Finding friction coefficient, car unbanked curve.

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Homework Help Overview

The discussion revolves around determining the minimum coefficient of friction required for a car to safely navigate an unbanked turn. The problem involves a car's mass, the driver's mass, the radius of the turn, and the velocity of the car.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between centripetal acceleration and friction, with one participant attempting to derive the coefficient of friction using the equation mu = (V^2/rg). Questions arise regarding the implications of a coefficient greater than 1 and whether this represents a minimum value.

Discussion Status

Participants are engaged in exploring the implications of a friction coefficient exceeding 1, questioning whether this indicates a stronger force than the normal force and discussing the nature of static friction. There is an ongoing examination of the concept of minimum versus maximum friction coefficients.

Contextual Notes

Participants reference external resources for friction coefficients and discuss the conditions under which coefficients greater than 1 may occur, such as with sticky tires. The discussion reflects on the assumptions related to the forces at play in the scenario.

blee2x2
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Homework Statement


What is the minimum coefficient of friction needed between the tires and the ground so you make the turn safely?[/B]

Car= 1394.7kg
You=75kg
Radius of turn = 40m
Velocity = 20m/s


Homework Equations


Fnet = MAc (Ac = centripetal Acceleration) = Force of friction
Force of Friction (Ff) = Mu(Fn) of Mu(mg)
MAc = (MV^2)/r[/B]

The Attempt at a Solution


(MV^2)/r = Mu(Mg) mass cancels out and i solve for mu ==> mu = (V^2/rg)

mu = 20^2 / (40*9.8)= 1.02

If that is correct is it possible for the friction coefficient to be more than 1?
 
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blee2x2 said:

Homework Statement


What is the minimum coefficient of friction needed between the tires and the ground so you make the turn safely?[/B]

Car= 1394.7kg
You=75kg
Radius of turn = 40m
Velocity = 20m/s


Homework Equations


Fnet = MAc (Ac = centripetal Acceleration) = Force of friction
Force of Friction (Ff) = Mu(Fn) of Mu(mg)
MAc = (MV^2)/r[/B]

The Attempt at a Solution


(MV^2)/r = Mu(Mg) mass cancels out and i solve for mu ==> mu = (V^2/rg)

mu = 20^2 / (40*9.8)= 1.02

If that is correct is it possible for the friction coefficient to be more than 1?
Yes, it can be between the tyre and the asphalt.
 
So since the Friction coefficient is more than 1... that just means it is stronger than the normal force? but is that the minimum?
 
blee2x2 said:
So since the Friction coefficient is more than 1... that just means it is stronger than the normal force? but is that the minimum?
What is minimum? The maximum of the static friction is μmg. Now μ is slightly greater than one. Imagine the tyre is sticky, you put some glue on it. In that case the "force of friction", the contact force between the tyre and asphalt can be quite big.You can find μ-s greater than 1 in this table. http://www.engineeringtoolbox.com/friction-coefficients-d_778.html
μ>1 means that you need greater force to pull the body on the horizontal force than to lift it vertically.
 

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