static friction!?

Question

http://img244.imageshack.us/img244/1603/untitled1pn0.jpg

Answer 1

The forces on the 3kg block are
(assume g = 10m/s^2)

1. tension in the cord. Parallel to ramp, upward
T = 1.5kg x 10m/s^2 = 15 N

2. Gravity. Downward.
W = 3kg x 10m/s^2 = 30 N

3. Reaction force of ramp. Normal to the ramp
R = 3 kg x 10m/s^2 x cos(30)

4. Static Friction force of ramp. Parallel to ramp, downward
F <= μ . R (<= is smaller or equal to)

We need to consider only the forces parallel to the ramp. Plus sign will point upward.
1. T = +15N
2. Wp = 3kg x 10m/s^2 x sin(30) = -15N
3.Rp = 0

The friction force is special. Its direction will oppose the net force exerted on the block so the sign cannot be computed yet.
4. |F |<= μ x 3 kg x 10m/s^2 x cos(30)
|F| <= μ x 26 N

Regardless since the forces on the block are already in equilibrium without the friction force (+15 - 15 = 0N), then any coefficient of friction will do.

Answer 2

The forces of the block on the ramp can be resolved as:
9.8 * 3.0 * cos30 = ~25 N force down on the ramp, and
9.8 * 3.0 * sin30 = ~15 N pulling down the ramp
1.5 * 9.8 = ~15 N pulling up the ramp

The system should be in static equilibrium regardless of the static friction.