A banked curve and coefficient of friction

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SUMMARY

The discussion focuses on calculating the minimum coefficient of friction required for vehicles to safely navigate a banked curve at a reduced speed due to rain. The curve is designed for a speed of 56 km/h with a radius of 203 m, while the actual speed during rain is 51 km/h. The user calculated the coefficient of friction as 0.021 using the formula: (gsin(x) - a*cos(x))/(gcos(x) + asin(x)), where x is the banking angle of 6.97°. Despite following the correct methodology, the user encountered discrepancies in their results, suggesting potential rounding errors in intermediate calculations.

PREREQUISITES
  • Understanding of banked curves in physics
  • Familiarity with trigonometric functions and their applications
  • Knowledge of basic kinematics, specifically circular motion
  • Ability to perform calculations involving coefficients of friction
NEXT STEPS
  • Review the derivation of the banking angle formula for circular motion
  • Learn about the effects of varying speeds on frictional forces in banked curves
  • Practice solving problems involving coefficients of friction in different weather conditions
  • Explore advanced topics in dynamics, such as centripetal acceleration and its impact on vehicle safety
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as engineers involved in road design and safety analysis.

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Homework Statement



A banked circular highway curve is designed for traffic moving at 56 km/h. The radius of the curve is 203 m. Traffic is moving along the highway at 51 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?

Homework Equations



tan x = v^2/(gr) with the v(designed)

a = v^2/r with the v(rainy)

coef of friction = (gsin(x) - acos(x))/(gcos(x)+asin(x))

The Attempt at a Solution



I have these values for my variables:
v(designed) = 15.6 m/s
v(rainy) = 14.2 m/s
x (the angle of the bank) = 6.97°
a = 14.17^2/203 = 0.989

so, coef of friction should be:
[9.8sin(6.97) - .989cos(6.97)] / [9.8sin(6.97) + 9.8cos(6.97)]
= 0.2075/9.84759
= 0.021

I used the exact same equation on my homework for three practice problems and they all came out right. I've tried 4 different variations of this with different decimal points or rounding the angle to 7° or acceleration to 1, but it still says it's wrong! grrrr...

I only get one more submission before I can't try anymore. Can someone please check my math on this? Thanks so much!
 
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I think it's just rounding error. Try keeping all the digits on intermediate calculations and rounding only the final result.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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