Solving Two Blocks Friction Homework

[magasiki]

Homework Statement

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 kinetic friction between the board and the box is, as usual, less than mu_s.

Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use F_f for the magnitude of the friction force between the board and the box.

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 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.

Homework Equations

F_f=mu_s*N
N=mg

The Attempt at a Solution

So i know that the force of friction is mu_s*mg, and i know that the force needs to exceed that, but i don't know what i have to do next to proceed to the answer the book wants. Please help, soon.

 

[Mindscrape]

I don't know what answer your book wants either, but it looks like it wants some sort of algebraic answer. I would say that Newton's second law is a good place to start. What conditions need to happen for the board to be pulled out?

 

[m2003]

I would first clarify the problem by restating the given information and the desired outcome. In this case, we have a small box (mass m_1) sitting on a board (mass m_2) with a length of L. The board is on a frictionless horizontal surface and has a coefficient of static friction (mu_s) with the box. We are asked to find the minimum constant force (F_min) needed to pull the board out from under the box, causing the box to fall off the other end.

To solve this problem, we can use Newton's second law, F=ma, where F is the net force on the system, m is the total mass of the system (m_1 + m_2), and a is the acceleration of the system. Since we are dealing with a system on a horizontal surface, we can assume that the acceleration is zero, and thus the net force must also be zero.

The net force in this system is the sum of the forces acting on the box and board. We have the force of gravity (mg) acting on both the box and board, and the force of friction (F_f) acting on the board due to the box. Using the equations listed in the problem, we can express these forces as:

F_f = mu_s*mg
mg = (m_1 + m_2)g

Since the box is not moving, we can also assume that the force of friction is equal to the applied force, F_min. Therefore, we can set up the following equation:

F_min - F_f = 0

Substituting the expressions for F_f and mg, we get:

F_min - mu_s*mg = 0

Solving for F_min, we get:

F_min = mu_s*mg

However, we are asked to express the answer in terms of mu_s, m_1, m_2, g, and L. We can do this by substituting the expression for mg and rearranging the equation:

F_min = mu_s*(m_1 + m_2)*g

Finally, we can substitute the given values for m_1, m_2, and g to get the final answer:

F_min = mu_s*(m_1 + m_2)*g = mu_s*(m_1*g + m_2*g) = mu_s*(m_1*g + m_2*g