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HW17-S07

# HW17-S07 - • Calculate the normal force on the left side...

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AEM 264-003/901 Homework #17 4/9/07 Due: Wednesday, 4/11/07 The six-step solution method, correct units and a reasonable number of significant digits (generally 3 or 4) are required for every problem solved in every homework assignment in this section of AEM 264. 1. The weight of the car shown is 4100 lbs. The coefficient of friction between the tires and the road surface is 0.85. Since the center of gravity is above the road surface, when the car accelerates, the vertical loads on the tires will change. The change in the vertical loads will change the maximum friction forces, which determine the maximum acceleration of the car. Determine the maximum acceleration if the car has rear-wheel drive Determine the maximum acceleration if the car has front-wheel drive Determine the maximum acceleration if the car has four-wheel drive 2. A 1996 Ford Ranger has a center of mass that is 24.6 inches from the ground as shown in the rear view to the right. The track width, t (distance between the centerlines of the left and right tires) is 57 inches. The Ranger weighs 3450 lbs.
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Unformatted text preview: • Calculate the normal force on the left side tires ( N L ) and right side tires ( N R ) as the Ranger goes around a flat left-hand curve (no banking or superelevation) of radius r C =350 feet at speeds of v=30 mph and v=60 mph . • Determine the speed at which the Ranger is just about to start a rollover, i.e., N L = 0. What coefficient of friction is required for a rollover to occur ( μ = F R /N R )? 3. An aluminum plate with 8 steel cylinders is arranged as shown. Determine the mass moment of inertia about the Z axis at the center of mass of the entire system . Express I G in units of lbf-ft-sec 2 and kg-m 2 . X Y ± Thickness of table = 0.80 cm = 0.31 in ± Table diameter = 20 cm = 7.9 in ± Table material = aluminum – ρ Al = 0.00265 kg/cm 3 = 0.096 lbm/in 3 ± Workpiece diameter = 3.0 cm = 1.18 in ± Workpiece height = 4.0 cm = 1.57 in ± Workpiece material = steel – ρ St = 0.0078 kg/cm 3 = 0.282 lbm/in 3 R N R F t h mg W = L N L F C r...
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