Frictional Force and Acceleration

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SUMMARY

The discussion focuses on calculating the kinetic frictional force acting on a 3.80-kg box in an elevator with a coefficient of kinetic friction of 0.500. When the elevator accelerates upward at 2.70 m/s², the normal force (Fn) increases due to the additional acceleration, resulting in a higher frictional force. Conversely, when the elevator accelerates downward at the same rate, the normal force decreases, leading to a lower frictional force. The correct approach involves incorporating gravitational acceleration (g) into the calculations for both scenarios.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of kinetic friction and its coefficient
  • Ability to calculate normal force in varying acceleration scenarios
  • Familiarity with basic physics equations, specifically F=ma
NEXT STEPS
  • Review the concept of normal force in non-inertial reference frames
  • Study the effects of acceleration on frictional forces in different contexts
  • Practice problems involving friction in elevators and other accelerating systems
  • Learn about the implications of gravitational force on objects in motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to friction and acceleration in non-inertial frames.

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).
 
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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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