text: # Homework Help: Friction problem involving 2 blocks sliding in 2 directions with 2 frictions
  
  1. Nov 9, 2008
  
  ### 2FAST4U8
  
  1. The problem statement, all variables and given/known data
  
  http://streetrodjohn.home.comcast.net/~streetrodjohn/physics.jpg [Broken]
  
  A 1-kg block is pushed against a 4-kg block on a horizontal surface of coefficient of friction 0.25, as shown in the figure. Determine the minimum force needed to ensure that the 1-kg block does not slip down. Assume that the coefficient of friction at the interface between the block is 0.4. Hint: The two blocks exert equal and opposite forces on each other.
  
  2. Relevant equations
  
  F=(A+B)a
  Ag=fs
  N=Ba
  Ag=$$\mu$$Ba
  
  3. The attempt at a solution
  
  In class we have done similiar problems, only with a frictionless horizontal surface. I don't know how to account for the horizontal coefficient of friction of .25 in this problem. Using the above equations and ignoring the horizontal coefficient of friction, I get this:
  
  a=(Ag)/($$\mu$$B
  
  F=(((A+B)Ag)/($$\mu$$B))
  
  F=(((4+1)(1)(9.8))/((.4)(4)))
  
  F=30.62N
  
  Now how do I account for the horizontal coefficient of friction of .25? Thanks in advance for any help you can provide.
  
  Last edited by a moderator: May 3, 2017
  2. Nov 9, 2008
  
  ### asleight
  
  Well, examining block one, we notice that we want $$\sum\vec{F}=m\vec{a}=0=\vec{F}_f+\vec{F}_g\rightarrow\vec{F}_f=-\vec{F}_g$$. From this, we can examine the frictional force, specifically:
  
  $$\vec{F}_f=\mu_i\vec{N}\rightarrow m\vec{g}/\mu_i=\vec{N}$$.
  
  What next? :)
  
  Last edited by a moderator: May 3, 2017
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