## ArchonX5 2 years ago A 600-kg car is going over a curve with a radius of 120 meters that is banked at an angle of 25 degrees with a speed of 30 meters per second. The coefficient of static friction between the car and the road is 0.3. What is the normal force exerted by the road on the car? a) 7240 N b) 1590 N c) 5330 N d) 3430 N e) 3620 N Note: The speed is not Vmax (roughly 32 m/s). I've been working under the assumption that mgcos(theta) will be increased by the frictional component acting downward. I simply can't seem to get the solution.

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1. ArchonX5

6488N is the correct answer (as given by teacher), but it was not worked for us. The link posted was very wrong, additionally you changed the value of gravity. Assume the car is driving on earth.

2. Vincent-Lyon.Fr

With N = 9.81 m/s², I find N = 7236 N There must be a mistake in the solution given to you.

3. Fifciol

I find also 7236 N

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5. dvsk000

@Fifciol and @Vincent, Guys is my approach right? I somehow don't get your results. Pls share ur knowledge.. The normal force|dw:1374731413407:dw|

6. Fifciol

@dvsk000 In your frame of reference you see an object moving at circumference so there must be centripetal force acting towards the center of a circle. But the object itself is in non inertial frame of reference (newton's second law doesn't hold unless you take into account fictitious forces) so it really feels centrifugal force acting backwards the center with the same magnitude as centripetal force. The free body diagram looks like this:|dw:1374741332658:dw So normal force is simply mgcos(theta) +mv^2/Rsin(theta) =7236 N

7. dvsk000

@Fifciol got it! Thank you. I had resolved the centripetal force wrongly. The vertical centripetal force should be mv^2/Rsin(theta). Thanks :)

8. VALERIAN34

Friction as no effect. Normal force is sum of force projected resulting of centripetal force and earth accelaration. m.g.cos\[m timesg \times \cos \alpha +m timesv ^{2}divR times sin alpha 5334+1902=7236 N