Centripetal force of a car on a curve

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Homework Help Overview

The discussion revolves around a physics problem involving a car navigating a banked curve, focusing on the forces acting on the car, including centripetal force, normal force, and friction. The problem specifies parameters such as the car's mass, curve radius, banking angle, and coefficient of static friction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate centripetal force to friction force and expresses confusion about the roles of these forces in the context of a free-body diagram. Other participants provide clarifications regarding the components of forces acting on the car and the direction of friction. There are questions about the placement of force arrows and their implications for the equations being used.

Discussion Status

Participants are actively engaging with the problem, offering insights and clarifications. Some guidance has been provided regarding the interpretation of forces, but there remains uncertainty about how to isolate the normal force in the equations presented by the original poster. Multiple interpretations of the force diagram are being explored.

Contextual Notes

The discussion includes considerations about the correct representation of forces in a free-body diagram and the potential confusion arising from multiple normal forces in the equations. There is also an acknowledgment of the complexity introduced by the banking angle and friction in the problem setup.

sheepy
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Homework Statement


A 600 kg car is going over a curve with radius 120m that is banked at angle of 25 degrees with speed of 30 m/s. The coefficient of static friction between car and road is 0.3. What is the normal force?

Homework Equations


F(centripetal)= (mv^2)/r
F(static) = Mu(normal force)

The Attempt at a Solution


I think that the centripetal force is equal to the friction force. So I can just substitute centripetal force for friction. But my teacher also said that is a parallel force but I'm not sure it is. From the free-body diagram, I will get..
F(c)= (Friction + Parallel force) Cos(angle)

I'm actually quite confused ~_~
 
Last edited:
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See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html#c1

The normal force means the normal force on the back which comes from the component of the weight point normal to the bank and the component of centrifugal force pointing normal to the bank.

The weight points downard (vertically) while the centrifugal/centripetal force points horizontally, but both have components pointing normal to the bank.
 
cool, thanks
it makes more sense now..
except how come the friction force is up there? usually my teacher draws it on the other side of the car, like on the same side as the centripetal force. does it matter? since its negative now instead of positive and etc?
 
The friction force operates at the contact surface between tires and bank (road surface). It doesn't matter where one draws the arrow, although it should be correctly drawn at the tire/road interface, but the direction should point down the bank since the friction opposes the force which try to push the car up the bank.
 
i got: x-component: Nsin25 + mu(N)(cos25) = (mv^2)/r
y-component: Nsin25 - mg - mu(N)(sin25) = 0

i'm not sure how to find N from that since, there's like 2 N's in each equation? and i can't seem to just factor it out...or did i just get the equation wrong entirely?
 

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