MEEN 221- Chapter+17

MEEN 221- Chapter+17 - ENGINEERING MECHANICS Dynamics W.F...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics W.F. Riley & LJD. Sturges 17-3 The 4-lb pendulum bob is released from 0 rest when 9 = 75 . Determine the work done on the pendulum bob By the cord tension T as the bob swings D D from 6 = 75 to 6 = o . By the gravitational force W as the bob swings from 9 = 750 to 9 = 0°. Solution a. The cord tension T always acts normal to the motion and hence does no work. UT = 0 lb'ft .......... ................................... Ans. b. The gravitational force W is constant in both magnitude and direction. As the 4-lb pendulum bob swings from 9 = O0 to 9 = 750 it drops (moves in the direction of the force an a distance of d = 25(cos Oo — cos 75°) = 18.52952 in. = 1.54413 ft and does work Fd = 4(1.54413) 6.18 lb'ft 833 3.7.7. 5 :m» w..." «Mme We..- 7., .,i_ __ "mp ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-5* A fully loaded Boeing 747 has a take-off weight of 660,000 lb and its engines develop a combined thrust of 200,000 lb. If the runway is flat and level, determine the work done on the aircraft a. By the engines as the aircraft moves 1000 ft down the runway. b. By the gravitational force W as the aircraft moves 1000 ft down the runway. Solution 3. The engine thrust is constant and acts directly in line with the motion. Therefore, the work done by the engine thrust is U: = Fd = 200,000(1000) = 200(106) lb'ft b. The weight force always acts in a vertical direction while the motion of the plane is always horizontal. Therefore, the weight force is always perpendicular to the motion and does no work UW = O lb'ft ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-7 A ZOO-lb block is pushed along a horizontal surface by a constant force F as shown. If the kinetic coefficient of friction between the block and the surface is ”k = 0.25, determine the work done on the block By the frictional force on the bottom of the block as the block moves 20 ft along the surface. By the gravitational force W as the block moves 20 ft along the surface. By the loo-lb force F as the block moves 20 ft along the surface. Solution Since the block has no motion in the vertical direction, it has no acceleration in that direction and the equations of motion give +T 2F = ma : N - 100 sin 34° - 200 = y y Therefore N 255.919 lb Eb 0.25N = 63.9798 lb 0 a. friction force is constant and acts 180 to the direction of motion of the block. Therefore U = Fd = -63.9798(20) u = -1280 lb‘ft ... ....... . ........ ............ ....... .. Ans. b. The weight force (vertical) is always perpendicular to the motion (horizontal), and therefore the weight force does no work ”W = 0 lb'ft ............................................. Ans. c. The force F is also constant. The component of F normal to the surface does no work since there is no motion in that direction. The work done by the component of F which acts along the surface (which is also constant) is 0 U = Fd = (100 cos 34 )(20) F = 1658 lb'ft ................................... 837 ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17—9* The pair of blocks shown are connected by a light inextensible cord and are released from rest when the spring is unstretched. The static and kinetic coefficients of friction are “5 = 0.2 and pk = 0.1, respectively. If the 10-lb block moves 9 in. to the right, determine the work done On the 5—lb block by the gravitational - mnmfi force. {if ff: On the 10-block by the frictional force acting on the bottom of the block. On the 10-lb block by the spring force F5 = 5(4 - to) where the spring constant is 5 = 20 lb/ft and the unstretched length of the spring is to = 12 in. Solution Since the 10-lb block has no motion in the direction normal to the surface, it has no acceleration in that direction and the equations of motion give +TEF =ma: N—1o=o y y Therefore N 10 lb F 0.1N = 1 lb a. The weight of the 5-lb block is constant and always acts vertically downward. As the 10—lb block moves 9 in. = 0.75 ft to the right, the 5-lb block will fall 0.75 ft and the weight force will do positive work on the block U5 Fd = 5(o.75) 3.75 lb'ft ...... ............ ...... ................ Ans. The friction force is constant and acts 180° to the direction of of the 10-lb block. Therefore (Problem 17-9 continues ...) 839 ENGINEERING-MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-10* The pair of blocks shown are connected by a light inextensible cord and are released from rest when the spring is stretched 600 mm. The static and kinetic coefficients of friction are NS = 0.3 and “k = 0.2, respectively. If the S-kg block moves 300 mm to the left, determine the work done a. 0n the S-kg block by the frictional force acting on the bottom of the block. 0n the 5-kg block by the spring force F; = £(£ - £0) where the spring constant is 5 = 1 kN/m and the unstretched length of the spring is to = 250 mm. 0n the lO—kg block by the gravitational force. 0n the lO—kg block by the frictional force acting on the bottom of the block. Solution Since the blocks have no motion in the direction normal to the incline, they have no acceleration in that direction and the equations of motion give +1‘ 2F = ma : N - 5(9.81) y y 5 + 2 = : — . o A Eh man N10 10(9 81) cos 50 Therefore N5 49.05 N F3 0.2N = 9.81 N NlO 63.0575 N F10 0.2N = 12.6115 N a. The friction force IE is constant and acts 180° to the direction of motion of the block. Therefore U = Fd = —9.81(0.3) F5 = -2.94 N'm = -2.94 J ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 17-12 The pair of blocks shown are connected by a light inextensible cord. The blocks are released from rest from the position shown with the spring stretched 150 mm. If the 2—kg block rises 100 mm, determine the work done On the 2—kg block by the gravitational force. On the 10-kg.block by the spring force F; = £(C - (b) where the spring constant is i = 1.2 kN/m. Solution a. The weight force W5 = 19.62 N is constant and acts vertically downward. As the 2-kg block rises 0.1 m, the work done by the weight force W5 is Uh = FE = -19.62(O.1) = —l.962 N°m = ‘1.962 J b. The length of the cord L = d + 52 + c is constant. Initially, 510 = 400 mm, d = 500 mm, and 52 = 300 mm. After the 2—kg block rises 100 mm, 52 = 200 mm, d = 600 mm, 510 = 519.615 mm, and the stretch in the spring has been reduces to 30.385 mm. The spring exerts a force F; 1200x (where x = 0 when the spring is unstretched) on the 10-kg block. If x increases to the right, then the spring force acts to the left (1800 to the motion), d0; = —1200x dx, and 0.030385 0‘ -f 1200x dx = -1200[ 0.150 2 0.030385 X 2 ] 0.150 = -600[(0.030385)2 — (0.150)2] = 12.95 N'm = 12.95 J ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-14* A 2-kg package slides on a frictionless floor and strikes the bumpers shown. The two linear springs are identical with spring constants of 5 = 1.5 kN/m. Determine the work done on the package by the springs as spring 1 is compressed 120 mm (and spring 2 is compressed 20 mm). Solution Each spring exerts a force F; = lSOOx (where x = 0 when the spring is unstretched) on the 2-kg block. If x increases to the right, then the spring forces act to the left (180° to the motion), dUa = -1500x dx, 0.120 0.020 -I 1500x dx - I 1500x dx 0 0 x 2 x 2 2 ]O.120 2 0.020 -1500[ - 1500[ ] 0 0 -750[(0.120)2 - O] - 750[(0.020)2 - 0] -11.10 N‘m = -11.10 J ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-23 A 7500—1b truck is traveling on the freeway at 65 mi/h when the driver suddenly notices a moose standing on the road 200 ft straight ahead. If it takes the driver 0.4 s to apply the brakes and the kinetic coefficient of friction between the tires and the road is “k = 0.5, Can the driver avoid hitting the moose without steering to one side? Where will the truck come to rest relative to the moose? If the driver must steer to one side, determine the speed of the truck as it passes the moose. Solution Initially, the truck is traveling at 65 mi/h = 95.33 ft/s = constant. Therefore, the truck will travel a distance of A51 = v At = (95.33)(o.4) = 38.13 ft before the driver applies the brakes. The normal force is +T 2F = ma : N - W = O Y Y N = W = 7500 lb and the friction force is F uN 0.5(7500) = 3750 lb and it is constant. Only the friction force does work and the work-energy equation gives 1 7500 2 ' —_ .— . _ A = 2 [ 32_2 ](95 33) 3750 52 0 A52 = 282.2 ft The total distance to stop is As1 + As 320.4 ft. Therefore The driver must swerve. Stop 120.4 ft past the moose Again use the work—energy equation to find the speed of the truck when it passes the moose at As = 200 - A51 = 161.9 ft of braking 1 7500 2 l 7500 2 2 [ 32.2 ](95.33) - (3750)(161.9) - —2_[—§2T2_]V 'v = 62.3 ft/s s 42.4 mi/h ......... ....... ...... Ans. 855 ember». .. "w, r“, -37...-e w __..i,. 1,. , ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-43 A 1.5—lb weight slides on a frictionless vertical rod as shown. The undeformed length of the spring is to = 8 in., the spring modulus is i = so lb/ft, and the distance d = 12 in. If the slider is released from rest when b = 9 in., determine the speed of the slider when b = 0. Solution The normal force N'does no work. As the slider rises 9 in. 0.75 ft, the constant weight force does work = =—.5o. UW Fd 1 ( 75) = -1.125 lb'ft The component of the spring force that is normal to the rod does no work; the work done by the component along the rod is U; = I —(£6 cos 9) db where 5 = fl — £0 = t - 8/12, cos 9 = b/fl, and b is positive downward. Then Ué = J0 -80[/ 1 + b2 - ——8—]-———— 75 O. 12 2 O -80 b - *—§— 1 + b2 = 9.16667 lb'ft 12 0.75 Therefore, the work-energy equation Ti + U = T f gives 0 — 1.125 + 9.16667 = 18.58 ft/s ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 17-57 The two blocks A and 3 shown weigh NA = 25 lb and WE = 50 1b, respectively. The blocks are at rest and the spring (5 = 25 lb/ft) is unstretched when the blocks are in the position shown. Determine The speed of block 3 when it is 1 ft below its initial position. The maximum distance that block 3 will drop. So1ution The tension is an internal force — its work will cancel out when the work-energy equations for the two blocks are added together. When block block B falls a distance d, block A rolls to the right the same distance. The spring force F5 = fix acts 180° to the motion of block A and the work done by the spring force and the weight of block 3 is d U = I -(25x) dx + 50d = 50d - 12.5d2 lb'ft 0 Therefore, the work-energy equation 11 + U = Tf 2 l 25 2 1 50 2 + _ . = __ .___—_ __ — (0 O) + (50d 12 5d ) 2 [ 32.2 ]v + 2 [ 32.2 ]V gives v2 = 42.9333d - 10.7333az2 a. When d = 1 ft, VB = v = 5.67 ft/s ¢ ................................... Ans. b. The maximum distance that B will fall corresponds to v = 0 0 42.9333d - 10.7333d2 d d = 4.00 ft ........................................ Ans. max ...
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