Coefficient of Static Friction?

Question

Application of Newton's Laws. A 4kg block resting on a 30 degrees incline is attached to a second block of mass m by a cord that passes over a smooth peg. The coefficient of static friction between the block and the incline is 0.4.

A. Find the range of possible valuesa for m for which the system will be in static equilibrium.

B. If m=1kg, the system will be in static equilibrium. What is the frictional force on the 4kg block in this case?

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here is the link on the picture of the blocks so that you easily understand the problem
(http://img508.imageshack.us/my.php?image=ps1nc2.jpg)

Answer 1

First do free body diagrams of both masses.

For the block on the plane define the +x direction as being up the plane, and the +y direction being perpendicular to the plane point toward the sky. Now the block has the tension force of teh rope pulling it up the plane, and the force of friction plus some poriton of its weight acting down the plane. It has the normal force acting in the +y direction and some poriton of its weight acting in the -y direction: The forces sum like:

y: 0 = N - Mg cos(q) where q = angle of plane with respect to horizontal

x: Ma = T - uN - Mgsin(q)

The other block has the rope tension T pulling up (+y) and its weight pulling down (-y)

ma = T - mg

Now part a wants the acceleration to be zero:

T = mg

N = Mg cos(q) so

0 = T - uN - Mgsin(q) = mg - Mg(u+sin(q)) -- m= M(u+sin(q))

m = 4*(.4+sin(30)) = 3.6 kg - largets value possible

Now for smaller m, friction force changes direction - the block is trying to slide down the plane. So

0 = T +uN - Mgsin(q) = mg + Mg(u-sin(q))
m = M(sin(q)-u) = 0.1*M = 0.4 kg

So 0.4 <= m <= 3.6 kg

b. Use 0 = mg + f -Mgsin(q) = 9.8 + f - 19.6 Nts

f = 19.6 - 9.8 = 9.6 Nts