δ . If the uniform links AB and CD each weigh W L , determine the magnitude of the applied couple moments M needed to maintain equilibrium. 1082 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 Given: θ 0 20 deg= a 1ft= W b 50 lb= b 4ft= δ 4in= c 1ft= W L 10 lb= d 2ft= k 2 lb in = Solution: θθ 0 = y 1 b 2 cos θ () = δ y 1 b− 2 sin θ () δθ = y 2 a 2 bcos θ () += δ y 2 b− sin θ () δθ = y 3 y 2 a 2 += δ y 3 δ y 2 = δ U 2− W L δ y 1 W b δ y 2 − 2k δδ y 3 − 2M δθ −= 0= δ U 2W L b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ () W b bsin θ () + 2k δ bsin θ () + 2M− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ δθ = 0= M W L W b + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ bk δ b+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ sin θ () = M 52.0 lb ft⋅= Problem 11-12 The spring is unstretched when θ = 0. Determine the angle θ for equilibrium. Due to the roller guide, the spring always remains vertical. Neglect the weight of the links. 1083 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 Given: P 8lb= k 50 lb ft = a 2ft= b 2ft= Solution: y 1 asin θ () = δ y 1 acos θ () δθ = y 2 ab+( )sin θ () a+ b+= δ y 2 ab+( )cos θ () δθ = δ Uk− y 1 δ y 1 P δ y 2 += k− asin θ () acos θ () Pa b+( ) cos θ () + ⎡ ⎣ ⎤ ⎦ δθ = 0= cos θ () Pa b+()ka 2 sin θ () − ⎡ ⎣ ⎤ ⎦ 0= There are 2 answers θ 1 acos 0()= θ 1 90deg= θ 2 asin Pa b+() ka 2 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = θ 2 9.207 deg= Problem 11-13 Determine the force P required to lift the block of mass M using the differential hoist. The lever arm is fixed to the upper pulley and turns with it. Given: a 800 mm= 1084 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 M 15 kg= b 150 mm= c 300 mm= g 9.81 m s 2 = Solution: δ U = 0; Pa δθ Mg 2 c− δθ b δθ + () + 0= Pa Mg 2 bc−()+ 0= P Mg 2 cb− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 13.8 N= Problem 11-14 Determine the magnitude of the applied couple moments M needed to maintain equilibrium at θ . The plate E has a weight W. Neglect the weight of the links AB and CD. Given: a 0.5 ft= d 2ft= b 1ft= c 2ft= W 50 lb= θ 20 deg= 1085 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 Solution: y m d cos θ () a 2 += δ y m d− sin θ () δθ = δ U 2− M δθ Wd− sin θ () δθ () ⎡ ⎣ ⎤ ⎦ −= Wdsin θ () 2 M− () δθ = 0= M 1 2 Wdsin θ () = M 17.1 lb ft⋅= Problem 11-15 The members of the mechanism are pin connected. If a horizontal force P acts at A, determine the angle θ for equilibrium.The spring is unstretched when θ = 90°. Units Used: kN 10 3 N= Given: a 0.5 m= k 20 kN m = P 400 N= Solution: x 1 acos θ () = δ x 1 a− sin θ () δθ = x 2 5acos θ () = δx 2 5− asin θ () δθ= δ UP5− asin θ ()() kacos θ () a− sin θ ()() − ⎡ ⎣ ⎤ ⎦ δθ = 0= sin θ () 5− Pkacos θ () + () 0= 1086 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 There are 2 equilibrium angles. θ 1 asin 0()= θ 1 0deg= θ 2 acos 5P ka ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 2 78.5 deg= Problem 11-16 Determine the force F needed to lift the block having weight W. Hint: Note that the coordinates S A and S B can be related to the constant vertical length l of the cord. Given: W 100 lb= Solution: ls A 2 s B += 0 δ s A 2 δ s B += δ s A 2− δ s B = δ UF δ s A W δ s B += 2− FW+() δ s B = 0= F W 2 = F 50lb= 1087 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 Problem 11-17 Each member of the pin-connected mechanism has a mass m 1 . If the spring is unstretched when θ = 0° determine the angle θ for equilibrium. Given: a 300 mm= k 2500 N m = m 1 8kg= M 50 N m⋅= Solution: xasin θ () = δ xacos θ () δθ = yacos θ () = δ ya− sin θ () δθ = F s kx= F s kasin θ () = δ Um 1 g 2 δ y 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ δ y+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ F s δ x− M δθ += 0= δ U 2− m 1 gasin θ () kasin θ () acos θ () − M+ () δ U= 0= Initial Guesses: θ 10 deg= Given 2− m 1 gasin θ () kasin θ () acos θ () − M+ 0= θ Find θ () = θ 10.7 deg= Now starting with a different guess we find anothe answer. θ 90 deg= Given 2− m 1 gasin θ () kasin θ () acos θ () − M+ 0= θ Find θ () = θ 89.3 deg= 1088 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 Problem 11-18 The bar is supported by the spring and smooth collar that allows the spring to be always perpendicular to the bar for any angle θ . If the unstretched length of the spring is l 0 , determine the force P needed to hold the bar in the equilibrium position θ . Neglect the weight of the bar. Solution: sasin θ () = δ s acos θ () δθ = ylsin θ () = δ y lcos θ () δθ = F s kasin θ () l 0 − () = δ UP δ yF s δ s−= 0= δ UPlcos θ () δθ kasin θ () l 0 − () acos θ () δθ −= 0= P ka l asin θ () l 0 − () = Problem 11-19 The scissors jack supports a load P . Determine the axial force in the screw necessary for equilibrium when the jack is in the position θ . Each of the four links has a length L and is pin-connected at its center. Points B and D can move horizontally. 1089 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 11 Solution: xLcos θ () = δ xL− sin θ () δθ = y 2L sin θ () = δ y 2L cos θ () δθ = δ UP− δ yF δ x−= P− 2Lcos θ () FLsin θ () + () δθ = 0= F 2P cot θ () = Problem 11-20 Determine the masses m A and m B of A and B required to hold the desk lamp of mass M in balance for any angles θ and φ . Neglect the weight of the mechanism and the size of the lamp. Given: M 400 gm= a 75 mm= b 75 mm= c 75 mm= d 300 mm= e 300 mm= g 9.81 m s 2 = 1090 © 2007 R. C. Hibbeler. Published by 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. [...]... means, without permission in writing from the publisher Engineering Mechanics - Statics V= Chapter 11 ⎛ γ π a2 h2 W3a ⎞ ⎜ ⎟ cos ( θ ) − 8 ⎠ ⎝ 2 ⎛ γ π a2 h2 W3a ⎞ ⎟ sin ( θ ) V = −⎜ − 8 ⎠ ⎝ 2 dθ d ⎛ γ π a2 h2 W3a ⎞ ⎟ cos ( θ ) V = −⎜ − 2 8 ⎠ ⎝ 2 dθ d 2 For neutral equilibrium we must have 2 2 γ πa h 2 − W3a =0 8 h = W3 4π γ a h = 3. 99 in Problem 1 1-4 8 Compute the force developed in the spring required... ⎟ ⎝ 4b ⎠ θ 2 = asin ⎛ ⎜ a 3 = π − θ2 θ 1 = 90 deg θ 2 = 36 .87 deg θ 3 = 1 43. 13 deg Check Stability V'' 1 = −a sin ( θ 1 ) − 4b cos ( 2θ 1 ) V'' 1 = 16 lb ft Stable V'' 2 = −a sin ( θ 2 ) − 4b cos ( 2θ 2 ) V'' 2 = −25.6 lb ft Unstable V'' 3 = −a sin ( θ 3 ) − 4b cos ( 2θ 3 ) V'' 3 = −25.6 lb ft Unstable Problem 1 1-2 9 If the potential energy for a conservative two-degree-of-freedom system is expressed... publisher Engineering Mechanics - Statics Chapter 11 a 60 + 60 - c b 60 Solution: a b = sin ( 60 deg) sin ( 60 deg − θ ) b=a sin ( 60 deg − θ ) sin ( 60 deg) ⎞ ⎛2 V = W⎜ d sin ( 60 deg) cos ( θ ) − b cos ( 30 deg − θ )⎟ 3 ⎝ ⎠ V= ( 2d cos (θ ) − 2a cos (2θ ) − a) W 2 3 d dθ V = W (−2d sin (θ ) + 8a sin ( θ ) cos ( θ )) = 0 2 3 θ 1 = asin ( 0) θ 1 = 0 deg d⎞ ⎟ ⎝ 4a ⎠ θ 2 = acos ⎛ ⎜ Problem 1 1-4 5 A homogeneous... ⎛ c1 ⎟ − ⎛ b ⎞ ⎛ c1 − c ⎟ ⎛ c1 + 3c ⎞ − ⎛ e ⎞ ⎛ c2 ⎟ ⎛ c2 ⎟ + ⎛ a ⎞ ⎛ c2 − c ⎟ ⎛ c2 + 3c ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠⎝ 4 ⎠ yc = 2 2 2 2 ⎞ ⎞ ⎞ ⎞ ⎛ d ⎞ ⎛ c1 ⎟ − ⎛ b ⎞ ⎛ c1 − c ⎟ − ⎛ e ⎞ ⎛ c2 ⎟ + ⎛ a ⎞ ⎛ c2 − c ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ 2⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝ 3 ⎠ hcr = yc hcr = 1 .32 ft If h > hcr then stable 1110... form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics At x = x2 Chapter 11 V'' 2 = 6a x2 + 2b V'' 2 = −12.2 lb ft V'' 2 < 0 Unstable Problem 1 1-2 8 If the potential energy for a conservative one-degree-of-freedom system is expressed by the relation V = asin(θ) + bcos(2θ), 0 deg ≤ θ ≤ 180 deg , determine the equilibrium positions and investigate the stability... Engineering Mechanics - Statics Chapter 11 Problem 1 1-4 3 Each bar has a mass per length of m0 Determine the angles θ and φ at which they are suspended in equilibrium The contact at A is smooth, and both are pin con-nected at B Solution: 1⎞ θ + φ = atan ⎛ ⎟ ⎜ ⎝ 2⎠ V=− d dθ l 3l l ⎞ ⎛ 3l ⎞ ⎛l⎞ ⎛ m0 ⎜ ⎟ cos ( θ ) − l m0 ⎜ ⎟ cos ( φ ) − m0 ⎜ l cos ( φ ) + sin ( φ )⎟ 4 2 2 ⎝ ⎝4⎠ ⎝ 2⎠ ⎠ V = Guess 9m0 l 8. .. in writing from the publisher Engineering Mechanics - Statics ∂ ∂ At ( 0 , 0) ∂x ∂ y Chapter 11 V =0 2 ⎤ ⎡⎛ ⎞⎜ ⎢ ∂ ∂ ⎞ ⎛ ∂2 ⎟ ⎛ ∂2 ⎞⎥ ⎜ ⎟ = − 4a b ⎢⎜ ∂x ∂ y V ⎟ − ⎜ x2 V ⎟ ⎜ y2 V ⎟⎥ ⎠ ⎝ ∂ ⎠ ⎝ ∂ ⎠⎦ ⎣⎝ At ( 0 , 0) 2 −4a b = −24 N m 2 0, then the vertical position is stable Problem 1 1 -3 8 If each of the three links of the mechanism has a weight W, determine the angle θ for equilibrium The spring, . publisher. Engineering Mechanics - Statics Chapter 11 M 15 kg= b 150 mm= c 30 0 mm= g 9 .81 m s 2 = Solution: δ U = 0; Pa δθ Mg 2 c− δθ b δθ + () + 0= Pa Mg 2 bc−()+ 0= P Mg 2 cb− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 13. 8. ft= Unstable V'' 3 a− sin θ 3 () 4bcos 2 θ 3 () −= V'' 3 25.6− lb ft= Unstable Problem 1 1-2 9 If the potential energy for a conservative two-degree-of-freedom system is expressed. 1− lb ft = c 3 lb= d 10 ft lb⋅= Solution: Vax 3 bx 2 + cx+ d+= Required Position: x V d d 3ax 2 2bx+ c+= 0= x 1 2− b 4b 2 43a()c−+ 23a() = x 1 0.59 ft= x 2 2− b 4b 2 43ac()−− 23a() = x 2 0.424−