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

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To determine the force required to move the middle block on a steel tabletop, the frictional force must be calculated using the coefficient of friction (u=0.7) and the normal force. The self-locking angle occurs when the applied force equals the maximum static friction, making it impossible to slide the block. A free-body diagram is essential for visualizing the forces acting on both blocks, including gravitational and frictional forces. Understanding the relationship between the angle of the string and the forces involved is crucial to solving the problem. The discussion emphasizes the importance of sketching the scenario for clarity in problem-solving.
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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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