# 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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