Question

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is μs. The coefficient of kinetic friction between the board and the box is, as usual, less than μs.
Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.

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Find Fmin, 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).

Express your answer in terms of some or all of the variables mu_s, m_1, m_2, g, and L. Do not include F_f in your answer.

Answer 1

Answer: Fmin = (m₁ + m₂) μsg

Explanation:

To begin, we would first define the parameters given in the question.

Mass of the box = m₁

Mass of the board = m₂

We have a Frictionless surface given that Fr is acting as the frictional force between the box and the board.

from our definition of force, i.e. the the frictional force against friction experienced by the box, we have

Fr = m₁a ...................(1)

Also considering the force between the box and the board gives;

Fr = μsm₁g .................(2)

therefore equating both (1) and (2) we get

m₁a = μsm₁g

eliminating like terms we get

a = μsg

To solve for the minimum force Fmin  that must be applied to the board in order to pull the board out from under the box, we have

Fmin  = (m₁ + m₂) a  ...........(3)

where a = μsg, substituting gives

Fmin = (m₁ + m₂) μsg

cheers i hope this helps