Tài liệu Engineering Mechanics - StaticsChapter 1Problem 1-1 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms (b) μkm (c) ks/mg (d) km⋅ μN Units Used: μN = 10−6N kmμkm = 109−6Gs = 10 s pptx

Engineering Mechanics - Statics Chapter Problem 1-1 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms (b) μkm (c) ks/mg (d) km⋅ μN Units Used: μN = 10 −6 μkm = 10 N −6 km Gs = 10 s ks = 10 s mN = 10 −3 −3 ms = 10 N s Solution: ( a) m 3m = × 10 ms s m km =1 ms s ( b) μkm = × 10 −3 m μkm = mm ( c) ks s = × 10 mg kg ks Gs =1 mg kg ( d) −3 km⋅ μN = × 10 mN km⋅ μN = mm⋅ N © 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 Problem 1-2 Wood has a density d What is its density expressed in SI units? Units Used: Mg = 1000 kg Given: d = 4.70 slug ft Solution: 1slug = 14.594 kg d = 2.42 Mg m Problem 1-3 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm (b) mN/μs (c) μm⋅ Mg Solution: ( a) Mg 10 kg 10 kg Gg = = = −3 mm m m 10 m Mg Gg = mm m ( b) mN μs mN μs ( c) = 10 −3 10 = N −6 = s 10 N kN = s s kN s ( −6 μm⋅ Mg = 10 )( ) m 10 kg = 10 −3 m⋅ kg μm⋅ Mg = mm⋅ kg © 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 Problem 1-4 Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/mm, (c) mN/( kg⋅ μs) Solution: ( a) Mg 10 kg 10 kg Gg = = = −3 ms s s 10 s Mg Gg = ms s N = mm ( b) 1N 10 −3 = 10 3N m m = kN m N kN = mm m mN ( c) kg⋅ μs 10 = mN kg⋅ μs −3 −6 10 = N = kg⋅ s kN kg⋅ s kN kg⋅ s Problem 1-5 Represent each of the following with SI units having an appropriate prefix: (a) S1, (b) S2, (c) S3 Units Used: kg = 1000 g −3 ms = 10 s kN = 10 N Given: S1 = 8653 ms S2 = 8368 N S3 = 0.893 kg Solution: ( a) S1 = 8.653 s © 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 ( b) S2 = 8.368 kN ( c) Chapter S3 = 893 g Problem 1-6 Represent each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, and (c) z Units Used: MN = 10 N μg = × 10 −6 gm kN = 10 N Given: x = 45320 kN ( y = 568 × 10 ) mm z = 0.00563 mg Solution: ( a) x = 45.3 MN ( b) y = 56.8 km ( c) z = 5.63 μg Problem 1-7 Evaluate ( a⋅ b)/c to three significant figures and express the answer in SI units using an appropriate prefix Units Used: μm = 10 −6 m Given: a = ( 204 mm) © 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 b = ( 0.00457 kg) c = ( 34.6 N) Solution: l = ab l = 26.945 c μm⋅ kg N Problem 1-8 If a car is traveling at speed v, determine its speed in kilometers per hour and meters per second Given: v = 55 mi hr Solution: v = 88.514 km hr m v = 24.6 s Problem 1-9 Convert: (a) S1 to N ⋅ m , (b) S2 to kN/m3, (c) S3 to mm/s Express the result to three significant figures Use an appropriate prefix Units Used: kN = 10 N Given: S1 = 200g lb⋅ ft S2 = 350g lb ft S3 = ft hr © 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 Solution: ( a) S1 = 271 N⋅ m ( b) S2 = 55.0 kN m S3 = 0.677 ( c) mm s Problem 1-10 What is the weight in newtons of an object that has a mass of: (a) m1, (b) m2, (c) m3? Express the result to three significant figures Use an appropriate prefix Units Used: Mg = 10 kg mN = 10 −3 N kN = 10 N Given: m1 = 10 kg m2 = 0.5 gm m3 = 4.50 Mg Solution: ( a) W = m1 g W = 98.1 N ( b) W = m2 g W = 4.90 mN ( c) W = m3 g W = 44.1 kN © 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 Problem 1-11 If an object has mass m, determine its mass in kilograms Given: m = 40 slug Solution: m = 584 kg Problem 1-12 The specific weight (wt./vol.) of brass is ρ Determine its density (mass/vol.) in SI units Use an appropriate prefix Units Used: Mg = 10 kg Given: lb ρ = 520 ft Solution: Mg ρ = 8.33 m Problem 1-13 A concrete column has diameter d and length L If the density (mass/volume) of concrete is ρ, determine the weight of the column in pounds Units Used: Mg = 10 kg kip = 10 lb Given: d = 350 mm L = 2m © 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 ρ = 2.45 Chapter Mg m Solution: d⎞ V = π⎛ ⎟ L ⎜ ⎝ 2⎠ W = ρ V V = 192.423 L W = 1.04 kip Problem 1-14 The density (mass/volume) of aluminum is ρ Determine its density in SI units Use an appropriate prefix Units Used: Mg = 1000 kg Given: ρ = 5.26 slug ft Solution: ρ = 2.17 Mg m Problem 1-15 Determine your own mass in kilograms, your weight in newtons, and your height in meters Solution: Example W = 150 lb m = W m = 68.039 kg W g = 667.233 N h = 72 in h = 1.829 m © 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 Problem 1-16 Two particles have masses m1 and m2, respectively If they are a distance d apart, determine the force of gravity acting between them Compare this result with the weight of each particle Units Used: G = 66.73 × 10 − 12 m kg⋅ s nN = 10 −9 N Given: m1 = kg m2 = 12 kg d = 800 mm Solution: F = G m1 m2 d F = 10.0 nN W1 = m1 g W2 = m2 g W1 W1 = 78.5 N W2 = 118 N = 7.85 × 10 W2 F = 1.18 × 10 F 10 Problem 1-17 Using the base units of the SI system, show that F = G(m1m2)/r2 is a dimensionally homogeneous equation which gives F in newtons Compute the gravitational force acting between two identical spheres that are touching each other The mass of each sphere is m1, and the radius is r Units Used: μN = 10 −6 N G = 66.73 10 − 12 m kg⋅ s © 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 Given: m1 = 150 kg r = 275 mm Solution: F = G m1 ( 2r) 2 F = 4.96 μN Since the force F is measured in Newtons, then the equation is dimensionally homogeneous Problem 1-18 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, (c) z Units Used: MN = 10 N kN = 10 N μm = 10 −6 m Given: x = ( 200 kN) y = ( 0.005 mm) z = ( 400 m) Solution: ( a) x = 0.040 MN ( b) y = 25.0 μm ( c) z = 0.0640 km 10 © 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 Equilibrium Guess θ = 30 deg Guess θ = 70 deg Chapter 11 Given M g⎛ ⎜ a + b⎞ ⎟ cos ( θ ) − ⎝ ⎠ a + b⎞ Given M g⎛ ⎜ ⎟ cos ( θ ) − ⎝ ⎠ k a sin ( 2θ ) = k a sin ( 2θ ) = θ = Find ( θ ) θ = Find ( θ ) Check Staibility V'' = −M g⎛ ⎜ a + b⎞ V'' = −M g⎛ ⎜ a + b⎞ ⎝ ⎝ 2 ⎟ sin ( θ 1) − k a cos ( 2θ 1) ⎠ ⎟ sin ( θ 2) − k a cos ( 2θ 2) ⎠ θ = 36.1 deg V'' = −1.3 kN⋅ m Unstable θ = 90.0 deg V'' = 0.82 kN⋅ m Stable Problem 11-36 Determine the angle θ for equilibrium and investigate the stability at this position The bars each have mass mb and the suspended block D has mass mD Cord DC has a total length of L Given: mb = kg mD = kg a = 500 mm L = 1m g = 9.81 m s Solution: V = 2mb g Equilibrium a sin ( θ ) − mD g( L + a − 2a cos ( θ ) ) = V = mb g a sin ( θ ) + 2mD g a cos ( θ ) − mD g( L + a) 1105 © 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 d dθ Chapter 11 V = mb g a cos ( θ ) − 2mD g a sin ( θ ) = tan ( θ ) = ⎛ mb ⎞ ⎟ ⎝ 2mD ⎠ mb θ = atan ⎜ 2mD θ = 12.095 deg Stability d V'' = dθ V = −mb g a sin ( θ ) − 2mD g a cos ( θ ) V'' = −mb g a sin ( θ ) − 2mD g a cos ( θ ) V'' = −70.229 N⋅ m Since V'' < the equilibrium point is unstable Problem 11-37 The bar supports a weight of W at its end If the springs are originally unstretched when the bar is vertical, , investigate the stability of the bar when it is in the vertical position Given: k1 = 300 lb ft k2 = 500 lb ft W = 500 lb a = ft Solution: V = W3a cos ( θ ) + 2 k1 ( a sin ( θ ) ) + k2 ( 2a sin ( θ ) ) 2 V = 3W a cos ( θ ) + a (k1 + 4k2) sin (θ )2 2 a V' = V = −3W a sin ( θ ) + (k1 + 4k2) sin(2θ ) dθ d V'' = d dθ V = −3W a cos ( θ ) + a ( k1 + 4k2 ) cos ( 2θ ) 2 1106 © 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 θ = deg At V'' = −3W a cos ( θ ) + a ( k1 + 4k2 ) cos ( 2θ ) Since V'' = 1.62 × 10 lb ft > 0, then the vertical position is stable Problem 11-38 If each of the three links of the mechanism has a weight W, determine the angle θ for equilibrium The spring, which always remains vertical, is unstretched when θ = 0° Solution: V= k ( a sin ( θ ) ) − 2W a sin ( θ ) − W( 2a) sin ( θ ) V= ka sin ( θ ) − 4W a sin ( θ ) 2 d dθ V = k a sin ( θ ) cos ( θ ) − 4W a cos ( θ ) = cos ( θ ) = θ = 90 deg sin ( θ ) = θ = asin ⎛ ⎜ 4W ka 4W ⎞ ⎟ ⎝ ka ⎠ Problem 11-39 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 θ and φ and the dimensions shown All members are pin connected 1107 © 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: φ = −θ + constant V = W2 a sin ( φ ) − W1 b cos ( θ ) V' = d dθ V = −W2 a cos ( φ ) + W1 b sin ( θ ) = ⎛ b sin ( θ ) ⎞ ⎟ ⎝ a cos ( φ ) ⎠ W2 = W1 ⎜ Problem 11-40 The uniform right circular cone having a mass m is suspended from the cord as shown Determine the angle θ at which it hangs from the wall for equilibrium Is the cone in stable equilibrium? Solution: ⎛ 3a cos ( θ ) + a sin ( θ )⎞ m g ⎟ ⎝2 ⎠ V = −⎜ ⎛ −3a sin ( θ ) + a cos ( θ )⎞ m g ⎟ ⎝ ⎠ V' = dV V'' = d V = −⎜ dθ dθ ⎛ −3a cos ( θ ) − a sin ( θ )⎞ m g ⎟ ⎝ ⎠ = −⎜ Equilibrium V' = sin ( θ ) = cos ( θ ) tan ( θ ) = 1⎞ θ = atan ⎛ ⎟ ⎜ ⎝ 6⎠ θ = 9.462 deg 1108 © 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 V'' = −⎛ ⎜ −3 ⎝2 cos ( θ ) − Chapter 11 sin ( θ )⎞ a m g ⎟ V'' = 1.521m g a ⎠ Stable Problem 11-41 The homogeneous cylinder has a conical cavity cut into its base as shown Determine the depth d of the cavity so that the cylinder balances on the pivot and remains in neutral equilibrium Given: a = 50 mm b = 150 mm Solution: yc = b d⎛1 ⎞ a π b − ⎜ π a d⎟ 4⎝3 ⎠ πa b − πa d V = ( yc − d) cos ( θ ) W d dθ V = −W sin ( θ ) ( yc − d) θ = deg d2 dθ (equilibrium position) V = −W cos ( θ ) ( yc − d) = d = yc Guess Given d = 10 mm d= d⎛1 ⎞ b a π b − ⎜ π a d⎟ 4⎝3 ⎠ πa b − πa d d = Find ( d) d = 87.868 mm 1109 © 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 11-42 The conical manhole cap is made of concrete and has the dimensions shown Determine the critical location h = hcr of the pick-up connectors at A and B so that when hoisted with constant velocity the cap is in neutral equilibrium Explain what would happen if the connectors were placed at a point h > hcr Given: a = ft b = 2.5 ft c = ft d = ft e = d − ( b − a) Solution: V = W( yc − h) cos ( θ ) d dθ V = W( h − yc) sin ( θ ) = Equilibrium at sin ( θ ) = θ = 0deg For neutral equilibrium require d dθ V = W( h − yc) cos ( θ ) = Thus yc = h Thus, A and B must be at the elevation of the center of gravity of the cap c1 = cd d−b c2 = 2 2 ⎞ ⎞ ⎞ ⎞ ⎞ ⎞ ⎛ d ⎞ ⎛ c1 ⎟ ⎛ c1 ⎟ − ⎛ b ⎞ ⎛ c1 − c ⎟ ⎛ c1 + 3c ⎞ − ⎛ e ⎞ ⎛ c2 ⎟ ⎛ c2 ⎟ + ⎛ a ⎞ ⎛ c2 − c ⎟ ⎛ c2 + 3c ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ yc = 2 2 ⎞ ⎞ ⎞ ⎞ ⎛ d ⎞ ⎛ c1 ⎟ − ⎛ b ⎞ ⎛ c1 − c ⎟ − ⎛ e ⎞ ⎛ c2 ⎟ + ⎛ a ⎞ ⎛ c2 − c ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ 2⎠ ⎝ ⎠ ⎝ 2⎠ ⎝ ⎠ ⎝ 2⎠ ⎝ ⎠ ⎝ 2⎠ ⎝ ⎠ hcr = yc hcr = 1.32 ft If h > hcr then stable 1110 © 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 ce e−a Engineering Mechanics - Statics Chapter 11 Problem 11-43 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 )⎟ 2 ⎝ ⎝4⎠ ⎝ 2⎠ ⎠ V = Guess 9m0 l sin ( θ ) − m0 l sin ( φ ) + θ = 10 deg Given cos ( φ ) = φ = 10 deg 1⎞ θ + φ = atan ⎛ ⎟ ⎜ ⎛θ⎞ ⎜ ⎟ = Find ( θ , φ ) ⎝φ⎠ m0 l ⎝ 2⎠ sin ( θ ) − sin ( φ ) + cos ( φ ) = 8 ⎛ θ ⎞ ⎛ 9.18 ⎞ ⎜ ⎟=⎜ ⎟ deg ⎝ φ ⎠ ⎝ 17.38 ⎠ Problem 11-44 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 θ for equilibrium 1111 © 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 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 )⎟ ⎝ ⎠ V= ( 2d cos (θ ) − 2a cos (2θ ) − a) W d dθ V = W (−2d sin (θ ) + 8a sin ( θ ) cos ( θ )) = θ = asin ( 0) θ = deg d⎞ ⎟ ⎝ 4a ⎠ θ = acos ⎛ ⎜ Problem 11-45 A homogeneous cone rests on top of the cylindrical surface Derive a relationship between the radius r of the cylinder and the height h of the cone for neutral equilibrium Hint: Establish the potential function for a small angle θ of tilt of the cone, i.e., approximate sin θ ≈ and cos θ ≈ 1−θ 2/2 1112 © 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: V= V app d dθ h⎞ ⎤ ⎟ cos ( θ ) + rθ sin ( θ )⎥ W 4⎠ ⎦ ⎡⎛r + ⎢⎜ ⎣⎝ ⎡⎛ h ⎞ ⎛ θ ⎞ ⎤ ⎟ + rθ 2⎥ W = ⎢⎜ r + ⎟ ⎜ − 2⎠ ⎣⎝ ⎠ ⎝ ⎦ ⎡⎛ ⎣⎝ V app = ⎢−⎜ r + dVapp dθ d ⎛ ⎝ = ⎜r − h⎞ ⎤ ⎟ θ + 2rθ⎥ W = 4⎠ ⎦ h⎞ ⎟ θW = 4⎠ h V = r− = app dθ Equilibrium θ = deg For neutral equilibrium: r= h Problem 11-46 The door has a uniform weight W1 It is hinged at A and is held open by the weight W2 and the pulley Determine the angle θ for equilibrium Given: W1 = 50 lb W2 = 30 lb 1113 © 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 a = ft b = ft Solution: V = W1 ⎛ ⎜ b⎞ 2 ⎟ sin ( θ ) + W2 a + b − 2a b sin ( θ ) ⎝ 2⎠ d dθ a b cos ( θ ) ⎞=0 ⎛ ⎟ cos ( θ ) − W2 ⎜ ⎟ 2 ⎝ 2⎠ ⎝ a + b − 2a b sin ( θ ) ⎠ V = W1 ⎛ ⎜ b⎞ Guess θ = 10 deg Given W1 ⎛ ⎜ a b cos ( θ ) ⎞=0 ⎛ ⎟ cos ( θ ) − W2 ⎜ ⎟ 2⎠ 2 ⎝ ⎝ a + b − 2a b sin ( θ ) ⎠ b⎞ θ = Find ( θ ) θ = 16.26 deg Problem 11-47 The hemisphere of weight W supports a cylinder having a specific weight γ If the radii of the cylinder and hemisphere are both a., determine the height h of the cylinder which will produce neutral equilibrium in the position shown Given: W = 60 lb a = in γ = 311 lb ft Solution: ⎛ h⎞ ⎟ cos ( θ ) + γ π a h⎜ ⎟ cos ( θ ) ⎝8⎠ ⎝ 2⎠ V = −W⎛ ⎜ 3a ⎞ 1114 © 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 V= Chapter 11 ⎛ γ π a2 h2 W3a ⎞ ⎜ ⎟ cos ( θ ) − ⎠ ⎝ ⎛ γ π a2 h2 W3a ⎞ ⎟ sin ( θ ) V = −⎜ − ⎠ ⎝ dθ d ⎛ γ π a2 h2 W3a ⎞ ⎟ cos ( θ ) V = −⎜ − ⎠ ⎝ dθ d For neutral equilibrium we must have 2 γ πa h − W3a =0 h = W3 4π γ a h = 3.99 in Problem 11-48 Compute the force developed in the spring required to keep the rod of mass Mrod in equilibrium at θ The spring remains horizontal due to the roller guide Given: k = 200 N m M = 40 N⋅ m a = 0.5 m θ = 30 deg Mrod = kg Solution: ⎛ a ⎞ sin ( θ ) + k ( a cos ( θ ) − δ ) ⎟ ⎝ 2⎠ V = Mθ + Mrod g⎜ d dθ ⎛ a ⎞ cos ( θ ) − k( a cos ( θ ) − δ ) a sin ( θ ) = ⎟ ⎝ 2⎠ V = M + Mrod g⎜ 1115 © 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 Guess Chapter 11 δ = 100 mm ⎛ a ⎞ cos ( θ ) − k( a cos ( θ ) − δ ) a sin ( θ ) = ⎟ ⎝ 2⎠ Given M + Mrod g⎜ δ = Find ( δ ) F = k( a cos ( θ ) − δ ) δ = −0.622 m F = 211.0 N Problem 11-49 Determine the force P acting on the cord which is required to maintain equilibrium of the horizontal bar CB of mass M Hint: First show that the coordinates sA and sB are related to the constant vertical length l of the cord by the equation 5sB − sA = L Given: M = 20 kg Solution: L = 4sB + ( sB − sA) L = 5sB − sA Δ L = 5Δ sB − Δ sA = Δ sA = 5ΔsB V = −M g sB + P sA Δ V = −M gΔ sB + P ΔsA = ( −M g + 5P) Δ sB = P = Mg P = 39.2 N 1116 © 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 11-50 The uniform bar AB has weight W If the attached spring is unstretched when θ = 90 deg, use the method of virtual work and determine the angle θ for equilibrium Note that the spring always remains in the vertical position due to the roller guide Given: W = 10 lb k = lb ft a = ft Solution: y = a sin ( θ ) δy = a cos ( θ ) δθ δU = ( −W + Fs) δ y = ⎡k( a − a sin ( θ ) ) − W⎤ a cos ( θ ) δθ = ⎣ ⎦ cos ( θ ) = sin ( θ ) = − θ = acos ( 0) W ka θ = asin ⎛ − ⎜ ⎝ θ = 90 deg W⎞ ⎟ ka⎠ θ = 30 deg Problem 11-51 The uniform bar AB has weight W If the attached spring is unstretched when θ = 90 deg, use the principle of potential energy and determine the angle θ for equilibrium Investigate the stability of the equilibrium positions Note that the spring always remains in the vertical position due to the roller guide Given: W = 10 lb k = lb ft a = ft 1117 © 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: V = W a sin ( θ ) + 1 2 k ( a − a sin ( θ ) ) = W a sin ( θ ) + k a ( − sin ( θ ) ) 2 Equilibrium d V = W a cos ( θ ) − k a ( − sin ( θ ) ) cos ( θ ) = dθ cos ( θ ) = sin ( θ ) = − θ = acos ( 0) Check Stability dθ ⎝ W⎞ ⎟ ka⎠ θ = 30 deg If V'' > the equilibrium point is stable If V'' < 0, then unstable = −W a sin ( θ ) + k a sin ( θ ) + k a cos ( 2θ ) d V V'' = θ = asin ⎛ − ⎜ W ka θ = 90 deg 2 V'' = −W a sin ( θ ) + k a sin ( θ ) + k a cos ( 2θ ) V'' = −40 lb⋅ ft V'' = −W a sin ( θ ) + k a sin ( θ ) + k a cos ( 2θ ) V'' = 60 lb⋅ ft 2 2 Problem 11-52 The punch press consists of the ram R, connecting rod AB, and a flywheel If a torque M is applied to the flywheel, determine the force F applied at the ram to hold the rod in the position θ = θ0 Given: M = 50 N⋅ m θ = 60 deg r = 0.1 m a = 0.4 m Solution: θ = θ0 Free Body Diagram: The system has only one degree of freedom defined by the independent coordinate θ When θ undergoes a positive displacement δθ, only force F and Moment M work a = x + r − 2x r cos ( θ ) 2 ⎛ x r sin ( θ ) ⎞ δθ ⎟ ⎝ r cos ( θ ) − x ⎠ = 2xδx − 2r cos ( θ ) δx + 2x r sin ( θ ) δθ δx = ⎜ ⎡ ⎛ x r sin ( θ ) ⎞ − M⎤ δθ = ⎟ ⎥ ⎣ ⎝ r cos ( θ ) − x ⎠ ⎦ δU = −F δx − Mδθ = ⎢−F ⎜ 1118 © 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 Guesses Given Chapter 11 F = 1N x = 0.1 m a = x + r − 2x r cos ( θ ) ⎛x⎞ ⎜ ⎟ = Find ( x , F) ⎝F⎠ 2 x = 0.441 m ⎛ x r sin ( θ ) ⎞ − M = ⎟ ⎝ r cos ( θ ) − x ⎠ −F ⎜ F = 512 N 1119 © 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 ( b) S2 = 8.368 kN ( c) Chapter S3 = 893 g Problem 1-6 Represent each of the following to three significant figures and express each answer in SI units using an appropriate. .. kg⋅ ? ?s −3 −6 10 = N = kg⋅ s kN kg⋅ s kN kg⋅ s Problem 1-5 Represent each of the following with SI units having an appropriate prefix: (a) S1 , (b) S2 , (c) S3 Units Used: kg = 100 0 g −3 ms = 10 s. .. figures and express each answer in SI units using an appropriate prefix: (a) a 1/b1, (b) a2b2/c2, (c) a3b3 Units Used: μm = 10 −6 m Mm = 10 m Mg = 10 gm kg = 10 gm −3 ms = 10 s Given: a1 = 684