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Mechanics}Centripetal Force


\vec{F}_air acts in the direction as shown and the centripetal acceleration acts in the direction of \vec{F}_A. Centripetal acceleration is a net force, however, and thus,

f_x is in the positive direction and f_y is in the negative direction. Thus the force of the road is \vec{F}_B.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
superPOOPERscooper
2013-10-17 08:50:18
I honestly think the tires are the most confusing part of this problem. There are two forces acting on the car: the drag force of air, indicated as such in the diagram, and the centripetal force, indicated by F_a in the diagram. Remember that centripetal force \frac{mv^2}{R} \emph{only} dictates the numerical value of the force -- it does NOT indicate which agency provides the force! In this light, we could replace the "tires" by anything: a string (where the tension would provide the centripetal force), a planet (where gravitation provides the centripetal force), or even a charged particle in a magnetic field (where Lorentz provides the centripetal force). Anyway, adding F_{air} to F_a gives choice B.Alternate Solution - Unverified
99percent
2008-11-06 06:20:59
Centripetal force is in the direction \vec{F_A}

Tires exert a force in the backward direction so that the car moves in the forward direction - By Newton's third law (Action-Reaction) - The road exerts an equal force in the forward direction, i.e., \vec{F_C}

the resultant of \vec{F_A} and \vec{F_C} will be in the direction \vec{F_B}

Bingo..!!
Alternate Solution - Unverified
Comments
enterprise
2018-04-01 13:56:41
Well , The total resultant force on the car is in the radial direction. Obviously you have a drag force backwards. You should ask , What should be the direction of the force such that when you add it to the drag force , you will get the resultant centripetal force ? Obviously , Only , F_B has a direction such that it\\\\\\\'s tangent component must cancel out with the the F_air , so that the only force that survives is in the radial direction. The tangential force must cancel out so that the speed is uniform. Otherwise , you will have a torque and angular momentum will change. NEC
superPOOPERscooper
2013-10-17 08:50:18
I honestly think the tires are the most confusing part of this problem. There are two forces acting on the car: the drag force of air, indicated as such in the diagram, and the centripetal force, indicated by F_a in the diagram. Remember that centripetal force \frac{mv^2}{R} \emph{only} dictates the numerical value of the force -- it does NOT indicate which agency provides the force! In this light, we could replace the "tires" by anything: a string (where the tension would provide the centripetal force), a planet (where gravitation provides the centripetal force), or even a charged particle in a magnetic field (where Lorentz provides the centripetal force). Anyway, adding F_{air} to F_a gives choice B.Alternate Solution - Unverified
sina2
2013-09-21 06:06:44
This is the way I figure out the correct answer. I think {F}_{air} has nothing to do with the force applied by the road to the tires. The car moves forward at the moment, we know the velocity is zero in point tires touch the road, hence tires make a {f}_{s} to the road in backward and oppositely, the road applies the same force at the forward to the tires. There's just another force we should think about it. In a rainy day, in defile of road, why we should reduce the speed, because it's possible we're thrown away of road toward outside. So, there is a centripetal force that keeps the car moving. So we have {F}_{C} and {F}_{A} and the result force would be {F}_{B}.NEC
tsharky87
2009-11-04 23:40:25
It seems to me that if we assume the car is back-wheel driven, then the force of the road on the back tires will be \vec{F_C}, and the force of the road on the front tires (which must be turned to the right) would be \vec{F_A}. So the driving tires are what makes it go forward and the turning tires are what pulls it toward the center, which makes sense. The resulting force of these is obviously \vec{F_B}.

But if we assume the car is front-wheel drive, then the back wheels are doing nothing but holding up the car, and the front wheels are the only force. Since the wheels must be turned to the right, the road is pushing the front wheels along where they are directed, \vec{F_B}.

I don't like this problem much and I got it wrong but that's just some insight that occurred to me.
NEC
Albert
2009-10-21 15:12:32
Thank you Yosun, for this wonderful site, I am learning a great deal!NEC
99percent
2008-11-06 06:20:59
Centripetal force is in the direction \vec{F_A}

Tires exert a force in the backward direction so that the car moves in the forward direction - By Newton's third law (Action-Reaction) - The road exerts an equal force in the forward direction, i.e., \vec{F_C}

the resultant of \vec{F_A} and \vec{F_C} will be in the direction \vec{F_B}

Bingo..!!
jmason86
2009-09-30 17:09:55
You said "The road exerts an equal force in the forward direction Fc".
The problem asks for "the horizontal force of the road on the car's tires."

It seems like by this logic, you should just go with (C). This isn't the right answer though...
pam d
2011-09-11 17:36:13
Just to be clear, you are implying here that the answer is B right?
postal1248
2013-10-03 16:37:51
I believe that by "horizontal" ETS means not the normal force, so technically all the forces in the diagram are horizontal because they lie in the horizontal plane. So the net horizontal force would be given by choice (B).
Alternate Solution - Unverified
isina
2008-10-16 14:45:53
I do not completely agree with the answer on the grounds that it is not the only solution. Even if there is no wind force there is still force in +x direction. The reason that there is forward friction force is as follows: as the car moves in +x direction, its tires rotate in the reverse direction (at the point of contact their velocity vector points -x). So the friction of the road on the tires is actually to the opposite direction of the cars tires movement and on the same direction as the movement of the car. This results in a +x directed friction (hence the car moves).
fpaslan
2011-10-12 07:21:47
But the velocity of the tires at the contact point is zero. So it is again static friction which prevents the tires from sliding like FA
NEC
evanb
2008-06-30 11:50:46
Just a small typo: you need braces for F_{air} so that it doesn't look like F_airTypo Alert!
yosun
2005-11-10 02:00:50
BWHB: thanks for the typo-alert; it has been corrected.
bcomnes
2011-10-09 17:57:55
Not in the PDF! I couldn't figure out what I was doing wrong! Apparently nothing.
NEC
BWHB
2005-11-10 01:38:49
All the work is right but the answer is wrong. It's FB. NEC

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