What Angle Is the Road Banked for a 1000kg Car on a 25m Radius Track at 30m/s?

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To determine the banking angle of a road for a 1000kg car on a 25m radius track at 30m/s, the centripetal force must equal the component of the gravitational force acting towards the center of the curve. The normal force and gravitational force can be analyzed using a free body diagram, which allows for the application of trigonometry to find the angle θ. The centripetal force equation Fc = mv²/r is crucial for this calculation. The challenge lies in expressing the forces in terms of θ, as the angle is initially unknown. Ultimately, solving for θ will provide the necessary banking angle for the track.
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Homework Statement


a 1000kg car negotiates a banked friction free track or radius 25m with speed 30m/s

to what angle to the horizontal is the road banked?


Homework Equations





The Attempt at a Solution


tried to calculate the normal force of the car, to find the angle between the weight vector and the normal force vector, which would be equal to the angle that the road is banked.
However, I could not find the angle fo the normal force
 
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If the angle of baking is θ, what is the component of the weight that points towards the center of the path?
 
mgsin theta, you don't know theta though..
 
superaznnerd said:
mgsin theta, you don't know theta though..

A free body diagram will allow you to use trigonometry to find theta.
 
superaznnerd said:
mgsin theta, you don't know theta though..

if that force provides the centripetal force, you can find θ.

Fc=mv2/r remember?
 
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