Newton's Laws with Friction; Inclines

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Homework Help Overview

The discussion revolves around a physics problem involving Newton's laws, specifically focusing on the effects of friction on a car's deceleration when transitioning from a level road to an inclined road at 13 degrees. The original poster is attempting to determine the deceleration of the car under these conditions while considering the coefficient of static friction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between force, mass, and acceleration, as well as the role of static versus kinetic friction. There are attempts to calculate the coefficient of static friction and its implications for the car's maximum acceleration on an incline.

Discussion Status

Participants are actively engaging with the problem, offering guidance on identifying the normal force and clarifying the distinction between static and kinetic friction. There is an ongoing exploration of how to apply Newton's second law to the scenario, with some calculations being shared and questioned.

Contextual Notes

There is a noted assumption that the coefficient of static friction remains constant, and participants are working under the constraint of not skidding while calculating the car's deceleration on the incline.

rvnt
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Homework Statement



A car can decelerate at -4.80m/s^2 without skidding when coming to rest on a level road. What would its deceleration be if the road were inclined at 13degrees uphill? Assume the same static friction coefficient.

Homework Equations


Ffr=µkFN


The Attempt at a Solution


I am having lots of trouble with friction related problems when the coefficient of static or kinetic friction is involved! But I was thinking...F=ma... F=m(-4.80)...
Ffr=µkFN= -4.8m=µk(9.80)...really not sure with these types..please help
 
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rvnt said:
But I was thinking...F=ma... F=m(-4.80)...
Ffr=µkFN= -4.8m=µk(9.80)...really not sure with these types..please help
You're on the right track for finding µs. (Note that it's static friction, not kinetic.)

Ff = ma
µsN = ma

What's the normal force, N? Then you can solve for µs.

Note that we're assuming the car is accelerating as quickly as possible without skidding, so it's the maximum value of static friction that we want.
 
Ffr=ma
µsN=ma
µsmg=ma
µs=4.8/9.80=0.4897? Than Sin(13degrees)*0.4897...??
 
rvnt said:
Ffr=ma
µsN=ma
µsmg=ma
µs=4.8/9.80=0.4897?
Good.
Than Sin(13degrees)*0.4897...??
No.

Now you have to use what you've found for µs to solve a new problem: what would the car's maximum acceleration be (without skidding) if it were going up that hill? Identify the forces and apply Newton's 2nd law.
 
µk(mgCos13deg)+mgSin13deg=a
(0.489)(9.8Cos13deg)+(9.8)(Sin13deg)=a
4.669+2.2045=6.87??
 
rvnt said:
µk(mgCos13deg)+mgSin13deg=a
(0.489)(9.8Cos13deg)+(9.8)(Sin13deg)=a
4.669+2.2045=6.87??

Looks good, except that it's µs because it's static friction and it's negative because it is deceleration.
 

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