What Is the Self-Locking Angle for a Block on a Steel Tabletop?

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SUMMARY

The discussion focuses on determining the self-locking angle for a block on a steel tabletop, specifically involving a 10kg block with a 3kg block on top. The coefficient of friction (u) is given as 0.7. To find the angle at which the middle block self-locks, it is essential to analyze the forces acting on both blocks using the equations of motion and friction. The self-locking angle is reached when the applied force equals the maximum static friction force, resulting in an infinite force requirement to slide the block.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of frictional forces (Ffr=uFn)
  • Ability to draw and interpret free-body diagrams
  • Familiarity with static and kinetic friction concepts
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  • Calculate the net force acting on the blocks using F=ma
  • Explore the implications of different angles on frictional forces
  • Study the effects of varying the coefficient of friction on self-locking angles
  • Learn about static versus kinetic friction in practical applications
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Students in physics or engineering courses, educators teaching mechanics, and anyone interested in understanding friction and motion dynamics in practical scenarios.

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Homework Statement


A 10kg block sits on a steel tabletop. A 3kg block sits on the first block. A string connects the top block to the table.

a) If the string makes an angle of 10 degrees with the horizontal, how much force is required to move the middle block?

b) At what angle will the block "self lock" (an infinite force is required to slide the middle block)?

u=0.7

Homework Equations



F=ma

Ffr=uFn

Fnet(first block)= Fapp-(Ffr+Ffr) ?

The Attempt at a Solution



I really have no idea.

I don't get the self lock thing at all.
 
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Sketch the problem first : table, block, the other block on top, string.
Draw the free-body diagram for both blocks.

ehild
 

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