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hw10 - PROBLEM 14.55 Three small identical spheres A B and...

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Unformatted text preview: PROBLEM 14.55 Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached to three strings, [125 ft. long, which are tied to a ring t1 Initially the spheres rotate counterclockwise . about the ring with a relative velocity of 2.5 fits and the ring moves along the )4: axis with a velocity v,J = [1.25 fitsli. Suddenly the ring breaks and the three spheres move freely in the my plane with A and 3 following paths parallel to the y axis at a distance a = [L433 ft from each other an C a path parallel to the x axis. Determine {a} the velocity of each sphere, to} the distance o'. 50 LUTIGN Use a frame of reference that is translating with the mass center G of the system. Let vLI be its velocity. v0 2 mi The initial velocities in this system are ({4} directed 120° apart. Thus, 0’ {v}; lc’ and lit/C )0, each having a magnitude of ire. They are W4)” + {Ville + [Vole = 0 Conservation of linear momentum: t‘ m[v_4]0+ ”(Vela + m{V'£-}D = ante,r — vfl} + m[vH — v”) + mivf - vfi} t} ={vr1j— vfli} + {—vflj — vni} + [vci — uni] Resolve into components. i: v, — 3v, = c w. = 3w, = {3]{125} = 3.75 fie j:v,,—vfi.=tl vfi.=v,I . . . . l 1 1 Initial kinetic energy: T. = {amt-5] + 3[Emi2w2) . . . l l l i F Inal klnette energy: It"2 = Emvi + Emvfi + Emvg = mvi + amt/3: Conservation of energy: 3'”: = l". Solve for vi. 321 2_12_s 2: at 2 v — I», + Etta} 2a,: _ 2{1.25} + 2{2.5} Etsss} = 4.5875 site, v, = 36511”th v, = 2.1? as Tit v3 = 2.1? as, to 'f'C = 3.35 fli'lls- —"‘ PROBLEM 14.59 The nozzle shown discharges a stream of water at a flow rate Q = 1.8 1113me with a velocity v of magnitude 2!] nits. The stream is split into two streams with equal flow rates by a wedge which is kept in a fixed position. Determine the components {drag and lift) of the Jforce exerted by the stream on the wedge. SDLU TIDN Let F be the force that the wedge exerts on the stream. Assume that the fluid speed is constant. v = 20 mt's. d . Mass flow rate: —m = {Q = [1mm kgi'mHMLS m'h'mm) = 1300 kg-"min = 3:] kgt's {it Velmity vectors: v = vi, vl = v[cos3[}°i + sin 3W” v1 = v[cos45°i — sin45‘l‘j} Impulse—momentum principle: (Emit; ._t start K + CW4 = 1; {antjv + Flight] = $11 + 3;: v1 amp 1 F=— _‘r| +_v'2 —Y hr 2 2 j—Tv|:%[oos3tl°i + sin30°jj ammo — 51.14503 — t] s {so kgt‘sHEfl Ws}[—fl.21343i— fl.1{}355j] = 4123.: up — [fillNH Force that the stream exerts on the wedge: —F={I28.1N}i+[52.lN}j drag=1sa.1N—~1 1tfi=sz_tt~t‘i PROBLEM 14.64 Tree limhe and branches are being feel at A at the rate cf 11'} lhfs inte a shredder which spews the resulting wee-ti chips at C with a velocity of 60 fits. Determine the hetizcntel ccmpcnent cf the three exerted by the 3hredder on the truck hitch at D. 50 LUTIDN Let F be the force exerted en the chipe. Apply the lmpulse—mcmenhtm principle tc the chips. Assume that the feed velocity is negligible. F:—-.\r..: h: '1 = [flhhmm + sinzse = {16.391b}i+{?.ETlh}j 291:0: fl_1.—Fx=fl c] =16.391h Force ch truck hitch at D: D, = [43.39 II: «It PROBLEM 14.112 Car A was traveling east at high speed when it collided at point 0 with car 3, which was traveling north at 1’2 lorvh. Car C, which was traveling west at 9B kmt'h, was II} 111 east and 3 rn north of point 0 at the time of the collision. Because the pavement was wet, the driver of car C could not prevent his car fi'om sliding into the other two cars. and the three cars, stuck together, kept sliding until they hit the utility pole P. Knowing that the masses of cars A, B, and C are, respectively, 1513:] kg, 13cc kg, and use kg, and neglecting the forces exerted on the cars by the wet pavement, solve the problems indicated. Knowing that the speed of car A was [29.6 kn'n'h and that the time elapsed from the first collision to the stop at P was 2.4 5, determine the coordinates of the utility pole P. SOLUTION Let t he the time elapsed since the first collision. No external forces in the xy plane act on the system consisting of cars A, B, and C“ during the impacts with one another. The mass center of the system moves at the velocity it had before the collision. Initial mass center E}: (m, + me + mciiioi 1' Eel-i = melxci + Fell a, = use = {name} = 3m, J7, = 0.33),: = [o.3]{3] = as: m Given velocities: v, = vAi, v, = {as nv's}j, v, = —[2s nv's}i velocity of mass center: {mg + m3 +11%]? = my” + mgvfi + mcvc v = {LETSvA + 9.325% + 13'ij Since the collided cars hit the pole at 1'? = It'd 1' Jr'i‘i lpi + ypj = fui + jflj + W Resolve into components. x: it}, = in + DJTSvAtF — 0.3}?ch y: yp = in + (1.325133%. v)1 2129.6kmi'h = 36 mils, tp = 2.4s IF = 3 + {fi.375}[36}{2.4} — (U.3}[25}[2.4} = 114 m y, = as + {o.sss)[2o}{2.4} = 16.5 m x, = 1?.4Dm4 y,. = rasamd ...
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