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What must be the friction?

Harry Hotrod rounds a corner in his sports car at 50 km/h. The friction force holds him on the road. If he has twice the speed, what must be the friction force to prevent him from skidding off the road?


The friction is proportional to the square of the speed. If the speed is doubled, then the friction force is quadripled. No other calculation is needed.

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John M. | John - Algebra TutorJohn - Algebra Tutor
Check Marked as Best Answer
The friction must counteract (and equal) the outward acceleration defined by:
eq 1)    a = v2/r
And the friction force can be found with
F= ma, subsituting eq 1)    F=mv2/r
For this problem mass and radius are constants. 
Let F1 equal the friction force at 50 km/h.
Let F2 equal the friction force at 100 km/h  (2x50)
If we set this up as a ratio:
F1        mv2/r
---- =  -------
F2        m(2v)2/r
the m, r and v2 cancel from the equation and we are left with 1/4. 
The friction force at 100 km/h must be 4 times larger than it was at 50 km/h.


If I were answering this I would be careful about saying there was "outward acceleration".
Part of the issue with circular motion of any kind is breaking a student's pre-conception and getting a student to see and believe that there are no forces or accelerations in the outward direction.  Conceptually they need to believe there it is only a centripetal force that causes a centripetal acceleration and circular motion.
Robert you are absolutely correct. The Physics behind why a passenger  
Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
Let f be the friction force. Apply Newton's second law towards to the center of the rotation.
f = mv^2/r, where f is the only force in centrilpetal direction.
If you double v, then f is quadripled.