Finding forces on a car rounding a curve

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    Car Curve Forces
AI Thread Summary
A car with a mass of 1200 kg is analyzed while turning with a radius of 40 m at a speed of 15 m/s, raising questions about the time to complete a half-circle, frictional forces, acceleration to 40 m/s, and banking angles for curves. The time to make a half turn is calculated to be approximately 16.75 seconds, while the maximum static frictional force is determined to be 6750 N. The discussion highlights uncertainty regarding the acceleration phase and the calculation of banking angles, with an initial angle estimate of 1.11 degrees deemed too low. The importance of maintaining consistent units throughout calculations is emphasized to avoid errors. Overall, the thread focuses on applying physics equations to solve real-world automotive dynamics problems.
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Homework Statement


A car with mass of 1200kg turns sharply with a radius of 40m and at 15/ms. The tires have a static friction of 0.9, rolling at 0.6 and kinetic at 0.3.

1) how long does it take to make a turn at half a circle.
2) what is the magnitude of frictional force on the tires in the turn
3) After the turn, the car accelerates to 40m/s, assuming the divers uses max acceleration from friction, how long does it take to reach the new speed.
4) the driver takes another curve at 40 m/s with a radius of 80 and skids. the roads are banked. what angles does the road need to be banked to prevent skidding.

Homework Equations


F=ma
mv^2/r

The Attempt at a Solution



a) t=\frac{2\Pi R}{v}= \frac{2 Pi (40)}{15} so 16.75 seconds

b) F=ma
F_{smax}=1200\frac{v^2}{r}
F_{smax}=1200\frac{15^2}{40}
=6750N
c) have no clue how to do C...im not sure how to go about it...

D) arctan ∅= \frac{v^2}{gR}

arctan ∅ \frac{40^2}{-9.8m/s*80}

I get theta 1.11° degrees but that seems kindsa low.
 
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1/a: half a circle, not a full circle.
2/b: I agree
3/c: Given the coefficient of friction, what is the maximal horizontal force between car and ground?
4/d: Did you draw a sketch?

In general, keeping the units everywhere can help to spot mistakes.
 
baird.lindsay said:
arctan ∅ \frac{40^2}{-9.8m/s*80}
I get theta 1.11° degrees but that seems kindsa low.
No, 1.11 radians.
 
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