2007-11-01

Today's nonsense
- Dakota rule- something about not having to bring cookies? Who came up with this nonsense?


Today we are doing two examples, and then we will get the homework that's due Monday, and tomorrow we may go over the quiz that was straight off of the review and the test and even then some of you failed it entirely so we're going to look at it.

The problem is this: In this case we need two free-body diagrams.

The total mass of the system is m_1 + m_2. The other part of this question is: what is the force of one block on the other.
a) What is the acceleration of the boxes?
F / (m_1 + m_2) = a

Let's look at the different free-body diagrams and see how this all works. So let's do the free-body diagram around m_1.
Sum of the forces = F_hand - F_(the force of block 2 on block 1) = and that net force is going to equal = m_(1)a_(1).
Sum of the forces on the second block = Sum of forces = F_(1,2) = m_(2)a_(2).
So do you realize that a_1=a_2 therefore it all equals a?

VectorF_(2,1) = -VectorF_(1,2)
F_(2,1) = F(1,2) (read this as "Force of 1 on 2")

So what are we going to do with this information? So can we do substitution? F - the force between the blocks = m1a, but the force between the blocks = m2a, so can we say
F - m_(2)a = m_(1)a
So we can solve for "a" by moving over m_(2)a and so on and we get the same exact conclusion as F / (m_1 + m_2) = a ... just like before.
This just confirms the answer for part A.


b) What is the magnitude of the force exerted by one box on the other?
F_(1,2) = m_(2)a = m_(2)F / ( m_(1) + m_(2))

#72 in the book or example #8.
ex 8) #72 pg 110 ch. 4. They give you a diagram, so if you draw the diagram, ...
Find a and T.
The acceleration's magnitude of block1 is going to be the same as block2. If one moves, the other will move. The tension in the rope at either locations will be the same. They can be called T_1 or T_2 but they're both the same thing. So the free body diagram, how do we begin?

Sum of F = net force = what causes acceleration = and this will only be on the x-axis = so we do not have to concern ourselves with the vertical =
Sum of F = T - you have congruent triangles in the diagram, you flip it and turn it
Sum of F = T_(1) - m_(1) * g * sin(theta) = m_(1) * a_(1)
-- sine(theta) because it's sine-theta times the adjacent to tell us the hypotanuse

For m_(2)?
Sum of F = T_(2) - m_(2) * g = m_(2) * a_(2)

If T_(1) = T_(2) then accelerations are equal and opposite (a_(1) = -a_(2))
If you say a_(1) = a_(2) then T_(1) = -T_(2)

m_(1) * a_(1) + m_(1) * g * sin(theta) = m_(2)*g + m_(2) * a_(2)
m_(1) * a + m_(1) * g * sin(theta) = m_(2) * g - m_(2) * a



Law of sines
The sine of an angle divided by the side opposite to that angle, is equal to the side of another angle ...
sin(A)/side A = sin(B) / sideB = sin(C) / sideC ... etc.

Law of Cosines = pythagorean theorem with some extras.