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Centripetal force and friction

  • Thread starter shwoll
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I have been having trouble with this problem for quite awhile now, and im wondering if anyone could offer some help...

I figured out part one...that the x direction forces must equal each other (mg sin theta = m*v^2/R) but i am completely lost on the second part. Number 6. Thanks!

5. [1pt]
Because of your physics background, you have been hired as a member of the team the state highway department has assigned to review the safety of freeways. This week you are studying a freeway in New Jersey which has a curve that is essentially 1/8 of a circle with a radius of 0.8 miles. The road has been designed with a banked curve so that the road makes an angle of 15 degrees to the horizontal throughout the curve. To begin the study, the head of your department asks that you calculate the maximum speed for a standard passenger car (2000 lbs) to complete the turn while maintaining a horizontal path along the road. He asks that you first consider the case of an ice-covered road with negligible friction. When you have completed that calculation, he wants you to do the case of a dry, clear road where the coefficient of kinetic friction is 0.6 and the coefficient of static friction is 0.7 between the tires and the road. This will give the team the two extremes of driving conditions on which to base the analysis.
Speed with icy road:

Correct, computer gets: 1.30E+02 mi/hr


6. [1pt]
Speed with dry road:
 

Answers and Replies

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Doc Al
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Newton's 2nd Law

shwoll said:
I figured out part one...that the x direction forces must equal each other (mg sin theta = m*v^2/R) but i am completely lost on the second part. Number 6. Thanks!
Not so fast. The forces in the x-direction must add up to equal [itex]m a_c = m v^2/r[/itex]. What forces act on the car? In the "no friction" case, there are only two forces. Friction adds a third force. (Which way does the friction act?)

You'll also need to realize that vertical forces are in equilibrium.
 

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