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1. Homework Statement
A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force ##\vec{F}##. The coefficient of static friction between all surfaces is 0.63 and the kinetic coefficient is 0.38. What is the minimum value of F needed to move the two blocks?
Homework Equations
##\|\vec{F}_{fs}\| = \mu_s \|\vec{N}\|##[/B]
The Attempt at a Solution
##\sum F_{bottom} = F_{external} - T - F_{fb}##
## F_{fb} = \mu_s(m_{bottom}+m_{top})g + m_{top}g ##
##\sum F_{top} = T - F_{ft}##
## F_{ft} = \mu_sm_{top}g ##
## \sum F = F_{external} - T - F_{fb} + T - F_{ft} = F_{external} - F_{fb} - F_{ft} = 0 ##
## F_{external} = F_{fb} + F_{ft} = \mu_sg((m_{bottom}+m_{top}) + 2m_{top} )##
After plugging in all the numbers:
## F_{external} = 86N ##
86N is the right answer, but it seems odd to me because I essentially counted the friction between the two surfaces twice. Is it just the case that the frictional force between two surfaces moving across each other in opposite directions is twice that of when one of the surfaces isn't moving?
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