Calculating Centripetal Force? (car moving in circle on a sloped road)

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Homework Help Overview

The discussion revolves around calculating centripetal force for a car moving in a circular path on a sloped road, as presented in an International Baccalaureate mechanics question. Participants are examining the relationship between forces acting on the car, particularly focusing on the reaction force and the horizontal component of that force.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the derivation of the reaction force value and questions the logic behind their own calculations compared to the markscheme. Other participants discuss the implications of increasing speed on centripetal force and the necessary friction to maintain equilibrium.

Discussion Status

Participants are actively engaging with the concepts, with some expressing clarity on the relationship between speed and required centripetal force. There is a recognition of the need for more friction as speed increases, but no consensus has been reached on the original poster's calculation errors.

Contextual Notes

Participants are working within the constraints of an IB mechanics question, and there is mention of a lack of access to a physics teacher for further clarification.

plat4m6
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Calculating Centripetal Force?!? (car moving in circle on a sloped road)

Hi guyz,
i have problem with this IB standard level mechanics question:

here is the link 2 the question: (it is from an International Bacalaureatte examination):

http://img494.imageshack.us/img494/4303/motionofcar5xa.png

The markscheme (answer sheet) has this as the way they got their answer:

horizontal component = R sin14;
= 8500 tan14;
= 2119 N --> which approx is 2100 N


my area of concern is how they arrived at the value for R, here is my calculation

x = 14 degrees, R = Reaction force of the road on the car, F = Horizontal component of the force R

http://www.rahmahwear.com/mymethod.doc [in this file]

i understand how they got their answr in the markscheme, but what's the error in my logic?

as for part d, i thought that if velocity increase, F will increase (i.e. for example.. instead of 2100 it become 3000 but it is still towards center of circle), therefore the car should slide DOWN the ramp?, they say that
"friction must supply larger force2ward center, car tends to slide UP he ramp"

sorry i couldn't ask my own physics teacher.. he's a great bludger.

Thanks in advance
 
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d) If the car speeds up than the centrafugal force needed to keep it at equilibrium is bigger. the weight stays the same.
 
hi daniel

but if car speeds up, the centripetal force will automatically increase since it means acceleration is increasing (a = v^2/r), hence if anything.. the car should remain where it is in same motion.. i think..

i mean.. when u wirl a steel ball attached to piece of string, the faster u spin it.. the Force will continue to piont to center no matter what..

what do u think?
 
plat4m6 said:
i mean.. when u wirl a steel ball attached to piece of string, the faster u spin it.. the Force will continue to piont to center no matter what..
what do u think?
Yes but you need more tension to keep the ball in place - If you spin it to fast then the string will rip and the ball will fly away.
Same here, as you move faster more friction is needed to hold you in place. If there isn't enough friction then like the ball you will fly away.
 
true that, now i understand that part, any suggestion with regard to why my calculation may be wrong?

thanks heaps daniel! lol it actually make sense to me now..
 

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