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xcortx November 3rd, 2014 01:57 PM

high school homework help??
A stunt driver for a movie needs to make a 2545kg car skid on a large, flat, parking lot surface. The force of friction between the tires and the concrete surface is 1.75x10^4N and he is driving at a speed of 24m/s. As he turns more sharply, what radius of curvature will he reach when the car just begins to skid?

skeeter November 3rd, 2014 04:45 PM

The force of static friction is providing the centripetal force on the car in the turn. For a large turn radius, the force of static friction is small in magnitude ... as the radius decreases (tighter turn) the max force of static friction is the point where the car just begins to skid.

$\displaystyle F_c \le f_{s \, max}$

$\displaystyle \frac{mv^2}{r} \le f_{s \, max}$

$\displaystyle \frac{mv^2}{f_{s \, max}} = r_{min}$

substitute the given mass, speed, and friction force to determine the minimum radius of curvature that the car can negotiate, just when the car starts to go into a skid

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