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Centripetal Forces

  • Thread starter raphile
  • Start date
  • #1
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Homework Statement



A minibus of mass 1.6 tonnes is rounding a flat bend of constant radius 25m at a steady speed of 72km/h. Find the centripetal force acting on the vehicle. What is the minimum co-efficient of sideways friction between the tyres and the road for this motion to take place without skidding?

The Attempt at a Solution



I thought I knew how to do this. Using mv^2/r, I got the centripetal force to be 25,600N (changing the speed into from km/h to m/s). Then, for the co-efficient of friction (denote as u), I used uR = u(mg) = mv^2/r and got, by taking g=9.81, an answer of u=1.631.

But I thought co-efficients of friction were supposed to be less than 1. An answer of 1.631 seems a bit high. Is it right?
 

Answers and Replies

  • #2
238
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Your answers are correct. [tex]\mu_[/tex] is not always <1. Since [tex]\mu>1[/tex], what do you think is happening?
 

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