Friction of an object on a moving board

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Homework Statement



A body of mass [itex]m_A=2 kg[/itex] is placed on a long board of mass [itex]m_B=8 kg[/itex] at distance [itex]d=1 m[/itex] from the rear edge of the board. The friction coefficient between the body and the board is [itex]μ=0.2[/itex]. A force of magnitude [itex]30 N[/itex] 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 [itex]F_f=F_n\cdot μ[/itex] 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: [itex]F=(m_A+m_B)a_1[/itex]

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

With these equations the problems doesn't come out right...
 

Answers and Replies

  • #2
Doc Al
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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.)
 
  • #3
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On the board: [itex]F[/itex] (and the weight of [itex]A[/itex] which is equilibrated by the board itself).

On the body: The force exerted by the board on the body, directed along the direction of [itex]F[/itex] and of a "certain" magnitude and the friction force.
 
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  • #4
Doc Al
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All we care about are the horizontal forces, since vertical forces will cancel.
On the board: [itex]F[/itex] (and the weight of [itex]A[/itex] 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 [itex]F[/itex] and of a "certain" magnitude and the friction force.
The only horizontal force on the body is the friction force from the board.
 
  • #5
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So the body exerts a horizontal force [itex]F_f[/itex] on the board in the opposite direction of [itex]F[/itex]? Then Newton's 2nd law for the board is
[itex]F-F_f=m_Ba_1[/itex]
and the Newton's law for the body would be:
[itex]F_f=m_Aa_2[/itex]?
 
  • #6
Doc Al
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So the body exerts a horizontal force [itex]F_f[/itex] on the board in the opposite direction of [itex]F[/itex]? Then Newton's 2nd law for the board is
[itex]F-F_f=m_Ba_1[/itex]
and the Newton's law for the body would be:
[itex]F_f=m_Aa_2[/itex]?
Right. And you also know how to calculate the friction force.
 

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