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Frictional Force and Acceleration

  • Thread starter LCB
  • Start date
  • #1
LCB
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A 3.80-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.500. Determine the kinetic frictional force that acts on the box when the elevator is

(a) accelerating upward with an acceleration whose magnitude is 2.70 m/s2, and

(b) accelerating downward with an acceleration whose magnitude is 2.70 m/s2.

For both, I started out with Ff=uFn
For (a), I did: Fn + 2.7 = 3.8. Solved for Fn, and of course, my answer was wrong.
For (b), I did: 3.8 + 2.7 = Fn. Again, wrong answer.
I did everything the way my teacher taught me to. I'm incredibly frustrated; I don't understand my mistake(s).
 

Answers and Replies

  • #2
LCB said:
A
(a) accelerating upward with an acceleration whose magnitude is 2.70 m/s2, and

For both, I started out with Ff=uFn
For (a), I did: Fn + 2.7 = 3.8. Solved for Fn, and of course, my answer was wrong.
I think gravity is working in the elevator, so you need it in your equations. If your elevator is going up, then it will press harder against the box, so the acceleration will be (2.7 m/s^2 + g). And if the elevator is going down, then it is pressing less, so the total acceleration will be less than g.

And then, you must use F=ma to get the normal force, once you have figured out the a.

I hope this helps you out.
Dot
 

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