Friction problem involving 2 blocks sliding in 2 directions with 2 frictionsby 2FAST4U8 Tags: force, friction, newtons law 

#1
Nov908, 09:24 PM

P: 1

1. The problem statement, all variables and given/known data
A 1kg block is pushed against a 4kg 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 1kg 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. 



#2
Nov908, 09:32 PM

P: 154

[tex]\vec{F}_f=\mu_i\vec{N}\rightarrow m\vec{g}/\mu_i=\vec{N}[/tex]. What next? :) 


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