Engineering Mechanics Summary and Solution, Quizzes of Engineering

Download Engineering Mechanics Summary and Solution and more Quizzes Engineering in PDF only on Docsity! Problem 14-2 The crate of weight W has a velocity v, when it is at A. Determine its velocity after it slides down the plane to s = s', The coefficient of kinetic friction between the crate and the plane is 44. Given: W = 20 lb a=3 ft = vg = 12 — aa s s' = 6 ft | Mk = 0.2 \ rin Solution: \ 6 = atan 7 b Guess vsi= s . (Ww 5 (Ww ft Given (2) v4" + Wsin(O)s' — Fs" = (*) ve v! = Find(v) f= 1772— & Ss  *Problem 14-12 The force F, acting in a constant direction on the block of mass M, has a magnitude which varies with position x of the block. Determine the speed of the block after it slides a distance d). When x = 0, the block is moving to the right at vy The coefficient of kinetic friction between the block and surface is 44.  W4 =31b nok | 3) al | 2) I Lng B= 8 lb * TF eee 1 ah Hk = 0.3 I f~ | A ft . _ vg =5— 7 ‘A Ny, a Solution: t We L=s4+2sp Guesses d=1ft Given 2 1 2 vA Wed - upW42d = —|Wav4 + Wel — 2g 2 d = Find(d) d=0.313ft  Problem 14-21 The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-absorbing material. The barrier stopping force is measured versus the vehicle penetration into the barrier. Determine the distance a car having weight W will penetrate the barrier if it is originally traveling at speed vy when it strikes the first barrel. Units Used: kip = 10° Ib Given: w W = 4000 Ib ft vg = 55 — s =< F ft g = 3225 e i s : Solution: 2 (wy) 2 205 10 15) 2 25 4") vo - Area = 0 Vehicle penetration (fi) 1( Ww 2 7 7 o Area = =| —|vo Area = 187.888 kip: ft | We must produce this much work with the barrels. Assume that 5 ft<x<15 ft Area = (2 fi)(9 kip) + (3 ft)(18 kip) + (x — 5 ft)(27 kip) Area — 72 kip: ft aa ttl x =9.292ft Check that the assumption is corrrect! 27 kip  *Problem 14-36 A block of weight WW rests on the smooth semicylindrical surface. An elastic cord having a stiffness k is attached to the block at B and to the base of the semicylinder at point C. If the block is released from rest at A(@= 0°), determine the unstretched length of the cord so the block begins to leave the semicylinder at the instant @ = 6,. Neglect the size of the block. Given: W=21b Ib k=2— ft 6; = 45 deg C a=1.5 ft m w g = 981 — t. 2 aN Solution: ft Guess 6=1ft w=l- s Given Ww vp Wsin()) = (z4 gja 1 2 1 2, Ww vp Lia 3)? — Life 6y)a~ 5]? wasin(oy) = (2 v1 . fl = Find(v7, 6) vy = 5.843 — 5 = 277K Oo s  Problem 14-75 The bob of the pendulum has a mass M and is released from rest when it is in the horizontal position shown. Determine its speed and the tension in the cord at the instant the bob passes through its lowest position. Given: M = 0.2 kg r= 0.75 m g= 9812 s Solution: Datum at initial position: T) +V,=T2+V2 ln 2 O40 = My2 — Mgr v2 = V2gr IF, = Ma, m v2 = 3.84 = T= 5.89 N ve ve -Nc - Mg = - ce Nc = ait Mg Nc = 16.825 kN Since Nc > 0 then the coaster successfully passes through loop C. Problem 14-85 The bob of mass M of a pendulum is fired from rest at position A. If the spring is compressed to a distance d and released, determine (a) its stiffness k so that the speed of the bob is zero when it reaches point B, where the radius of curvature is still r, and (b) the stiffness & so that when the bob reaches point C the tension in the cord is zero. Units Used: kN = 10° N Given: 4 m M = 0.75 kg g = 9.81 > s fo 6=50mm r=06m , ay Solution: j \ AtB: { —— 1 =| 1 2 Be | iki =k5 = Mgr | = 2 i" f 2M, kN p= EE 423.53 —— f Se m ALC: :  Problem 14-91 The ride at an amusement park consists of a gondola which is lifted to a height / at A. If it is released from rest and falls along the parabolic track, determine the speed at the instant y = d. Also determine the normal reaction of the tracks on the gondola at this instant. The gondola and passenger have a total weight W. Neglect the effects of friction. Given: W = 500 Ib h = 120 ft d d= 20 ft a = 260 ft