I think that static friction happens only when the object doesn't move. And if the maximum static friction=applied horizontal force, the object still remains at rest. I have already posted a topic about friction yesterday. But there is a problem here:
Two crates are stacked on top of each other on a horizontal floor. The coefficient of static friction between Box A and Box B is 0.6. The coefficient of static friction between Box A and the floor is 0.2. A force of 100 N is applied parallel to the floor on Box B. The boxes move together along the floor. What's the ratio of the mass of Box A to Box B.
My teacher said if there is net force 0 acted on an object, the object is either at rest or moves with constant velocity. So she said the force 100 = the static friction for Box B. I have two questions here: If an object is at rest, then zero net force will still make it rest, right? But is there a possibility that it will move with constant velocity according to what my teacher said? Second, if the applied force = static friction, then the object is not supposed to move? Only if the force is larger than the static friction, then it will move? I have few confusions. Hope you can help me out.
Let's try to get into this one step at a time (Mr_Homm, where are you? 
).
About the "zero net force" and constant motion or no motion. First, consider that "no motion" is just the case of a constant motion of zero. Recall that Newton's first law says essentially, "Without force, there is no change." And by "force" he of course meant net force or more accurately the vector sum of all forces acting on the object.
So, an object either at rest or in a constant straight path is a giveaway that the net force is zero (and, as an aside, there is no change of momentum going on).
As to your second question about static friction, yes, if the force being applied is less than or equal to that required to "break" static friction the object will not move. Now, the thing about the so-called "coefficient of static friction" is that it is an upper bound. The actual friction counter force will be exactly the same and opposite to the applied force until the maximum static resistance is reached, then motion can start (now under the rule of sliding friction). In other words, say you calculate the static friction for something is 10 newtons. Well, that 10 newtons of counterforce is present only if you push with at least 10 newtons: any less and the counterforce is accordingly less.
Now to the problem ... here's where we are taking our chances on my math
First of all, we know that 100 newtons gets both boxes going ... we don't know if they are accelerating or in "constant" motion, but it really won't matter.
We also know that the 100 newtons isn't enough "push" to topple box B off the top of box A.
Starting with the fundamental equation F = u N, where "u" is coefficient of friction, "N" is "normal force", we know for the top box B that for mass of B "m_b" and gravitational acceleration "g":
100N < 0.6 * m_b * g ... Equation 1
In other words, 100 newtons is less than the force needed to push B off of A. (Recall that the "normal force" is essentially the object's weight: mass*g).
However, since the whole system (m_a + m_b) is in motion, we also know that:
100N >= 0.2 * (m_a + m_b) * g ... Equation 2
Or, the 100 newtons force is greater than that countered by the static friction of the combined mass.
Let's look at the boundary case where things just balance: not quite enough force to topple B but enough to get everything moving. So we have two equations in two unknowns to solve. To do it let's subtract equation 2 from equation 1:
100N - 100N = 0.6 * m_b * g - (0.2 * (m_a + m_b) * g)
or,
0 = (0.6 * m_b * g) - (0.2 * m_a * g) - (0.2 * m_b * g)
or,
0 = (0.4 * m_b * g) - (0.2 * m_a * g)
then,
(0.2 * m_a * g) = (0.4 * m_b * g)
so,
0.2 * m_a = 0.4 * m_b
therefore,
m_a = 2 * m_b
Thus mass A must be twice that of B so A:B = 2:1.
... hopefully
).About the "zero net force" and constant motion or no motion. First, consider that "no motion" is just the case of a constant motion of zero. Recall that Newton's first law says essentially, "Without force, there is no change." And by "force" he of course meant net force or more accurately the vector sum of all forces acting on the object.
So, an object either at rest or in a constant straight path is a giveaway that the net force is zero (and, as an aside, there is no change of momentum going on).
As to your second question about static friction, yes, if the force being applied is less than or equal to that required to "break" static friction the object will not move. Now, the thing about the so-called "coefficient of static friction" is that it is an upper bound. The actual friction counter force will be exactly the same and opposite to the applied force until the maximum static resistance is reached, then motion can start (now under the rule of sliding friction). In other words, say you calculate the static friction for something is 10 newtons. Well, that 10 newtons of counterforce is present only if you push with at least 10 newtons: any less and the counterforce is accordingly less.
Now to the problem ... here's where we are taking our chances on my math
First of all, we know that 100 newtons gets both boxes going ... we don't know if they are accelerating or in "constant" motion, but it really won't matter.
We also know that the 100 newtons isn't enough "push" to topple box B off the top of box A.
Starting with the fundamental equation F = u N, where "u" is coefficient of friction, "N" is "normal force", we know for the top box B that for mass of B "m_b" and gravitational acceleration "g":
100N < 0.6 * m_b * g ... Equation 1
In other words, 100 newtons is less than the force needed to push B off of A. (Recall that the "normal force" is essentially the object's weight: mass*g).
However, since the whole system (m_a + m_b) is in motion, we also know that:
100N >= 0.2 * (m_a + m_b) * g ... Equation 2
Or, the 100 newtons force is greater than that countered by the static friction of the combined mass.
Let's look at the boundary case where things just balance: not quite enough force to topple B but enough to get everything moving. So we have two equations in two unknowns to solve. To do it let's subtract equation 2 from equation 1:
100N - 100N = 0.6 * m_b * g - (0.2 * (m_a + m_b) * g)
or,
0 = (0.6 * m_b * g) - (0.2 * m_a * g) - (0.2 * m_b * g)
or,
0 = (0.4 * m_b * g) - (0.2 * m_a * g)
then,
(0.2 * m_a * g) = (0.4 * m_b * g)
so,
0.2 * m_a = 0.4 * m_b
therefore,
m_a = 2 * m_b
Thus mass A must be twice that of B so A:B = 2:1.
... hopefully
Thank you, your method and understanding seemed to be really easy and clear for me. I got it now. But just one question relating to math that why boundary case where things just balance, in other words, why can you just subtract equation 2 from equation 1 in two inequality equations?