16th October 2007 - 03:00 AM
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)
0 = (0.6 * m_b * g) - (0.2 * m_a * g) - (0.2 * m_b * g)
0 = (0.4 * m_b * g) - (0.2 * m_a * g)
(0.2 * m_a * g) = (0.4 * m_b * g)
0.2 * m_a = 0.4 * m_b
m_a = 2 * m_b
Thus mass A must be twice that of B so A:B = 2:1.