Minimum amount of friction in a turn

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    Friction Minimum
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To determine the minimum coefficient of friction for a 3700 kg truck navigating a 25 m unbanked curve at 45 km/h, the correct formula is k(min) = V²/(gr). The initial calculation yielded an incorrect coefficient of 7.62 due to improper unit conversion and substitution. The speed must be converted to meters per second, resulting in 12.5 m/s, leading to the correct evaluation of V². Participants clarified that squaring the speed is necessary for the formula, and confusion arose regarding the use of exponents in calculations. Accurate calculations are essential for determining the required friction for safe navigation of the curve.
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Homework Statement


What minimum coefficient of friction between the tires and the road will allow a 3700 kg truck to navigate an unbanked curve of radius 25 m at a speed of 45 km/h?



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The Attempt at a Solution


The minimum coefficient is 7.62. To arrive at this i used the formula k(min)=(V2/(gr)). This will be substituted to 252/(3700/45).
 
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Your formula is correct, but your substitution is not. Hint: Convert the speed to standard units of m/s.
 
so instead put in 12.5, and then 625/296 = 2.11 ?
 
Last edited:
TheronSimon said:
so instead put in 12.5, and then 625/296 = 2.11 ?
No. Where do the numbers 625 and 296 come from?
 
well 25^2 is 625 and 3700/12.5 is 296
 
TheronSimon said:
well 25^2 is 625 and 3700/12.5 is 296
You need to evaluate: (V2)/(gr)

No need to square 25 or involve 3700.
 
you lost me now, why would i square V^2? :(
 
TheronSimon said:
you lost me now, why would i square V^2? :(
See your first post. (You need to evaluate V^2, which is V squared.)
TheronSimon said:
To arrive at this i used the formula k(min)=(V2/(gr)).
 
but i thought it would only SQRT if both sides have an exponent of 2 to get rid of the 2, because what i do to one side i have to do to the other do i not?
 
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