Question

how large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 125 m at a speed of 26.4 m/s?

Answer 1

The coefficient of static friction be between the tires and the road is 0.568.

Step-by-step explanation:

Equation for Centripetal Force

`F_c=(mv^2)/(r)`

We know that the maximum static friction (at which the tires roll but do not slip) is
`u_s*N`, where
`u_s` is the static coefficient of friction and
`N` is the normal force. The normal force equals the car’s weight on level ground, so
`N=mg`. Thus the centripetal force in this situation is

`F_c=u_smg`

Now we have a relationship between centripetal force and the coefficient of friction. Therefore we can say

`(mv^2)/(r)=u_smg`

Lets solve for
`u_s`

Divide both sides by
`mg`.

`((mv^2)/(r))/(mg) =(u_smg)/(mg)`

Simplify the right side by cancelling the common factor of
`m`.

`((mv^2)/(r))/(mg) =(u_sg)/(g)`

Simplify the right side by cancelling the common factor of
`g`.

`((mv^2)/(r))/(mg) =u_s`

Simplify the left side by multiplying the numerator by the reciprocal of the denominator.

`(mv^2)/(r)*(1)/(mg) =u_s`

Combine the fractions.

`(mv^2*1)/(rmg) =u_s`

Simplify the left side by cancelling the common factor of
`m`.

Finally we get

`u_s=(v^2)/(rg)`

We are given

`r=125`

`v=26.4`

`g=9.81`

Substitute these values into our equation.

`u_s=(26.4^2)/(125*9.81)`

`u_s=(696.96)/(125*9.81)`

`u_s=(696.96)/(1226.25)`

`u_s=0.568`