ehunger1
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Homework Statement
The given diagram is an inclined plane with a 15 degree angle. There are three boxes, A, B, and C all weighing 4 kg. The three packages are touching, but B is in the middle and larger than A and C, which are the same size.
The coefficients of friction for A and C are static=.30 and kinetic=.20
The coefficients of friction for B are static=.10 and kinetic=.08
Determine which, if any, of the packages will move and the friction force acting on each box.
Homework Equations
SUM Fx=0
SUM Fy=0
The Attempt at a Solution
I've determined the weight of each box is 39.2 N (F=ma)
For box A (closest to the angle of the inclined plane)
tan(15)=.268 < .3=static coefficient, therefore, block A will not move on its own
SUM Fx=f1-39.2sin(15)=0
SUM Fy=N1-39.2cos(15)=0
f1=friction force=10.15 N
N1=Normal Force from plane=37.9N
I don't know how to treat the blocks being in contact with each other. I know there should be a normal force there, and I know that for block B, tan(.15)=.268>.1, therefore on its own, block B would move. Block C will not move because it is identical to block A and there is no force acting on it to push it down the ramp.
How do I know if block A and B move?