(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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=f_{s}

N=Ba

Ag=[tex]\mu[/tex]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=(A_{g})/([tex]\mu[/tex]_{B}

F=(((A+B)A_{g})/([tex]\mu[/tex]_{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.

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# Friction problem involving 2 blocks sliding in 2 directions with 2 frictions

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