What Is the Minimum Force Needed to Move a Box on a Frictionless Board?

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To determine the minimum force needed to move a box on a frictionless board, the key is understanding the relationship between the static friction and the forces acting on both the box and the board. The box, with mass m_1, experiences a maximum static friction force of μ_s * m_1 * g, which dictates its acceleration. The board, with mass m_2, will accelerate based on the net force applied minus the friction force acting against it. The required force must exceed the static friction threshold to ensure the box slips off the board. The solution involves calculating the force that results in the maximum acceleration of the box without exceeding static friction limits.
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Please help! Finding minimum force!

Homework Statement



A small box of mass m_1 is sitting on a board of mass m_2 and length L (Intro 1 figure) . The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is mu_s. The coefficient of kinetic friction between the board and the box is, as usual, less than mu_s.


Homework Equations



acceleration of box = μs *g
acceleration of board = ((F- gμs(m1)) / (m2)


The Attempt at a Solution


my solution was a= μs *g*m1
but it was wrong
 
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What is the question?
 


Find , the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).
 


Sketch. Show F = umg pulling the box forward and the board backward.
For the box the sum of forces is umg = ma, a = ug is the maximum acceleration because umg is the maximum pulling force that can be applied to it. Here u is the static coefficient of friction. If the board accelerates the tiniest bit more than that, it will slip.
 
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