Minimum amount of friction in a turn

1. Mar 18, 2012

TheronSimon

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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).

2. Mar 18, 2012

Staff: Mentor

Your formula is correct, but your substitution is not. Hint: Convert the speed to standard units of m/s.

3. Mar 18, 2012

TheronSimon

so instead put in 12.5, and then 625/296 = 2.11 ???

Last edited: Mar 18, 2012
4. Mar 18, 2012

Staff: Mentor

No. Where do the numbers 625 and 296 come from?

5. Mar 18, 2012

TheronSimon

well 25^2 is 625 and 3700/12.5 is 296

6. Mar 18, 2012

Staff: Mentor

You need to evaluate: (V2)/(gr)

No need to square 25 or involve 3700.

7. Mar 18, 2012

TheronSimon

you lost me now, why would i square V^2? :(

8. Mar 18, 2012

Staff: Mentor

See your first post. (You need to evaluate V^2, which is V squared.)

9. Mar 19, 2012

TheronSimon

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?