Friction of an object on a moving board

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A 2 kg body is placed on an 8 kg board, 1 meter from the rear edge, with a friction coefficient of 0.2. A force of 30 N is applied to the front of the board, causing the body to move towards the rear. The discussion focuses on applying Newton's laws to analyze the forces acting on both the body and the board separately, emphasizing that they should not be treated as a single system due to their relative motion. Key equations involve the friction force and the accelerations of both the board and the body. The analysis concludes that understanding the interaction forces is crucial for solving the problem accurately.
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Homework Statement



A body of mass m_A=2 kg is placed on a long board of mass m_B=8 kg at distance d=1 m from the rear edge of the board. The friction coefficient between the body and the board is μ=0.2. A force of magnitude 30 N is applied to the front edge of the board and the body start moving towards the rear edge. How much time will it take to fall off the board?


Homework Equations



The force of friction is given by F_f=F_n\cdot μ where F_n is the normal force exerted by the object on the surface.


The Attempt at a Solution



I tried to write down Newton's equation of motion (on the x-axis) for the body and the board as follows:

Board: F=(m_A+m_B)a_1

Body: F\frac{m_A}{m_A+m_B}-F_f=m_A a_2 where F_f=m_Agμ.

With these equations the problems doesn't come out right...
 
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What forces act on the body? Apply Newton's 2nd law.
What forces act on the board? Apply Newton's 2nd law.

(Don't treat 'board + body' as a single system, since parts are in relative motion.)
 
On the board: F (and the weight of A which is equilibrated by the board itself).

On the body: The force exerted by the board on the body, directed along the direction of F and of a "certain" magnitude and the friction force.
 
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All we care about are the horizontal forces, since vertical forces will cancel.
grusini said:
On the board: F (and the weight of A which is equilibrated by the board itself).
You are missing the horizontal force of the body on the board. (Newton's 3rd law.)
On the body: The force exerted by the board on the body, directed along the direction of F and of a "certain" magnitude and the friction force.
The only horizontal force on the body is the friction force from the board.
 
So the body exerts a horizontal force F_f on the board in the opposite direction of F? Then Newton's 2nd law for the board is
F-F_f=m_Ba_1
and the Newton's law for the body would be:
F_f=m_Aa_2?
 
grusini said:
So the body exerts a horizontal force F_f on the board in the opposite direction of F? Then Newton's 2nd law for the board is
F-F_f=m_Ba_1
and the Newton's law for the body would be:
F_f=m_Aa_2?
Right. And you also know how to calculate the friction force.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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