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Unformatted text preview: 3.18. CHAPTER 3, PROBLEM 18 171 3.18 Chapter 3, Problem 18 Problem: An automobile of mass m traveling at constant speed V encounters a depression in the road, which has a radius of curvature, R. The automobile maintains constant speed in the depression. We wish to determine the reaction force, N , exerted by the road on the automobile at the lowest point of the depression in terms of m, V , R and gravitational acceleration, g . (a) Find N using natural coordinates. (b) Find N using cylindrical coordinates. (c) Compute N for mg = 2800 lb, V = 30 mph and R = 50 ft. mg . . . . . . . . . . . . . ... .. . .. . . .......................................... .. . . . . . ..... . . . . .. . . . ........................................... ...... .......... . .. . . ... . . ............... . .............................. ............................................................... ........................................... ..................................................... .................................................................................... ................. . ....................................................................................... ............ .. . . . . .................................................................................... . . .... . . . . . . ... ... . . .... . ... ............... . .................................................................................................. . . ......................................................... . ...... .................. . . . ...... . .. ..................................................................................................................... .. .. . ............................................................................................................................................................................................................................................................................................. .............................................................................................................................................................................................. .. . . .......... . . ....... . ......................................................................................................................................................................................................................... . ... . .... .......... . .......... . ... .... . .. . .. .... . .. . . .. ... . .. . . . ....... . . ....... .......... .. .. ............................................................................................................................................................................................. .. ............................................................................. ........................................................................... . . ..... .................................................................................................................................. . ............................................................................ . . ...... . . . . . . . . . . . V N Solution: For both natural and cylindrical coordinates, the origin of the coordinate system is located a distance R directly above the lowest point of the road’s depression. (a) The normal component of Newton’s Second Law in natural coordinates is Fni = m i V2 R The forces acting are the automobile’s weight, mg , which acts downward and the reaction force from the road, N , which acts upward. Since the center of curvature is above the automobile, the acceleration, V 2 /R, is positive upward. Thus, Newton’s Second Law becomes N − mg = m V2 R =⇒ N = mg 1 + V2 gR (b) In terms of cylindrical coordinates, the radial component of Newton’s Second Law is ˙ Fri = m r − rθ2 ¨ i ˙ Since the motion occurs on a circular arc of radius r = R, necessarily r = 0 and Rθ = V . So, this ¨ equation simplifies to 2 V V2 Fri = −mR = −m R R i For cylindrical coordinates, r increases downward so that the automobile’s weight acts in the positive direction and the reaction force acts in the negative direction. Hence, mg − N = −m V2 R =⇒ N = mg 1 + V2 gR 172 CHAPTER 3. FORCE AND ACCELERATION (c) We are given mg = 2800 lb, V = 30 mph and R = 50 ft. Since 30 mph = 44 ft/sec, the reaction force is ⎡ ⎤ 2 (44 ft/sec) ⎦ = 6168 lb N = (2800 lb) ⎣1 + 2 32.174 ft/sec (50 ft) ...
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