physics help needed badly...im going insane bc of this one?

Question

major help required....this might be complicated it you dont kno what an Atwood machine is tho so heres a pic
http://spiff.rit.edu/classes/phys311/workshops/w5b/slide_block.html

ok so if M1=2kg and the static and kinetic friction between m1 and the inclined plane are .3 and .2, respectively...(and the angle of the plane is 37 degrees)

A-what is m2 if both objects are to remain at rest?
b-what is m2 if both masses are moving at a constant velocity?



the book says a is between .72kg and 1.7kg
and b is between .88kg and 1.5 kg

Answer 1

A) If both objects are to remain at rest, the net force on them must be zero. The tension, T, in the cord must be
T = m2*g
to hold m2 up. So the net force on m2 is zero.

So m1 has an upslope tension of m2*g. Opposing that:
There is a downslope force of m1*g*sin(37) and
There is static friction, fs where
fs = .3*N = .3*m1*g*cos(37)
Assume that m2 is almost large enough to pull m1 upslope. In this case, the friction is fighting going upslope so this force is downslope. The tension on m1 is the only upslope force.
So T = m1*g*sin(37) + .3*m1*g*cos(37)
This will give you m2 at the limit where the friction at m1 can just barely prevent going upslope.

Now assume friction and m2 are barely enough to hold m1 from going downslope. In this case the friction at m1 is the same value as before, but it's fighting going downslope, so the fs is a force acting upslope. Therefore change the sign on it and redo the last equation above.

B) If both masses are moving at a constant velocity (meaning zero acceleration) the net force on them must be zero. Do the same as for A except use the other coefficient of friction.

I wonder if you quoted your book accurately. In this case, only the 2 masses given, not any mass between, should yield constant velocity.