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

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Homework Help Overview

The discussion revolves around determining the minimum force required to move a box resting on a frictionless board, which is itself on a frictionless surface. The problem involves concepts of static and kinetic friction, as well as the dynamics of two bodies interacting under these conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the applied force, the coefficients of friction, and the resulting accelerations of both the box and the board. There are attempts to clarify the question regarding the nature of the force needed and its implications on the motion of the box and board.

Discussion Status

Some participants have provided insights into the forces acting on the box and board, while others seek clarification on the problem statement. There is an ongoing exploration of the dynamics involved, with various interpretations of the forces and accelerations being discussed.

Contextual Notes

Participants note the importance of the coefficients of static and kinetic friction in determining the conditions under which the box will slip off the board. The setup assumes a frictionless environment, which raises questions about the applicability of typical friction equations.

lvelar
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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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