Solve the coefficient of friction in a problem involving circular motion

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SUMMARY

The discussion focuses on calculating the coefficient of static friction required for a car to navigate a banked curve without skidding. Given a curve radius of 88 meters and a car speed of 95 km/hr, the user initially miscalculated the coefficient, arriving at an incorrect value of 9015. The correct approach involves using the equation µ = v²/(gr), where v is the speed converted to meters per second, g is the acceleration due to gravity (9.8 m/s²), and r is the radius of the curve. The final answer, as confirmed by the textbook, is a coefficient of static friction of 0.22.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Knowledge of free body diagrams
  • Familiarity with the equations of motion, specifically F = ma
  • Basic trigonometry, particularly tangent functions
NEXT STEPS
  • Study the derivation of the centripetal force equation in circular motion
  • Learn about the effects of banking angles on vehicle dynamics
  • Explore the relationship between speed, radius, and friction in circular motion
  • Practice converting units, specifically from km/hr to m/s, for physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of applying friction concepts in real-world scenarios.

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Homework Statement


If a curve with a radius of 88m is perfectly banked for a car traveling 75km/hr, what must be the coefficient of static friction for a car not to skid when traveling at 95km/hr?


Homework Equations


i guess relevant equations would be tan θ = v^2/rg, but its says next to friction not needed?
f=ma or fr=mar
and Ffr = µs x Fn


The Attempt at a Solution



I made a free body diagram and since it said it was banked i set that in the y-axis there's no acceleration, so following the book (Giancoli), I got Fn=mg/cos θ. I solved for that and got 95^2/(88)(9.8) = .40 then inverse tan 10.46 or 84.5 for the angle

then in the x-axis i got Fn sin θ - Ffr = mar for the sum of the forces. I substituted Ffr for µs x Fn and Fn = mg cos θ. I
I then had Fn sin θ - µs mg cos θ = mar
I divided mass from both sides and divided both sides by m.
I then for ar substituted v^2/r
Then I isolated µs by dividing cos θ and subtracting Fn sin θ
So I then had µs = V^2/r x cos θ - Fn sin θ
I plugged everything in and I got 9015, I know this wrong just by looking at it, the back of the book says .22, I am completely clueless, so if anyone could help, I'd greatly appreciate it. Sorry if it is a little confusing.
 
Physics news on Phys.org
The centripetal force of the car is caused by friction

The car is accelerating at a = \frac{v^2}{r}

Equating the force of friction and the centripetal force, we have

F_f = \mu F_N = ma

\mu * mg = \frac{mv^2}{r}

\mu = \frac{v^2}{gr}
 
thank you, I forgot to convert 95km/hr to m/s too, anyways thanks again!
 

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