Calculating the Force of Static Friction on a 2000kg Car

AI Thread Summary
To calculate the force of static friction on a 2000kg car making a left turn with a radius of 6.5 m in 2 seconds, the initial velocity is determined to be 3.25 m/s. The centripetal force is calculated using the formula Fc = (m*v^2)/r, resulting in a value of 3.2 kN. The distance traveled during the turn is identified as a quarter of a circle, leading to a recalculated speed of 5.1 m/s. Participants emphasize the importance of correctly squaring the velocity in calculations. The discussion highlights the need for careful attention to detail in physics problems.
yankees26an
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Homework Statement


A 2000kg car makes a left run of radius 6.5 m. If this turn is made in 2 seconds, what isthe force of static friction of the car during the turn?


Homework Equations



a = v^2/r, F = (m*v^2)/r ; d = v*t;

The Attempt at a Solution



d/t = v = 3.25 m/s

Fc = (m*v^2)/r = 3250

Fs(static friction) <= Fc
 
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yankees26an said:
d/t = v = 3.25 m/s
What distance is traveled in making the turn? How would you calculate it?
 
Oh whoops the distance would be 2*pi*r/4 since it goes a quarter of a circle.

That would make d/t = 5.1 m/s

Fc = 3.2 kN. Still stuck from there
 
yankees26an said:
Oh whoops the distance would be 2*pi*r/4 since it goes a quarter of a circle.

That would make d/t = 5.1 m/s
Good.
Fc = 3.2 kN.
Don't forget to recalculate Fc using the corrected speed.
 
Doc Al said:
Good.

Don't forget to recalculate Fc using the corrected speed.

Yea I did but I forgot to square! :-p Got it now thanks
 
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