Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
A 600kg 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.
 10 months ago
 10 months ago
A 600kg 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.
 10 months ago
 10 months ago

This Question is Open

ArchonX5Best ResponseYou've already chosen the best response.0
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.
 10 months ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
With N = 9.81 m/s², I find N = 7236 N There must be a mistake in the solution given to you.
 10 months ago

sadiBest ResponseYou've already chosen the best response.0
can u tell me friends how can i download books related to physics
 9 months ago

dvsk000Best ResponseYou've already chosen the best response.0
@Fifciol and @Vincent, Guys is my approach right? I somehow don't get your results. Pls share ur knowledge.. The normal forcedw:1374731413407:dw
 8 months ago

FifciolBest ResponseYou've already chosen the best response.1
@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
 8 months ago

dvsk000Best ResponseYou've already chosen the best response.0
@Fifciol got it! Thank you. I had resolved the centripetal force wrongly. The vertical centripetal force should be mv^2/Rsin(theta). Thanks :)
 8 months ago

VALERIAN34Best ResponseYou've already chosen the best response.0
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
 23 days ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.