homework4 - Problem 14.9 Aerodynamic drag exerts a force on...

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Unformatted text preview: Problem 14.9 Aerodynamic drag exerts a force on the skier in Problem 14.8 of magnitude 0.6v2, where v is the magnitude of his velocity. Determine the skier’s velocity when he has gone 20 in down the slope, and compare your answer with the answer to part (b) of Problem 14.8. Strategy: After using Newton’s second law to deter- mine the skier’s acceleration in the direction parallel to the slope, use the chain rule to express it as the product of the velocity and the derivative of the velocity with respect to position. Solution: m = 80 kg g = 9.31 m/s ZFX : —f— D+mgsin25° =ma 2F}, = N —mgcos25° = 0 D = 0.6112 f = MN = 0.08 N .‘.f = 0.08 mg cos 25° Substituting for D and F in the first equation, we get ma = ~0.08 mg cos 25° + mg sin 25° — 0.61)2 Rewriting, we have a:———~—-—v.._ dt _ dx dt _ dx Also a = Cg sin 25° — 0.08 gcos 25°) — v2 m Let b2 = (g sin 25° — 0‘08 gcos25°) (‘2 = 0.6/m Calculating, b = 1.85, c = 0.0866 v = 13.4 m/s 1 20¢2 z —5 In —40€2 = ln 0 Solving, [Gr—vi Problem 14.16 A 2—kg cart containing 8 kg of water y r is initially stationary (Fig. a). The center of mass of the “object” consisting of the cart and water is at x = 0. The cart is subjected to the time-dependent force shown in Fig. b, where F0 = 5 N and to = 2 5. Assume that no water spills out of the cart and that the horizontal forces exerted on the wheels by the floor are negligible. (a) Do you know the acceleration of the cart during the period 0 < t < to? (b) Do you know the acceleration of the center of mass of the “object” consisting of the cart and water during the period 0 < t < to? (c) What is the x—coordinate of the center of mass of the “object” when t > 2to? (b) Solution: (a) No, the internal dynamics make it impossible to determine the acceleration of just the cart. (b) Yes, the entire system (cart + water) obeys Newton‘s 2“d Law. . _ _ 5N _ 2 ZF.(5N)—(10kg)a => 11— 10kg—0.5n1/s (c) The center of mass moves as a “super particle”. F0r0<t<t0 5N N:10k =—=o.5m/2 5 ( g)a=>a 10kg s v = (0.5 m/s2)t, s = (0.25 m/Sz)t2 Att=tU:2s, v=l.0m/s, s=1.0m For to < t < 2m, — 5 N = (10 kg)a, a = —0.5 m/52, v = —(0.5 m/s2)(t — to) + 1.0 m/s 3 = —(0.25 m/s2)(t — to)2 + (1.0 m/s)(t — to) + 1.0 m Problem 14.22 Consider the 11,000—kg airplane shown in Problem 14.2]. At the instant shown, the airplane’s velocity is v = 300i (m/s). The rate of change of the magnitude of the velocity is dv/dt = 5 m/sz. The radius of curvature of the airplane’s path is 4500 m, and the y axis points toward the concave side of the path. The thrust is T = 120,000 N. Determine the lift L and drag D. Solution: ax = 5 m/s2 m = [1000 kg g = 9.81 m/s2 |V| = 300 m/s T = 120000 N ‘ i p = 4500 m - 2sz TcoslS°~D—mgsin15°=max ZFy: L+Tsinls°—mgcos15°:mu_V “y = Vz/P Solving, D = 33.0 kN, L := 293 kN Path Horizontal Problem 14.36 The 100-lb crate is initially stationary. The coefficients of friction between the crate and the inclined surface are ,ux = 0.2 and ,uk = 0.16. Determine how far the crate moves from its initial position in 2 s if the horizontal force F = 90 1b. Solution: Denote w : IOO lb, g = 32.17 ft/sz, F = 901b, and 0 = 30°. Choose a coordinate system with the positive x axis parallel to the inclined surface. (See free body diagram next page.) The normal force exerted by the surface on the box is N = F sin0 + Wcos0 = 131.6 lb. The sum of the non-friction forces acting to move the box is Fe 2 Fcost9 - Wsin0 = 27.9 lb. Slip only occurs if chl a lNusl which 27.9 > 26.32 (lb), s0 slip occurs. The direction of slip is determined from the sign of the sum of the non friction forces: Fe > 0, implies that the box slips up the surface, and F6 < 0 implies that the box slips down the surface (if the condition for slip is met). Since Fc > 0 the box slips up the surface. After the box slips, the sum of the forces on the box parallel to the surface is 2 FX : Fc — sgn(Fc)p.kN, where sgn(Fc) = F W 176'. From Newton‘s second law, 2 Fx = (~) (1, from which a = c 3 %(Fc -—sgn(Fc);tkN) : 2.125 ft/sz. The velocity is v(t) =at ft/s, since 11(0) 2 0. The displacement is s = gtz ft, since 3(0) = 0. The position after 2 s is s (2) = 4.43 ft up the inclined surface. Problem 1456* In Problem 1455, how long does the collar take to go from A to B if the coefficient of kinetic friction between the collar and the bar is M = 0.2? Strategy: This problem is almost the same as prob- lem 14.55. The major difference is that now we must calculate the magnitude of the. normal force, N, and then must add a term uk |N| to the forces tangent to the bar (in the direction from B toward A ——opposing the motion). This will give us a new acceleration, which will result in a longer time for the collar to go from A to B. Solution: We use the unit vector e” from Problem 14.55. The free body diagram for the collar is shown at the right. There are four forces acting on the collar. These are the applied force F, the weight force W = —mgj = —58.86 j (N), the force N which acts normal to the smooth bar, and the friction force I“ = ~M|N|eA3. The normal force must be equal and opposite to the components of the forces F and W which are perpendicular (not parallel) to AB. The friction force is parallel to AB. The magnitude of IF + W| is calculate by adding these two known forces and then finding the magnitude of the sum. The result is that IF + W| = 57.66 N. From Problem 14.55, we know that the component of IF + WI tangent to therbar is |FAB| = 20.76 N. Hence, knowing the total force and its component tangent to the bar, we can find the magnitude of its component normal to the bar. Thus, the magnitude of the component of IF + W| normal to the bar is 53.79 N. This is also the magnitude of the normal force N. The equation of motion for the collar is ZFWHM = F + W + N — uklNleAB = ma. In the direction tangent to the bar, the equation is (F + W) ~eA3 — [LkINI 2 mm. The force tangent to the bar is FAB = (F + W) ~eAB — [LklNl = 10.00 N. The acceleration of the 6-kg collar caused by this force is a, = 1.667 m/sz. We now only need to know how long it takes the collar to move a distance of 0.3 m, starting from rest, with this aCCel- eration. The kinematic equations are u, 2 cm, and s, = a,r2/2. We set s, = 0.3 m and solve for the time. The time required is t I: 0.600 s Problem 14.76 In Problem 14.75, determine the mag— nitude of the velocity of the mass and the angle 9 if the tension in the string is 50 lb. Solution: EFT : Tcose —mg :0 (T2 — In2g2)L Tm Solving we find 6 = cos‘1 , v 2 Using the problem numbers we have 1 slug 32.2 ms2 _ ~I ______ = o 0 — cos ( 50 lb ) 49.9 _ [(50 lb)2 _ (l slug 32.2 ft/S2)2]4 ft _ v _ W _10.8 ft/s Problem 14.93 The combined mass of the motorcy- cle and rider is 160 kg. The motorcycle starts from rest at t = 0 and moves along a circular track with 400-m radius. The tangential component of acceleration as a function of time is a, = 2 + 0.2t m/sz. The coefficient of static friction between the tires and the track is ,us 2 0.8. How long after it starts does the motorcycle reach the limit of adhesion, which means its tires are on the verge of slipping? How fast is the motorcycle moving when that occurs? Strategy: Draw a free-body diagram showing the tan— gential and normal components of force acting on the motorcycle. ” Solution: m = 160 kg R = 400 m Along track motion: a, 2 2 + 0.21 m/s2 v, = V =2t+0.1t2 m/s 3 =t2 +0.1:3/3 m Forces at impending slip |F + f1 = ukN at impending slip |F+f|=‘/F2+f2 since f_LF Force eqns. ZFl:F=ma, R=400m ZFnzfzmvz/R m=l60 kg ZFZ:N—mg=0 g=9.81m/32 “5:08 F2+f2=usN a, =2+0.2t v=2t+0.lt2 Six eqns, six unknowns (F, f, v, (1,, N, t) Solving, we have t=14.4s F=781N v = 49.6 m/s f = 983 N N = 1570 N a, 2 4.88 m/s2 (at t = 14.4 S) [as]:=0.25 = [(4 cos 60(20) + 2(—80sin 0)(t)(20)(t)],=0_25 = -5214 ft/sz. rad From Newton‘s second law for transverse component F9 — far],=o.25 = [— 80 sin 0 — (1600 cos 0)t2 mg cos 0 = ( as, from vnhich F9 = —0.2025 lb (4 cos 6)(20t)2]:=o,25 = —209 ft/sz. Problem 14.108* In Problem 14.107, suppose that the _ \ unstretched length of the spring is r0. Determine the ‘ smallest value of the spring constant k for which the pin will remain on the surface of the cam. Solution: The spring force holding the pin on the surface of the cam is Fr = k(r — r0) = k(m(2 —— cos6) — m) = kro(l -— cos 6). This force acts toward the origin, which by definition is the same direction as the radial acceleration, from which Newton‘s second law for the \ pin is 2 F = ker — cos 6) = —ma,.. From the solution to Problem l4.l07, kr0(l - cos 6) = —2mrwfi(cos0 — 1). Reduce and solve: k = mefi. Since cos(-) 5 l, ker — cos 6) 3 0, and mewgmosd — 1) 5 0. If k > 2mg, Define Fcq = ker - cos a) + 2mrw3(cos6 — 1). If ch > 0 the spring force dominates over the range of 0, so that the pin remains on the cam surface. If k < meS, ch < 0 and the radial acceleration dominates over the range of 6, so that the pin will leave the cam surface at some value of (9. Thus is the minimum value of the spring constant required to keep the pin in contact with the cam surface. ...
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homework4 - Problem 14.9 Aerodynamic drag exerts a force on...

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