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
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 is [tex]\mu_{s} g({m_{1}+m_{2}})[/tex].
Given data:
The mass of small box is, [tex]m_{1}[/tex].
The mass of board is, [tex]m_{2}[/tex].
The length of board is, L.
The coefficient of static friction between the board and box is, [tex]\mu_{s}[/tex].
The linear force acting between the box and the board provides the necessary friction to box. Therefore,
[tex]F=F_{f}\\F=\mu_{s}m_{1}g\\m_{1} \times a = \mu_{s} \times m_{1}g\\a= \mu_{s} \times g[/tex]
a is the linear acceleration of board.
Then, the minimum force applied on the board is,
[tex]F_{min}=({m_{1}+m_{2}})a\\F_{min}=({m_{1}+m_{2}})(\mu_{s} \times g)\\F_{min}=\mu_{s} g({m_{1}+m_{2}})[/tex]
Thus, 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 is
[tex]\mu_{s} g({m_{1}+m_{2}})[/tex].
Learn more about the frictional force here:
https://brainly.com/question/4230804?referrer=searchResults
