Asked by Anonymous
If the coefficient of static friction between tires and pavement is 0.7, calculate the minimum torque that must be applied to the 63-cm-diameter tire of a 925-kg automobile in order to "lay rubber" (make the sheels spin, slipping as the car accelerates). Assume each wheel [is a disk that] supports an equal share of the weight.
Answers
Answered by
Damon
weight on each tire = 945*9.8/4 = 2315 Newtons on each tire
force of friction at slip = .7*2315 = 1621 N
times radius (0.63/2) to get torque
force of friction at slip = .7*2315 = 1621 N
times radius (0.63/2) to get torque
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