Static Friction: Finding F_min for Box on Board

[Spectre32]

Alright folks i have a rather simple problem. I trying to figure this problem out: A small box of mass m_1 is sitting on a board of mass m_2 and length L . 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 sliding friction between the board and the box is, as usual, less than mu_s .

Throughout the problem, use G for the acceleration due to gravity.
(also there is a F vector of force indicating that something is pulling on the board to the right)

Ok and then the question asks: Find F_min, 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 F_min in terms of m_1, m_2, g mu_s and/or L.


Now i have figured out the formuls for the accerleration for the box. The accerleration would be F_f/M_1. The max acceleration would be mu_s*g. The only problem that i am having is figureing out the acceleration of the board. I was hinting at something along the lines of mu_s-F/m_1+m_2g Reason being that friction acts in the opposite direction of the force, but i think something is not right with my logic. Any help would be appericated.

 

[gnome]

I'm not sure what you were trying to write here: mu_s-F/m_1+m_2g.

But look -- you already solved for the acceleration. That is the maximum acceleration that the static friction between the board and the box can sustain. That is the critical point at which the box starts to slip. Now it's simple. What force must be applied to the board to achieve that rate of acceleration, using the combined masses. Just think, what if the box was nailed to the board? How much force would be needed then? Same force applies to your situation.

 

[M98Ranger]

Hi there,

Your approach is on the right track. In order to find the minimum force required to move the board, we need to consider the forces acting on both the box and the board.

Let's start with the box. The only force acting on it is the force of static friction, which is equal to mu_s*m_1*g. This force must be equal to the force of sliding friction, which is given by mu_k*m_1*g. This is because the box is not moving, so the two forces must cancel each other out. Therefore, we can set these two forces equal to each other and solve for mu_k:

mu_s*m_1*g = mu_k*m_1*g
mu_k = mu_s

Now, let's look at the forces acting on the board. We have the force of static friction acting on one end of the board, and the force of the applied force (F_min) acting on the other end. The board is not moving, so these two forces must also cancel each other out. Therefore, we can set them equal to each other and solve for F_min:

mu_s*m_2*g = F_min
F_min = mu_s*m_2*g

So we have our answer: F_min = mu_s*m_2*g. This makes sense because the minimum force required to move the board is directly proportional to the mass of the board (m_2) and the coefficient of static friction (mu_s).

I hope this helps! Let me know if you have any other questions.