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Newton's Laws with Friction; Inclines

  • Thread starter rvnt
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  • #1
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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
 

Answers and Replies

  • #2
Doc Al
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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.
 
  • #3
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Ffr=ma
µsN=ma
µsmg=ma
µs=4.8/9.80=0.4897? Than Sin(13degrees)*0.4897...??
 
  • #4
Doc Al
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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.
 
  • #5
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µk(mgCos13deg)+mgSin13deg=a
(0.489)(9.8Cos13deg)+(9.8)(Sin13deg)=a
4.669+2.2045=6.87??
 
  • #6
thrill3rnit3
Gold Member
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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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