(790) Problem i 0: A car with mass m-1000 kg completes a turn of radius r-660 m

Question

(790) Problem i 0: A car with mass m-1000 kg completes a turn of radius r-660 m at a constant speed of v 26 m/s. As the car goes around the turn, the tires begin to slip. Assume that the turn is on a level road, i.e, the road is not banked at an angle. Randomized Variables r= 660 m v=26 m/s what is the numeric value of the coefficient of static friction, , between the road and tires? Grade S Potential Submiss sin0 cotanO asin0 acos0 atan0 acotan0 sinho cos0 ted ted ted ted ted 4 5 6 Degrees Radian

Explanation / Answer

Here

centripetal Force = static frictional Force

m*v^2/R = mu_s*m*g

v^2 = mu_s*R*g

mu_s = v^2/(R*g)

mu_s = 26^2/(660*9.8) = 0.104

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