MEN 201 Problems For Chapter 17 with Solutions, Spring 2012

Engineering Mechanics - Dynamics Chapter 17 Problem 17-1 The right circular cone is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the cone. The cone has a constant density ρ. Solution: m 0 h x ρ π rx h ⎛⎜⎝ ⎞⎟⎠ 2 ⌠ ⎮ ⎮ ⌡ d = 1 3 hρ πr2 = ρ 3m hπr2 = Ix 3 = 10mr2 Ix 3m hπr2 0 h x 1 2π rx h ⎛⎜⎝ ⎞⎟⎠ 2 rx h ⎛⎜⎝ ⎞⎟⎠ 2 ⌠ ⎮ ⎮ ⌡ d = Problem 17-2 Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m. Solution: m 0 2π θ ρR ⌠ ⎮⌡ d = = 2π ρR ρ m 2πR = Iz m 2πR 0 2π θ R R2 ⌠ ⎮⌡ d = = mR2 Iz mR2 = Problem 17-3 The solid is formed by revolving the shaded area around the y axis. Determine the radius of gyration ky. The specific weight of the material is γ. Given: γ 380 lb ft3 = a = 3 in b 3 in = 504 Engineering Mechanics - Dynamics Chapter 17 Solution: m 0 b y γ π a y3 b3 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d = m 0.083 slug = Iy 0 b y γ 1 2π a y3 b3 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ 2 a y3 b3 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d = Iy 0.201 slug in2 ⋅ = k Iy m = k 1.56 in = *Problem 17-4 Determine the moment of inertia Ix of the sphere and express the result in terms of the total mass m of the sphere.The sphere has a constant density ρ. Solution: m r − r x ρ π r2 − x2 ( ) ⌠ ⎮⌡ d = 4 = 3 r3 ρ π ρ 3m 4πr3 = Ix 3m 4πr3 r − r x π 2 r2 − x2 ( ) 2 ⌠ ⎮ ⎮⌡ d = Ix 2 5mr2 = Problem 17-5 Determine the radius of gyration kx of the paraboloid. The density of the material is ρ. Units Used: Mg 106 gm = Given: h 200 mm = r 100 mm = ρ 5 Mg m3 = 505 Engineering Mechanics - Dynamics Chapter 17 Solution: M 0 h x ρ π xr2 h ⎛⎜ ⎝ ⎞⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d = M 15.708 kg = Ix 0 h x 1 2ρ π xr2 h ⎛⎜ ⎝ ⎞⎟ ⎠ 2 ⌠⎮ ⎮ ⎮⌡ d = Ix 0.052 kg m2 ⋅ = kx Ix M = kx 57.7 mm = Problem 17-6 Determine the moment of inertia of the semiellipsoid with respect to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density ρ. Solution: m 0 a x ρ πb2 1 x2 a2 ⎛⎜ − ⎜⎝ ⎞⎟ ⎟⎠ ⌠⎮ ⎮ ⎮⌡ d = 2 = 3 aρ πb2 ρ 3m 2aπb2 = Ix 3m 2aπb2 0 a x 1 2π b2 1 x2 a2 ⎛⎜ − ⎜⎝ ⎞⎟ ⎟⎠ ⎡⎢ ⎢⎣ ⎤⎥ ⎥⎦ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d = Ix 2 5mb2 = Problem 17-7 Determine the radius of gyration kx of the body. The specific weight of the material is γ. Given: γ 380 lb ft3 = h 8 in = r 2 in = 506 Engineering Mechanics - Dynamics Chapter 17 Solution: M 0 h x γ π r x h ⎛⎜⎝ ⎞⎟⎠ 1 3 ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⎮ ⎮⌡ d = M 0.412 slug = Ix 0 h x 1 2γ π r x h ⎛⎜⎝ ⎞⎟⎠ 1 3 ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 4 ⌠ ⎮ ⎮ ⎮ ⎮⌡ d = Ix 0.589 slug in2 ⋅ = kx Ix M = kx 1.20 in = *Problem 17-8 Determine the moment of inertia of the ellipsoid with respect to the x axis and express the result in terms of the mass m of the ellipsoid. The material has a constant density ρ. Solution: m a − a x ρ πb2 1 x2 a2 ⎛⎜ − ⎜⎝ ⎞⎟ ⎟⎠ ⌠⎮ ⎮ ⎮⌡ d = 4 3 aρ πb2 = ρ 3m 4aπb2 = Ix 3m 4aπb2 a − a x 1 2π b2 1 x2 a2 ⎛⎜ − ⎜⎝ ⎞⎟ ⎟⎠ ⎡⎢ ⎢⎣ ⎤⎥ ⎥⎦ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d = Ix 2 5mb2 = Problem 17-9 Determine the moment of inertia of the homogeneous pyramid of mass m with respect to the z axis.The density of the material is ρ. Suggestion: Use a rectangular plate element having a volume of dV = (2x)(2y) dz. 507 Engineering Mechanics - Dynamics Chapter 17 Solution: x a h − z ( ) 2h = m 0 h z ρ a h − z ( ) h ⎡⎢⎣ ⎤⎥⎦ 2 ⌠ ⎮ ⎮ ⌡ d = 1 3 hρa2 = ρ 3m ha2 = Iz 3m ha2 0 h z 1 6 a h − z ( ) h ⎡⎢⎣ ⎤⎥⎦ 4 ⌠ ⎮ ⎮ ⌡ d = Iz 1 10ma2 = Problem 17-10 The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of inertia Iy. The specific weight of concrete is γ. Given: a 6 in = b 4 in = h 8 in = γ 150 lb ft3 = Solution: Iy 0 h y γ 1 2π a + b ( )4 1 2π a2 y h ⎛⎜⎝ ⎞⎟⎠ 2 − ⎡⎢ ⎣ ⎤⎥ ⎦ ⌠ ⎮ ⎮ ⌡ d = Iy 2.25 slug ft2 ⋅ = Problem 17-11 Determine the moment of inertia of the thin plate about an axis perpendicular to the page and passing through the pin at O. The plate has a hole in its center. Its thickness is t, and the material has a density of ρ. 508 Engineering Mechanics - Dynamics Chapter 17 Given: a 1.40 m = r 150 mm = t 50 mm = ρ 50 kg m3 = Solution: IO ρa2t 2a2 3 ⎛⎜ ⎝ ⎞⎟ ⎠ ρ πr2t r2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ ρ πr2t 2a 2 ⎛⎜⎝ ⎞⎟⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − = IO 6.227 kg m2 ⋅ = *Problem 17-12 Determine the moment of inertia Iz of the frustum of the cone which has a conical depression. The material has a density ρ. Given: r1 0.2 m = r2 0.4 m = h1 0.6 m = h2 0.8 m = ρ 200 kg m3 = Solution: h3 r2 h2 r2 − r1 = h4 h3 − h2 = Iz ρ πr2 2h3 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 10 ⎛⎜⎝ ⎞⎟⎠ r2 2 ρ πr1 2h4 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 10 ⎛⎜⎝ ⎞⎟⎠ r1 2 − ρ πr2 2h1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 10 ⎛⎜⎝ ⎞⎟⎠ r2 2 − = Iz 1.53 kg m2 ⋅ = 509 Engineering Mechanics - Dynamics Chapter 17 Problem 17-13 Determine the moment of inertia of the assembly about an axis which is perpendicular to the page and passes through the center of mass G. The material has a specific weight γ. Given: a = 0.5 ft r1 1 ft = b = 0.25 ft r2 2 ft = c = 1 ft γ 90 lb ft3 = Solution: IG 1 2γ πc r2 + a ( )4 1 2γ π c − b ( ) r2 4 − 1 2γ πbr1 4 − = IG 118slug ft2 ⋅ = Problem 17-14 Determine the moment of inertia of the assembly about an axis which is perpendicular to the page and passes through point O. The material has a specific weight γ. Given: a 0.5 ft = b 0.25 ft = c 1 ft = r1 1 ft = r2 2 ft = γ 90 lb ft3 = Solution: IO 3 2γ πc r2 + a ( )4 1 2γ π c − b ( ) r2 4 γ π c − b ( ) r2 2 r2 + a ( )2 + ⎡⎢⎣ ⎤⎥⎦ − 1 2γ πbr1 4 γ πbr1 2 r2 + a ( )2 + ⎡⎢⎣ ⎤⎥⎦ − + ... = IO 283slug ft2 ⋅ = 510 Engineering Mechanics - Dynamics Chapter 17 Problem 17-15 The wheel consists of a thin ring having a mass M1 and four spokes made from slender rods, each having a mass M2. Determine the wheel’s moment of inertia about an axis perpendicular to the page and passing through point A. Given: M1 10 kg = M2 2 kg = r 500 mm = Solution: IG M1 r2 4M2 r2 3 ⎛⎜ ⎝ ⎞⎟ ⎠ + = IA IG M1 + 4M2 ( )r2 + = IA 7.67 kg m2 ⋅ = Problem 17-16 The slender rods have a weight density γ. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point A. Given: γ 3 lb ft = a 2 ft = b 1 ft = c 1.5 ft = Solution: IA γ a + b ( ) a + b ( )2 3 ⎡⎢ ⎣ ⎤⎥ ⎦ γ 2c ( ) (2c )2 12 + γ 2c ( ) a2 + = IA 2.17 slug ft2 ⋅ = Problem 17-17 Each of the three rods has a mass m. Determine the moment of inertia of the assembly about an axis which is perpendicular to the page and passes through the center point O. 511 Engineering Mechanics - Dynamics Chapter 17 Solution: I0 3 1 12ma2 m a ( )sin 60 deg ( ) 3 ⎡⎢⎣ ⎤⎥⎦ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ = I0 1 2ma2 = Problem 17-18 The slender rods have weight density γ. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through the pin at A. Given: γ 3 lb ft = a 2 ft = b 1.5 ft = Solution: IA 1 3γa3 1 12γ 2b ( )3 + γ 2b ( ) a2 + = IA 1.58 slug ft2 ⋅ = Problem 17-19 The pendulum consists of a plate having weight Wp and a slender rod having weight Wr. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O. Given: Wp 12 lb = Wr 4 lb = a 1 ft = b 3 ft = c 2 ft = 512 Engineering Mechanics - Dynamics Chapter 17 Solution: IO 1 3 b b c + ⎛⎜⎝ ⎞⎟⎠ Wrb2 1 3 c b c + ⎛⎜⎝ ⎞⎟⎠ Wrc2 + 1 + 6Wp a2 Wp b a ⎛⎜⎝ + 2 ⎞⎟⎠ 2 + = IO 4.921 slug ft2 ⋅ = kO IO Wr + Wp = kO 3.146 ft = *Problem 17-20 Determine the moment of inertia of the overhung crank about the x axis. The material is steel having a density ρ. Units Used: Mg 103 kg = Given: ρ 7.85 Mg m3 = c 180 mm = a = 20 mm d 30 mm = b = 50 mm e 20 mm = Solution: Ix 2 ρ π 2 e 2 ⎛⎜⎝ ⎞⎟⎠ 2 b e 2 ⎛⎜⎝ ⎞⎟⎠ 2 ρ π e 2 ⎛⎜⎝ ⎞⎟⎠ 2 b c − 2d 2 ⎛⎜⎝ ⎞⎟⎠ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ ρadc 12 d2 + c2 ( ) + = Ix 3.25 10 3 − × kg m2 ⋅ = Problem 17-21 Determine the moment of inertia of the overhung crank about the x' axis. The material is steel having a density ρ. Units Used: Mg 103 kg = 513 Engineering Mechanics - Dynamics Chapter 17 Given: ρ 7.85 Mg m3 = a 20 mm = b 50 mm = c 180 mm = d 30 mm = e 20 mm = Solution: Ix 2 ρ π 2 e 2 ⎛⎜⎝ ⎞⎟⎠ 2 b e 2 ⎛⎜⎝ ⎞⎟⎠ 2 ρ π e 2 ⎛⎜⎝ ⎞⎟⎠ 2 b c − 2d 2 ⎛⎜⎝ ⎞⎟⎠ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ ρadc 12 d2 + c2 ( ) + = Ix' Ix 2ρ π e 2 ⎛⎜⎝ ⎞⎟⎠ 2 b + ρadc ⎡⎢ ⎣ ⎤⎥ ⎦ c − 2d 2 ⎛⎜⎝ ⎞⎟⎠ 2 + = Ix' 7.19 10 3 − × kg m2 ⋅ = Problem 17-22 Determine the moment of inertia of the solid steel assembly about the x axis. Steel has specific weight γst. Given: a 0.25 ft = b 2 ft = c 3 ft = γst 490 lb ft3 = Solution: Ix 1 2π 2a ( )2c 2a ( )2 3 10 1 3π 2a ( )2 2b ( ) 2a ( )2 + 3 10 1 3πa2ba2 − ⎡⎢⎣ ⎤⎥⎦ γst = Ix 5.644 slug ft2 ⋅ = 514 Engineering Mechanics - Dynamics Chapter 17 Problem 17-23 The pendulum consists of two slender rods AB and OC which have a mass density ρ1. The thin plate has a mass density ρ2. Determine the location y' of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G. Given: ρ1 3 kg m = a = 0.4 m c 0.1 m = ρ2 12 kg m2 = b = 1.5 m r 0.3 m = Solution: y' ρ1 b b 2 ⎛⎜⎝ ⎞⎟⎠ ρ2π r2 − c2 ( ) b + r ( ) + ρ1 b + 2a ( ) ρ2π r2 − c2 ( ) + = y' 0.888 m = IO 1 12ρ1 2a ( )3 1 + 3ρ1 b3 ρ2 2 ⎛⎜⎝ ⎞⎟⎠ πr4 + ρ2πr2 r ( + b )2 + ρ2 2 ⎛⎜⎝ ⎞⎟⎠ πc4 ρ2πc2 r ( + b )2 + ⎡⎢⎣ ⎤⎥⎦ − = IG IO ρ1 2a + b ( ) ρ2π r2 − c2 ( ) + ⎡⎣ ⎤⎦y'2 − = IG 5.61 kg m2 ⋅ = *Problem 17-24 Determine the greatest possible acceleration of the race car of mass M so that its front tires do not leave the ground or the tires slip on the track. The coefficients of static and kinetic friction are μs and μk respectively. Neglect the mass of the tires. The car has rear-wheel drive and the front tires are free to roll. Given: M = 975 kg μs 0.8 = a = 1.82 m μk 0.6 = b = 2.20 m g 9.81 m s2 = h 0.55 m = 515 Engineering Mechanics - Dynamics Chapter 17 Solution: First assume that the rear wheels are on the verge of slipping FB μsNB = Guesses NA = 1 N NB = 1 N aG 1 m s2 = Given μsNB = M aG NA + NB − M g 0 = −NA a NB b − a ( ) + − μsNBh 0 = NA NB aG1 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , aG ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 326 − ⎛⎜⎝9891 ⎞⎟⎠ N = aG1 8.12 m s2 = Next assume that the front wheels lose contact with the ground NA 0 = Guesses NB = 1 N FB = 1 N aG 1 m s2 = Given FB = M aG NB − M g = 0 NB b − a ( ) − FBh 0 = NB FB aG2 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NB FB , , aG ( ) = NB FB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 9565 ⎛⎜⎝6608 ⎞⎟⎠ N = aG2 6.78 m s2 = Choose the critical case aG min aG1 aG2 , ( ) = aG 6.78 m s2 = Problem 17-25 Determine the greatest possible acceleration of the race car of mass M so that its front tires do not leave the ground nor the tires slip on the track. The coefficients of static and kinetic friction are μs and μk respectively. Neglect the mass of the tires. The car has four-wheel drive. Given: M = 975 kg μs 0.8 = a = 1.82 m μk 0.6 = b = 2.20 m g 9.81 m s2 = h 0.55 m = 516 Engineering Mechanics - Dynamics Chapter 17 Solution: First assume that all wheels are on the verge of slipping FA μsNA = FB μsNB = Guesses NA = 1 N NB = 1 N aG 1 m s2 = Given μsNB + μsNA = M aG NA + NB − M g 0 = −NA a NB b − a ( ) + − μsNBh − μsNAh 0 = NA NB aG1 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , aG ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 261 − ⎛⎜⎝9826 ⎞⎟⎠ N = aG1 7.85 m s2 = Next assume that the front wheels lose contact with the ground NA 0 = Guesses NB = 1 N FB = 1 N aG 1 m s2 = Given FB = M aG NB − M g = 0 NB b − a ( ) − FBh 0 = NB FB aG2 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NB FB , , aG ( ) = NB FB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 9565 ⎛⎜⎝6608 ⎞⎟⎠ N = aG2 6.78 m s2 = Choose the critical case aG min aG1 aG2 , ( ) = aG 6.78 m s2 = Problem 17-26 The bottle of weight W rests on the check-out conveyor at a grocery store. If the coefficient of static friction is μs, determine the largest acceleration the conveyor can have without causing the bottle to slip or tip. The center of gravity is at G. Given: W 2 lb = μs 0.2 = 517 Engineering Mechanics - Dynamics Chapter 17 b 8 in = c 1.5 in = g 32.2 ft s2 = Solution: Assume that bottle tips before slipping x c = Guesses aG 1 ft s2 = FB = 1 lb NB = 1 lb Fmax 1 lb = Given FB W g ⎛⎜⎝ ⎞⎟⎠ aG = NB − W 0 = FBb − NBx = 0 Fmax μsNB = aGt FB NB Fmax ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find aG FB , , NB , Fmax ( ) = FB NB Fmax ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 0.375 2 0.4 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb = aGt 6.037 ft s2 = If FB = 0.375 lb < Fmax = 0.4 lb then we have the correct answer. If FB = 0.375 lb > Fmax = 0.4 lb then we know that slipping occurs first. If this is the case, FB μsNB = Given FB W g ⎛⎜⎝ ⎞⎟⎠ aG = NB − W = 0 FBb − NBx 0 = aGs NB x ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find aG NB , , x ( ) = NB = 2lb x = 1.6 in aGs 6.44 ft s2 = As a check, we should have x = 1.6 in < c = 1.5 in if slipping occurs first In either case, the answer is aG min aGs aGt , ( ) = aG 6.037 ft s2 = Problem 17-27 The assembly has mass ma and is hoisted using the boom and pulley system. If the winch at B 518 Engineering Mechanics - Dynamics Chapter 17 draws in the cable with acceleration a, determine the compressive force in the hydraulic cylinder needed to support the boom. The boom has mass mb and mass center at G. Units Used: Mg 103 kg = kN 103 N = Given: ma 8 Mg = mb 2 Mg = a 2 m s2 = b 4 m = c 2 m = d 6 m = e 1 m = f 2 m = θ 60 deg = g 9.81 m s2 = Solution: 2T − ma g ma a 2 = T 1 2 ma g ma a 2 + ⎛⎜⎝ ⎞⎟⎠ = T 43.24 kN = 2 − T b + c + d ( ) mb g b + c ( ) − + FCDb ⎡⎣ ⎤⎦cos θ( ) 0 = FCD 2T b + c + d ( ) mb g b + c ( ) + b = FCD 289kN = *Problem 17-28 The jet aircraft has total mass M and a center of mass at G. Initially at take-off the engines provide thrusts 2T and T'. Determine the acceleration of the plane and the normal reactions on the nose wheel and each of the two wing wheels located at B. Neglect the mass of the wheels and, due to low velocity, neglect any lift caused by the wings. Units Used: Mg 103 kg = 519 Engineering Mechanics - Dynamics Chapter 17 kN 103 N = Given: b = 2.5 m M 22 Mg = c = 2.3 m T 2 kN = d = 3 m T' 1.5 kN = e = 6 m g 9.81 m s2 = f 1.2 m = Solution: Guesses aG 1 m s2 = By = 1 kN Ay 1 kN = Given T' + 2T = M aG 2By + Ay − M g 0 = −T' b − 2T c − M gd Ay d + e ( ) + −M aG f = aG By Ay ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find aG By , , Ay ( ) = aG 0.250 m s2 = Ay By ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 72.6 ⎛⎜⎝71.6 ⎞⎟⎠ kN = Problem 17-29 The lift truck has mass mt and mass center at G. If it lifts the spool of mass ms with acceleration a, determine the reactions of each of the four wheels on the ground. The loading is symmetric. Neglect the mass of the movable arm CD. Given: a 3 m s2 = d 0.4 m = b = 0.75 m e 0.7 m = c = 0.5 m g 9.81 m s2 = mt = 70 kg ms 120 kg = 520 Engineering Mechanics - Dynamics Chapter 17 Guesses NA = 1 N NB 1 N = Given 2 NA + NB ( ) mt + ms ( )g − msa = 2 − NA b + c ( ) + msge + mt gc −ms ae = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find NA NB , ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 568 ⎛⎜⎝544 ⎞⎟⎠ N = Problem 17-30 The lift truck has mass mt and mass center at G. Determine the largest upward acceleration of the spool of mass ms so that no reaction of the wheels on the ground exceeds Fmax. Given: mt = 70 kg b 0.75 m = ms = 120 kg c 0.5 m = Fmax = 600 N d 0.4 m = g 9.81 m s2 = e 0.7 m = Solution: Assume NA Fmax = Guesses a 1 m s2 = NB 1 N = 521 Solution: Engineering Mechanics - Dynamics Chapter 17 2 NA + NB ( ) mt + ms ( )g − msa = 2 − NA b + c ( ) + msge + mt gc −ms ae = a ⎛⎜NB ⎝ ⎞⎟ ⎠ Find a NB ( , ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 600 ⎛⎜⎝569.529 ⎞⎟⎠ N = a 3.96 m s2 = Check: Since NB = 570 N < Fmax = 600 N then our assumption is good. Problem 17-31 The door has weight W and center of gravity at G. Determine how far the door moves in time t starting from rest, if a man pushes on it at C with a horizontal force F. Also, find the vertical reactions at the rollers A and B. Given: W = 200 lb c 5 ft = t = 2 s d 12 ft = F = 30 lb e 6 ft = b = 3 ft g 32.2 ft s2 = Solution: Guesses a 1 ft s2 = NA = 1 lb NB 1 lb = Given F W g ⎛⎜⎝ ⎞⎟⎠ a = NA + NB − W 0 = F c − b ( ) + NBe − NAe 0 = 522 Given Engineering Mechanics - Dynamics Chapter 17 a NA NB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find a NA , , NB ( ) = a 4.83 ft s2 = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 105.0 ⎛⎜⎝ 95.0 ⎞⎟⎠ lb = d 1 = 2at2 d 9.66 ft = *Problem 17-32 The door has weight W and center of gravity at G. Determine the constant force F that must be applied to the door to push it open a distance d to the right in time t, starting from rest. Also, find the vertical reactions at the rollers A and B. Given: W = 200 lb c 5 ft = t = 5 s d 12 ft = d = 12 ft e 6 ft = b = 3 ft g 32.2 ft s2 = Solution: a 2 d t2 ⎛⎜ ⎝ ⎞⎟ ⎠ = a 0.96 ft s2 = Guesses F = 1 lb NA = 1 lb NB 1 lb = Given F W g ⎛⎜⎝ ⎞⎟⎠ a = NA + NB − W 0 = F c − b ( ) + NBe − NAe 0 = F NA NB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find F NA , , NB ( ) = F NA NB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 5.96 100.99 99.01 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb = Problem 17-33 The fork lift has a boom with mass M1 and a mass center at G. If the vertical acceleration of the boom is aG, determine the horizontal and vertical reactions at the pin A and on the short link BC when the load M2 is lifted. 523 Engineering Mechanics - Dynamics Chapter 17 Units Used: Mg = 103 kg kN 103 N = Given: M1 = 800 kg a 1 m = b = 2 m M2 1.25 Mg = c 1.5 m = d = 1.25 m aG 4 m s2 = Solution: Guesses Ax = 1 N FCB 1 N = Ay 1 N = Given −FCB + Ax 0 = Ay M1 + M2 ( )g − M1 + M2 ( )aG = FCB c − M1 ga M2 g a + b ( ) − M1 aGa M2 aG a + b ( ) + = Ax Ay FCB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find Ax Ay , , FCB ( ) = Ax Ay FCB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 41.9 28.3 41.9 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN = Problem 17-34 The pipe has mass M and is being towed behind the truck. If the acceleration of the truck is at, determine the angle θ and the tension in the cable. The coefficient of kinetic friction between the pipe 524 Engineering Mechanics - Dynamics Chapter 17 and the ground is μk. Units Used: kN 103 N = Given: M = 800 kg r 0.4 m = at 0.5 m s2 = φ 45 deg = μk = 0.1 g 9.81 m s2 = Solution: Guesses θ = 10 deg NC = 1 N T 1 N = Given T cos φ( ) − μk NC M at = T sin φ( ) − M g + NC 0 = T sin φ ( − θ )r − μk NC r 0 = θ NC T ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find θ NC , , T ( ) = NC = 6.771 kN T = 1.523 kN θ 18.608 deg = Problem 17-35 The pipe has mass M and is being towed behind a truck. Determine the acceleration of the truck and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is μk. 525 Engineering Mechanics - Dynamics Chapter 17 Units Used: kN 103 N = Given: M = 800 kg r 0.4 m = θ = 30 deg φ 45 deg = μk = 0.1 g 9.81 m s2 = Solution: Guesses at 1 m s2 = NC 1 N = T 1 N = Given T cos φ( ) − μk NC M at = T sin φ( ) − M g + NC 0 = T sin φ ( − θ )r − μk NC r 0 = at NC T ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find at NC , , T ( ) = NC = 6.164 kN T = 2.382 kN at 1.335 m s2 = *Problem 17-36 The pipe has a mass M and is held in place on the truck bed using the two boards A and B. Determine the acceleration of the truck so that the pipe begins to lose contact at A and the bed of the truck and starts to pivot about B. Assume board B will not slip on the bed of the truck, and the pipe is smooth. Also, what force does board B exert on the pipe during the acceleration? 526 Engineering Mechanics - Dynamics Chapter 17 Units Used: kN 103 N = Given: M 460 kg = a 0.5 m = b 0.3 m = c 0.4 m = d 0.1 m = g 9.81 m s2 = Solution: θ asin b a ⎛⎜⎝ ⎞⎟⎠ = Guesses NBx = 1 N NBy = 1 N at 1 m s2 = Given NBx = M at NBy − M g = 0 NBx a ( ) cos θ( ) − NBy b 0 = NBx NBy at ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NBx NBy , , at ( ) = NBx NBy ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3.384 ⎛⎜⎝4.513 ⎞⎟⎠ kN = at 7.36 m s2 = NBx NBy ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 5.64 kN = Problem 17-37 The drop gate at the end of the trailer has mass M and mass center at G. If it is supported by the cable AB and hinge at C, determine the tension in the cable when the truck begins to accelerate at rate a. Also, what are the horizontal and vertical components of reaction at the hinge C? Given: kN 103 N = M 1.25 × 103 kg = a 5 m s2 = θ 30 deg = 527 Engineering Mechanics - Dynamics Chapter 17 b = 1.5 m φ 45 deg = c = 1 m g 9.81 m s2 = Solution: Guesses T = 1 N Cx = 1 N Cy 1 N = Given T − cos φ ( − θ ) + Cx −M a = T − sin φ ( − θ ) − M g + Cy 0 = T sin θ( ) b + c ( ) − M gbcos φ( ) M absin φ( ) = T Cx Cy ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find T Cx , , Cy ( ) = T Cx Cy ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 15.708 8.923 16.328 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN = Problem 17-38 The sports car has mass M and a center of mass at G. Determine the shortest time it takes for it to reach speed v, starting from rest, if the engine only drives the rear wheels, whereas the front wheels are free rolling. The coefficient of static friction between the wheels and the road is μs. Neglect the mass of the wheels for the calculation. If driving power could be supplied to all four wheels, what would be the shortest time for the car to reach a speed of v? Given: M 1.5 × 103 kg = v 80 km hr = μs 0.2 = b 1.25 m = c 0.75 m = d 0.35 m = g 9.81 m s2 = 528 Engineering Mechanics - Dynamics Chapter 17 (a) Rear wheel drive only Guesses NA = 1 N NB = 1 N aG 1 m s2 = Given NA + NB − M g = 0 μsNB M aG = M gc NA b + c ( ) − M aGd = NA NB aG ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , aG ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 5.185 103 × 9.53 103 × ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ N = aG 1.271 m s2 = trw v aG = trw 17.488 s = (b) Four wheel drive Guesses NA = 1 N NB = 1 N aG 1 m s2 = Given NA + NB − M g = 0 μsNB + μsNA M aG = M gc NA b + c ( ) − M aGd = NA NB aG ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , aG ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 5.003 103 × 9.712 103 × ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ N = aG 1.962 m s2 = trw v aG = trw 11.326 s = Problem 17-39 The crate of mass m is supported on a cart of negligible mass. Determine the maximum force P that can be applied a distance d from the cart bottom without causing the crate to tip on the cart. 529 Solution: Engineering Mechanics - Dynamics Chapter 17 Solution: Require Nc to act at corner B for tipping. P d mg b 2 ⎛⎜⎝ ⎞⎟⎠ − maG h 2 ⎛⎜⎝ ⎞⎟⎠ = P maG = P d mg b 2 ⎛⎜⎝ ⎞⎟⎠ − P h 2 ⎛⎜⎝ ⎞⎟⎠ = Pmax mgb 2 d h ⎛⎜⎝ − 2 ⎞⎟⎠ = *Problem 17-40 The car accelerates uniformly from rest to speed v in time t. If it has weight W and a center of gravity at G, determine the normal reaction of each wheel on the pavement during the motion. Power is developed at the front wheels, whereas the rear wheels are free to roll. Neglect the mass of the wheels and take the coefficients of static and kinetic friction to be μs and μk respectively. Given: a 2.5 ft = v 88 ft s = μk 0.2 = t = 15 s b 4 ft = W = 3800 lb c 3 ft = μs = 0.4 g 32.2 ft s2 = Solution: Assume no slipping aG v = t aG 5.867 ft s2 = Guesses NB = 1 lb NA = 1 lb FA 1 lb = Given 2NB + 2NA − W 0 = 2FA W g ⎛⎜⎝ ⎞⎟⎠ aG = 2 − NB b + c ( ) + W c W − g ⎛⎜⎝ ⎞⎟⎠ aGa = 530 Engineering Mechanics - Dynamics Chapter 17 NA NB FA ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , FA ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 962 ⎛⎜⎝938 ⎞⎟⎠ lb = FA 346lb = Fmax = μsNA Fmax 385lb = Check: Our no-slip assumption is true if FA = 346lb < Fmax = 385lb Problem 17-41 Block A has weight W1 and the platform has weight W2. Determine the normal force exerted by block A on B. Neglect the weight of the pulleys and bars of the triangular frame. Given: W1 50 lb = W2 10 lb = P 100 lb = Solution: P W1 + W2 ( ) − W1 W2 + g ⎛⎜⎝ ⎞⎟⎠ aG = aG g P − 1 W1 1 W2 − W1 W2 + ⎛⎜ ⎝ ⎞⎟ ⎠ = aG 21.47 ft s2 = Guesses R = 25 lb T 22 lb = Given 2T + R − W1 W1 g ⎛⎜⎝ ⎞⎟⎠ aG = 2T − R − W2 W2 g ⎛⎜⎝ ⎞⎟⎠ aG = R ⎛⎜⎝T ⎞⎟⎠ Find R T , ( ) = R = 33.3 lb T 25lb = Problem 17-42 The car of mass M shown has been “raked” by increasing the height of its center of mass to h This was done by raising the springs on the rear axle. If the coefficient of kinetic friction 531 Engineering Mechanics - Dynamics Chapter 17 between the rear wheels and the ground is μk , show that the car can accelerate slightly faster than its counterpart for which h = 0. Neglect the mass of the wheels and driver and assume the front wheels at B are free to roll while the rear wheels slip. Units Used: Mg = 103 kg kN 103 N = Given: M = 1.6 Mg a 1.6 m = μk = 0.3 b 1.3 m = h = 0.2 m g 9.81 m s2 = h1 0.4 m = Solution: In the raised position Guesses aG 1 m s2 = NA = 4 N NB 5 N = Given μk NA M aG = NA + NB − M g 0 = −M ga NB a + b ( ) + −M aG h ( + h1 ) = aGr NA NB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find aG NA , , NB ( ) = aGr 1.41 m s2 = In the lower (regular) position Given μk NA = M aG NA + NB − M g = 0 −M ga NB a + b ( ) + −M aGh1 = aGl NA NB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find aG NA , , NB ( ) = aGl 1.38 m s2 = Thus the advantage in the raised position is aGr − aGl 0.03 m s2 = 532 Engineering Mechanics - Dynamics Chapter 17 Problem 17-43 The forklift and operator have combined weight W and center of mass at G. If the forklift is used to lift the concrete pipe of weight Wp determine the maximum vertical acceleration it can give to the pipe so that it does not tip forward on its front wheels. Given: W = 10000 lb b = 5 ft d 6 ft = Wp = 2000 lb c = 4 ft g 32.2 ft s2 = Solution: It is required that NB = 0 Wp b − W c Wp − g ab = a W c Wp b − Wp ⎛⎜ ⎝ ⎞⎟ ⎠ g b = a 96.6 ft s2 = *Problem 17-44 The forklift and operator have combined weight W and center of mass at G. If the forklift is used to lift the concrete pipe of weight Wp determine the normal reactions on each of its four wheels if the pipe is given upward acceleration a. 533 Engineering Mechanics - Dynamics Chapter 17 Units Used: kip 103 lb = Given: Wp 2000 lb = W 10000 lb = a 4 ft s2 = b 5 ft = c 4 ft = d 6 ft = g 9.81 m s2 = Solution: Guesses NA 1 lb = NB 1 lb = Given 2NA + 2NB − W − Wp Wp g ⎛⎜⎝ ⎞⎟⎠ a = 2NAb W b + c ( ) − 2NB b + c + d ( ) + 0 = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find NA NB , ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4.686 ⎛⎜⎝1.438 ⎞⎟⎠ kip = 534 Engineering Mechanics - Dynamics Chapter 17 Problem 17-45 The van has weight Wv and center of gravity at Gv. It carries fixed load Wl which has center of gravity at Gl. If the van is traveling at speed v, determine the distance it skids before stopping. The brakes cause all the wheels to lock or skid. The coefficient of kinetic friction between the wheels and the pavement isμk.. Compare this distance with that of the van being empty. Neglect the mass of the wheels. Given: Wv = 4500 lb b 2 ft = Wl = 800 lb c 3 ft = v 40 ft s = d 4 ft = μk = 0.3 e 6 ft = f = 2 ft g 32.2 ft s2 = Solution: Loaded Guesses NA = 1 lb NB = 1 lb a 1 ft s2 = Given NA + NB − Wv − Wl 0 = μk NA + NB ( ) Wv Wl + g ⎛⎜⎝ ⎞⎟⎠ a = −NB f + b + c ( ) Wl b + c ( ) + + Wv c Wl g ⎛⎜⎝ ⎞⎟⎠ ae Wv g ⎛⎜⎝ ⎞⎟⎠ ad + = NA NB a ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , a ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3777 ⎛⎜⎝1523 ⎞⎟⎠ lb = a 9.66 ft s2 = dl v2 = 2a dl 82.816 ft = Unloaded Wl 0 lb = Guesses NA = 1 lb NB = 1 lb a 1 ft s2 = Given NA + NB − Wv − Wl 0 = 535 Engineering Mechanics - Dynamics Chapter 17 μk NA + NB ( ) Wv Wl + g ⎛⎜⎝ ⎞⎟⎠ a = −NB f + b + c ( ) Wl b + c ( ) + + Wv c Wl g ⎛⎜⎝ ⎞⎟⎠ ae Wv g ⎛⎜⎝ ⎞⎟⎠ ad + = NA NB a ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , a ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3343 ⎛⎜⎝1157 ⎞⎟⎠ lb = a 9.66 ft s2 = dul v2 2a = dul 82.816 ft = The distance is the same in both cases although the forces on the tires are different. Problem 17-46 The “muscle car” is designed to do a “wheeley”, i.e., to be able to lift its front wheels off the ground in the manner shown when it accelerates. If the car of mass M1 has a center of mass at G, determine the minimum torque that must be developed at both rear wheels in order to do this. Also, what is the smallest necessary coefficient of static friction assuming the thick-walled rear wheels do not slip on the pavement? Neglect the mass of the wheels. Units Used: Mg 103 kg = kN 103 N = Given: M1 1.35 Mg = a 1.10 m = b 1.76 m = c 0.67 m = d 0.31 m = g 9.81 m s2 = Solution: Guesses aG 1 m s2 = FA = 1 N NA = 1 N M = 1 Nm μs 0.1 = 536 Engineering Mechanics - Dynamics Chapter 17 Given FA M1 aG = NA − M1 g 0 = M1 ga M1 aGc = −M + FAd 0 = FA μsNA = aG FA NA M μs ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ Find aG FA , , NA , M , μs ( ) = FA NA ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 21.7 ⎛⎜⎝13.2 ⎞⎟⎠ kN = M 6.74 kN m ⋅ = μs = 1.642 aG 16.11 m s2 = Problem 17-47 The bicycle and rider have a mass M with center of mass located at G. If the coefficient of kinetic friction at the rear tire is μB, determine the normal reactions at the tires A and B, and the deceleration of the rider, when the rear wheel locks for braking. What is the normal reaction at the rear wheel when the bicycle is traveling at constant velocity and the brakes are not applied? Neglect the mass of the wheels. Given: M = 80 kg μB 0.8 = a = 0.55 m b 0.4 m = c 1.2 m = Solution: Deceleration: Guesses aG 1 m s2 = NB = 1 N NA 1 N = Given μBNB = M aG NA + NB − M g 0 = −NB a + b ( ) + M ga M aGc = 537 Engineering Mechanics - Dynamics Chapter 17 aG NB NA ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find aG NB , , NA ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 559 ⎛⎜⎝226 ⎞⎟⎠ N = aG 2.26 m s2 = Equilibrium Given NA + NB − M g 0 = −NB a + b ( ) + M ga 0 = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find NA NB , ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 330 ⎛⎜⎝454 ⎞⎟⎠ N = *Problem 17-48 The bicycle and rider have a mass M with center of mass located at G. Determine the minimum coefficient of kinetic friction between the road and the wheels so that the rear wheel B starts to lift off the ground when the rider applies the brakes to the front wheel. Neglect the mass of the wheels. Given: M 80 kg = a 0.55 m = b 0.4 m = c 1.2 m = Solution: NB 0 = Guesses aG 1 m s2 = μk 0.1 = NA 1 N = Given μk NA M aG = NA − M g 0 = M ga M aGc = 538 Engineering Mechanics - Dynamics Chapter 17 aG μk NA ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find aG μk , , NA ( ) = NA 785 N = aG 4.50 m s2 = μk 0.458 = Problem 17-49 The dresser has a weight W and is pushed along the floor. If the coefficient of static friction at A and B is μs and the coefficient of kinetic friction is μk , determine the smallest horizontal force P needed to cause motion. If this force is increased slightly, determine the acceleration of the dresser. Also, what are the normal reactions at A and B when it begins to move? Given: W 80 lb = μs 0.3 = μk 0.2 = a 1.5 ft = b 2.5 ft = c 4ft = Solution: Impending Motion Guesses P 1 lb = NA = 1 lb NB 1 lb = Given NA + NB − W 0 = μsNA + μsNB − P 0 = P c + W a NA 2a ( ) − 0 = 539 Engineering Mechanics - Dynamics Chapter 17 P NA NB ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find P NA , , NB ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 72 ⎛⎜⎝ 8 ⎞⎟⎠ lb = P 24lb = Motion Guesses aG 1 ft s2 = Given NA + NB − W = 0 μk NA + μk NB − P −W g ⎛⎜⎝ ⎞⎟⎠ aG = P c − b ( ) μk NA + NB ( )b + + NBa − NAa 0 = NA NB aG ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find NA NB , , aG ( ) = NA NB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 65.3 ⎛⎜⎝14.7 ⎞⎟⎠ lb = aG 3.22 ft s2 = Problem 17-50 The dresser has a weight W and is pushed along the floor. If the coefficient of static friction at A and B is μs and the coefficient of kinetic friction is μk , determine the maximum horizontal force P that can be applied without causing the dresser to tip over. Given: W 80 lb = μs 0.3 = μk 0.2 = a 1.5 ft = b 2.5 ft = c 4 ft = Solution: Dresser slides before tipping occurs Guesses NA 1 lb =