+ ] B A BA / t BA / n. B G BG / t BG / n. a = (5)(4) = 80 in./s. A G AG / t AG / n. ] + [48 in./s ]

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Download "+ ] B A BA / t BA / n. B G BG / t BG / n. a = (5)(4) = 80 in./s. A G AG / t AG / n. ] + [48 in./s ]"

Darcy Briggs
6 years ago
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PROLEM 15.113 3-in.-radius drum is rigidly attached to a 5-in.-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum.

1 PROLEM in.-radius drum is rigidly attached to a 5-in.-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that at the instant shown end D of the cord has a velocity of 8 in./s and an acceleration of 30 in./s, both directed to the left, determine the accelerations of Points,, and C of the drums. SOLUTION Velocity analysis. v = v = 8 in./s Instantaneous center is at Point. v = ( ) ω, 8 = (5 3) ω D ω = 4 rad/s cceleration analysis. a = [ a ] for no slipping. α = α a = [30 in./s ] [( a ) n a = [ a ] G G + ] a = a + ( a ) + ( a ) / t / n [ a ] = [30 ] + [( a ) n ] + [(5 3)α ] + [5 3)ω ] Components : 0 = 30 + a a = 15 rad/s a = a + ( a ) + ( a ) G G / t G / n [ a ] = [ a G ] + [5α ] + [5ω ] Components : : 0 = a + 5a a = 5a = 75 in./s G G a = (5)(4) = 80 in./s a = a + ( a ) + ( a ) G G / t G / n a = 80.0 in./s = [75 ] + [3α ] + [3ω ] = [75 ] + [45 ] + [48 ] = [30 in./s ] + [48 in./s ] a = 56.6 in./s 58.0

2 PROLEM (Continued) a = a + ( a ) + ( a ) C G CG / t CG / n = [75 ] + [5α ] + [5ω ] = [75 ] + [75 ] + [80 ] = [155 in./s ] + [75 in./s ] a C = 17. in./s 5.8

PROLEM 15.13 The disk shown has a constant angular velocity of 500 rpm counter-clockwise. Knowing that rod D is 10 in. long, determine the acceleration of collar D when (a) θ = 90, (b) θ = 180.

3 PROLEM The disk shown has a constant angular velocity of 500 rpm counter-clockwise. Knowing that rod D is 10 in. long, determine the acceleration of collar D when (a) θ = 90, (b) θ = 180. SOLUTION Disk. ω = 500 rpm α = 0, ( ) = in. = 5.36 rad/s v a = ( ) ω = ()(5.36) = in./s = ( ) ω = ()(5.36) = in./s (a) θ = 90. v = m/s, v D = v D v D and v are parallel. in. sin β = = 0.4 β = in. ω = 0 D a = in./s, a D = ad, α D = αd α D/ = [( D) α D = [10 α D β ] + 0 a = a + a D D β] + [( D) ωd β ] / Resolve into components. : 0 = (10 cos βα ) D a = rad/s D : ad = 0 (10 sin β )( 598.6) + 0 = in./s a = ft/s D

4 PROLEM (Continued) (b) θ = 180. v = in./s, v D = v D 6 in. sin β = = 0.6 β = in. v = in./s, v D = v D Instantaneous center of bar D lies at Point C. v ωd = = = rad/s ( D) 10 cos β a = in./s, a D = ad, α D = αd α / = [( D) α D D β] + [( D) ωd β ] = [10α D β ] + [ β ] a = a + a D D / Resolve in components. : 0 = 0 + (10 cos βα ) sinβ a = rad/s D : a = (10 sin β)( 18.51) cos β D D = in./s a = 635 ft/s D

5 PROLEM The 5-lb double pulley shown is at rest and in equilibrium when a constant 3.5 lb ft couple M is applied. Neglecting the effect of friction and knowing that the radius of gyration of the double pulley is 6 in., determine (a) the angular acceleration of the double pulley, (b) the tension in each rope. SOLUTION Given: mp = slugs, m = slugs, m = slugs M = 3.5 ft-lbs, k = 6 in P Free ody Diagrams: 4 8 Kinetics: Σ M0 = T T 1 1 Kinematics: 5 6 = I α = α a = ra = a 1 8 a = ra = a 1

6 PROLEM (Continued) Kinetics: : : 10 Σ F = 10 T= a 3. 5 Σ F = T 5 = a 3. 10a 4 5a = 0.5 g 1 g 1 g Elimination: ( ) a α g α α. = 6.5 g g (a) a = rad/s (b) T = 8.78 lb, T = 6. lb

PROLEM 16.119 40-lb ladder rests against a wall when the bottom begins to slide out.

. For θ = 40, determine (a) the angular acceleration of the ladder, (b) the forces at and. SOLUTION Given: 40 lb m = = 1.

81 ft Free ody Diagram: Kinetics: M G = Ia F x = ma µ N N = ma k x ( ) ( ) x ( 1) µ knc Nc+ µ knd + Nd = Ia µ d c N + µ c+ d

7 PROLEM lb ladder rests against a wall when the bottom begins to slide out. The ladder is 30 ft long and the coefficient of kinetic friction between the ladder and all surfaces is 0.. For θ = 40, determine (a) the angular acceleration of the ladder, (b) the forces at and. SOLUTION Given: 40 lb m = = 1.4 slugs 3. ft/s l = 30 ft, µ k = 0., θ = 40 1 ω = 0, I = ml 1 l l Geometry: c = cos 40 = ft, d = sin 40 = 4.81 ft Free ody Diagram: Kinetics: M G = Ia F x = ma µ N N = ma k x ( ) ( ) x ( 1) µ knc Nc+ µ knd + Nd = Ia µ d c N + µ c+ d N = Ia (3) k k F y = ma N + µ N mg = ma y k y ( ) Kinematics: a = a ω r G / G/ ai+ ak ( ci dj) ( a da) i caj = + = + + a r (4) a = a ω / / ( ) ( a) a i = a j+ ak ci dj a i = dai + a c j (5) r + a r

8 PROLEM (Continued) Solving the i components of (5): a = da (6) Substitute into (6) (5): a = dai+ caj (7) Substitute into (7) (1): N = µ N mda (8) Substitute into (7), (8) (): N + µ µ N mda mg = mca (9) mca + g+ µ kda Solve (9) for N ): N (10) = 1+ µ ca + g + µ kda Substitute (10)-(8): N (11) = m µ k da 1+ µ k Substitute into (10), (11) (3): ( ) k ( ) k k ( ) ( ) k ( ) ( ) m ca + g + µ kda ca + g + µ kda µ kd c + ( µ kc+ d) m µ k da Ia = 1+ µ k 1+ µ k Solve for α: mg mg ( µ kd c) + ( µ kc+ d) µ k 1+ µ k 1+ µ k a = m( c+ µ kd) ( c+ µ kd) I ( µ kd c) ( µ kc+ d) m µ k d 1+ µ k 1+ µ k (a) Substitute in known values: Substitute known values into (10): Substitute known values into (11): N = lb. µ N = 6.4 lb. k N = 13.5 lb. µ N =.70 lb. k a = rad/s (b) F = lb F = lb

PROLEM 17.14 The double pulley shown has a mass of 15 kg and a centroidal radius of gyration of 160 mm. Cylinder and block are attached to cords that are wrapped on the pulleys as shown.

9 PROLEM The double pulley shown has a mass of 15 kg and a centroidal radius of gyration of 160 mm. Cylinder and block are attached to cords that are wrapped on the pulleys as shown. The coefficient of kinetic friction between block and the surface is 0.. Knowing that the system is at rest in the position shown when a constant force P 00 N is applied to cylinder, determine (a) the velocity of cylinder as it strikes the ground, (b) the total distance that block moves before coming to rest. SOLUTION Kinematics. Let r be the radius of the outer pulley and r that of the inner pulley. v rc r v rc v r s r C r s s r Use the principle of work and energy with position 1 being the initial rest position and position being when cylinder strikes the ground. T U T 1 1 : where T1 0 and with T m v m v I C C 5 kg, 15 kg, C C C (15 kg)(0.160 m) kg m m m I m k 1 mr I C T m v r r 1 (15 kg)(0.150 m) kg m 5kg (0.50 m) (0.50 m) (8.7 kg) v v Principle of work and energy applied to the system consisting of blocks and and the double pulley C. Work. U1 Ps mg s FFs mg s sin 30 where s 1m

10 PROLEM (Continued) and s r m s (1 m) 0.6 m r 0.50 m To find F f use the free body diagram of block. 60 F 0: N m gcos 30 0 U N F 1 m gcos30 (15 kg)(9.81 m/s)cos N N (0.)(17.44 N) N f k (00 N)(1 m) (5 kg)(9.81 m/s )(1 m) (5.487 N)(0.6 m) (15 kg)(9.81 m/s )(0.6 m)sin J Work-energy: J (8.7 kg) v (a) Velocity of. v m/s v 4.79 m/s when the cylinder strikes the ground, r m v v ( m/s).876 m/s r 0.50 m v m/s C rad/s r 0.50 m fter the cylinder strikes the ground use the principle of work and energy applied to a system consisting of block and double pulley C. Let T 3 be its kinetic energy when strikes the ground. 1 1 T3 mv ICC 1 1 (15 kg)(.876 m/s) (0.384 kg m )( rad/s) J When the system comes to rest, T4 0 U3 4 (5.487 N) s (15 kg)(9.81 m/s )( s sin 30 ) (99.06 N) s where s is the additional travel of block. T3 U34 T4 : J (99.06 N) s 0 s m (b) Total distance: s s m

PROLEM 17.39 The ends of a 9-lb rod are constrained to move along slots cut in a vertical plate as shown. spring of constant k 3 lb/in.

11 PROLEM The ends of a 9-lb rod are constrained to move along slots cut in a vertical plate as shown. spring of constant k 3 lb/in. is attached to end in such a way that its tension is zero when 0. If the rod is released from rest when 50, determine the angular velocity of the rod and the velocity of end when 0. SOLUTION v v 1 L L x LLcos50 (5 in.)(1 cos50 ) in. Position 1. L 1 V1 W sin 50 kx1 5 in. 1 V1 (9 lb) sin 50 (3 lb/in.)( in.) in. lb.787 ft lb. T 0 Position. V ( Vg) ( Ve) T mv I ml 1 L 1 1 m lb 5in. ml

12 PROLEM (Continued) Conservation of energy: T1 V1 T V ft lb rad/s ω 3.71 rad/s 5 in. Velocity of : v L (3.713 rad/s) ft/s v 7.74 ft/s

PROBLEM 16.4 SOLUTION

PROBLEM 16.4 SOLUTION PROBLEM 16.4 The motion of the.5-kg rod AB is guided b two small wheels which roll freel in horizontal slots. If a force P of magnitude 8 N is applied at B, determine (a) the acceleration of the rod, (b)