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Suppose the coefficient of kinetic firxxtion between mass a and a plane is .10 and mass A and B is 2.7kg. As Mass B moves down, determine the magnitude of mass A and B given an angle of 31 degrees. What is the smallest value of kinetic friction to keep the system from accelerating?

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I've figured out wha the acceleration is (2.0m/s^2 \[(2.7kg)(9.80m/s^2)(1-\sin31-.10\cos31)/5.4kg=2.0m/s^2\]
I also know the component forms of box A is Fy=22.68N and Fx=13.62N
both boxes are in contact?

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Other answers:

Both boxes are connected to a string on a pulley, Box B is hanging from the pulley and box A is on an incline.
u can draw the free-body diagram showing the forces acting on the box A and B?
fig is|dw:1333223131753:dw|so we have three equation:\[m _{B}g-T=m_{B}a\]\[m_{A}gsin \theta-\mu_{k}m_{A}gcos \theta=m_{A}a\]\[m_{A}+m_{B}=2.7\]so we can solve for mB & mA
for 2nd part i think : we get 2nd formula equal to 0 and solve for mu k
^Sorry about the wait, but yeah it looks just like the image hosein drew
sorry i forgot this: theta oughtbe 60 degrees in formulas

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