Newton's Laws with Friction; Inclines

[rvnt]

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

 

[Doc Al]

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 µ_s. (Note that it's static friction, not kinetic.)

F_f = ma
µ_sN = ma

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

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.

 

[rvnt]

Ffr=ma
µsN=ma
µsmg=ma
µs=4.8/9.80=0.4897? Than Sin(13degrees)*0.4897...??

 

[Doc Al]

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.

 

[rvnt]

µk(mgCos13deg)+mgSin13deg=a
(0.489)(9.8Cos13deg)+(9.8)(Sin13deg)=a
4.669+2.2045=6.87??

 

[thrill3rnit3]

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.