Why N·sin(θ) ≠ (m·v²)/r when a car is stationary on a banked curve

[annamal]

If there is a car resting on a banked curve with angle theta, velocity v = 0, but N (normal)*sin(theta) > 0. So N*sin(theta) =/= (m*v^2)/r with v = 0. But my physics textbook just defined N*sin(theta) = (m*v^2)/r in banked curve. What is going on here?

 

[Andrew Mason]

$\vec N\sin\theta$ is the horizontal component of the normal force. The Normal force, N, is the component of $\vec F_g = -m\vec g$ that is perpendicular to the surface of the road.

This component of the normal force provides contributes to the centripetal force required to keep the car turning with the road. So the bank reduces the need for friction between the road and tires to supply all the force needed for the car to make the turn successfully.

AM

 

[annamal]

$\vec N\sin\theta$ is the horizontal component of the normal force. The Normal force, N, is the component of $\vec F_g = -m\vec g$ that is perpendicular to the surface of the road.

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This component of the normal force provides contributes to the centripetal force required to keep the car turning with the road. So the bank reduces the need for friction between the road and tires to supply all the force needed for the car to make the turn successfully.

AM

Yes, I understand that part, but my question is if the car were at rest. v = 0, but N*sin(theta) > 0

 

[Lnewqban]

There is a horizontal component trying to slide the car down the slope, but that is from the weight of the stopped car, not from the centrifugal effect.

 

[Andrew Mason]

$\frac{mv^2}{r}$ is the centripetal force that the designers want the horizontal component of normal force to provide for the expected or recommended car speed rounding that curve. The designers first determine what the appropriate range of speed should be for cars rounding that curve. The coefficient of static friction factors into this appropriate range because they don't want cars to slide up or down when rounding the curve because they are traveling in the upper or lower parts of that range. Then they determine what $\theta$ should be such that for v in the middle of that appropriate range $F_N\sin\theta=mv^2/r$.

AM

 

[nasu]

If the car is at rest on the slope there must be some friction. Otherwise the car will slide down the slope. You should include the friction components in the fee body diagram and the Newton's laws. And of course that there is no $$ mv^2/r $$ term as there is no centripetal acceleration in this case. You should realize that the normal force depends on the situation, there is no universal formula that gives the normal force for any system and any state of motion.