When Does a Book Slide Off a Tilting Shelf?

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SUMMARY

The discussion revolves around a physics problem involving a book sliding off a tilting shelf. The shelf is 6.1 meters long, and the coefficient of static friction for the book is μS = 11/60. After 5.5 seconds of the librarian lifting one end of the shelf at a constant rate of 0.20 m/s, the book begins to slide. Participants suggest using the relationship a = g sin(θ) to derive the acceleration and emphasize the importance of integrating this function to find the time it takes for the book to fall off the shelf.

PREREQUISITES
  • Understanding of static and kinetic friction, specifically μS = 11/60.
  • Knowledge of basic calculus, particularly differentiation and integration.
  • Familiarity with the concepts of angular velocity and acceleration in physics.
  • Ability to analyze free-body diagrams and apply Newton's laws of motion.
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  • Learn how to derive equations of motion for objects on inclined planes.
  • Study the principles of energy conservation in dynamic systems.
  • Explore the relationship between angular velocity and linear speed in physics.
  • Investigate the effects of varying coefficients of friction on motion.
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Students studying physics, particularly those focusing on dynamics and motion, as well as educators looking for practical examples of applying calculus to physical problems.

  • #31
NihalSh said:
@Simon Bridge
I know about that technicality, but I was trying to avoid it. You already know this is a basic Physics question. The poster hasn't even studied rotation yet. I believe the question was set with a mindset that the student will apply conservation of energy (although you know why this cannot be applied). If someone has a better method to go about it, I would really appreciate it.
Pls take a look at my post #24.
 
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  • #32
haruspex said:
Pls take a look at my post #24.

yeah, looks like OP would have to solve that differential to get the precise result and would also have to get familiar with rotational variables!

Edit: If we use this method (which we should), the breakaway condition would change:

$$μ.g.Cos(ω.t)+(3.05).ω^2=g.Sin(ω.t)$$
 
Last edited:
  • #33
NihalSh said:
yeah, looks like OP would have to solve that differential to get the precise result and would also have to get familiar with rotational variables!

Edit: If we use this method (which we should), the breakaway condition would change:

$$μ.g.Cos(ω.t)+(3.05).ω^2=g.Sin(ω.t)$$
Good point.
 

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