Circular motion question about a car banked

[F|234K]

if a curve with a radius of 60 meter is properly banked for a car traveling at 60km/h, what must be the coefficient of static friction for a car not to skid when traveling at 90km/h?

i know that in order to solve the question, one needs to find the angle first, and i found the angle to be 25.3 degrees(not sure to be correct)...

and the answer is 0.39...i just can't seem to get the answer...

thanks in advance

 

[Doc Al]

i know that in order to solve the question, one needs to find the angle first, and i found the angle to be 25.3 degrees(not sure to be correct)...

That's correct.

and the answer is 0.39...i just can't seem to get the answer...

Since you didn't show any work, there is no way to tell where you got stuck. The basic idea is exactly the same as in the no friction case: the only difference is the addition of the friction force on the car ($\mu N$ acting down the incline).

 

[xanthym]

if a curve with a radius of 60 meter is properly banked for a car traveling at 60km/h, what must be the coefficient of static friction for a car not to skid when traveling at 90km/h?

i know that in order to solve the question, one needs to find the angle first, and i found the angle to be 25.3 degrees(not sure to be correct)...

and the answer is 0.39...i just can't seem to get the answer...

Your calculated road banking angle of 25.3 deg relative to horizontal is correct (also indicated by the previous msg).

The required static friction coefficient for 90 km/h along the same banked road is calculated to be 0.39 in agreement with the provided answer (but apparently not in line with your calculations).

You may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system.

A typical formulation based on the friction horizontal component only is:

$$:(1): \ \ \ \ \frac {v^2} {rg} = Tan(\theta) + \mu_s Cos(\theta)$$

The above equation yields a static friction coefficient of 0.65 for this case. The formulation based on fully coupled horizontal and vertical components is:

$$:(2): \ \ \ \ \frac {v^2} {rg} = \frac {Sin(\theta) + \mu_s Cos(\theta)} {Cos(\theta) - \mu_s Sin(\theta) }$$

This latter formulation yields 0.39 for Coefficient of Static Friction.

The complete system of equations for horizontal and vertical components is given below:

$$:(3): \ \ \ \ \frac{mv^2} {r} = N Sin(\theta) + \mu_s N Cos(\theta)$$

$$:(4): \ \ \ \ 0 = N Cos(\theta) - \mu_s N Sin(\theta) - mg$$

where N is the force Normal to the road surface, (theta) the bank angle, and the other variables defined per usual conventions.


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[F|234K]

thank you guys very much

"You're may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system."

and yes, i got my answer to be 0.59, thank you xanthym for your detailed and consummate reply which enlightens me on the vertical component of friction. (i only calculated the horizontal part of friction.)