Understanding Friction: Solving for Forces on Boxes A and B

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Applying an 18N force to box A is the minimum required to overcome static friction and initiate movement. At this force, box A is on the verge of moving, and once it begins to move, kinetic friction takes effect, causing it to accelerate. Box B experiences two horizontal forces: the kinetic friction from box A and the friction between box B and the ground. The total of these forces is used in the equation F = ma for box B. Understanding these frictional forces is crucial for solving the problem accurately.
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Homework Statement



[PLAIN]http://img510.imageshack.us/img510/3893/58096739.jpg

Homework Equations



F = ma, friction...

The Attempt at a Solution


I just want to understand the problem so I can do it.

1a. If I apply 18N to box A, does it actually move? or it just reaches the max friction and still stay put?

1b. how many forces should be on box B? I see 2 forces, the kinetic friction given by A on B and the static (or kinetic?) friction between box B and the ground
 
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Hi Iva! :smile:
IvaNMK said:
1a. If I apply 18N to box A, does it actually move? or it just reaches the max friction and still stay put?

The 18 N applied force in the question is stated to be the minimum force needed to get box A to move (ie, "to overcome the maximum static friction").

There's two answers to your question! …
i] no, it's only "on the point of moving" or "about to move", to quote various exam questions on static friction
ii] yes, it moves imperceptibly slowly at first, but then the coefficient of kinetic friction takes over, so it accelerates. :wink:

1b. how many forces should be on box B? I see 2 forces, the kinetic friction given by A on B and the static (or kinetic?) friction between box B and the ground

Yes, only 2 (horizontal) forces on B …

their total is the F in F = ma for B. :smile:
 
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