![]() |
Work & Energy 02 |
OPENING QUESTIONS:
Be prepared to brief your classmates on any or all of these upon request ═══════════════════════════ LEARNING OBJECTIVES: 1) I will be able to explain the concept of a 'dot' product to a bright, articulate algebra 2 student after today's class. WORDS O' THE DAY:
FORMULAE OBJECTUS:
WORK O' THE DAY: We had a question on variable forces -- here's an interesting example. There *appears* to be a problem with the logic here.... see if you can identify it and see if you can see how the author works around it. I'm gonna call a wee bit of an audible here... for your second homework problem, calculate the normal force acting on the box (that should be interesting). Take a few moments and make some written comments on what you learned from doing the homework. Be prepared to speak articulately to the class on this point. ═══════════════════════════ Let's start recording our formulae objectus. You MUST be able to use and recite (on deman) any and all formulae we have used THIS SEMESTER!!! Recall that our definition of work for an object undergoing a constant force was: W = Fxcos(θ) Where F = the force being the applied force and x = the displacement the object experiences NOTE: The book sometimes uses 'r' to represent the displacement of the object for expedience.
We will start using 'r' now as we move away from motion in just the 'x' or 'y' directions. So... W = Frcos(θ) ═══════════════════════════ Notice also that the definition of a scalar ("dot") product is: ABcos(θ) It is certainly no accident that the definition of a scalar product is the same form as the definition of work (since work is a scalar quantity) ═══════════════════════════ Let's take a look at dot product math HERE Now let's take a look at how our book introduces dot products: (example 7.2 on page 182)
Now let's take a look at part 2 of that problem which gets a tad more difficult:
It's fairly straightforward to calculate the the lengths of the hypotenuses' of each vector since the 'i' and 'j' quantities are basically the two legs of a triangle: root(22 + 32) = root(13) Similarly root(-12 + 22) = root(5) Here's where (I think at least) it gets a tad tricky: Consider the definition of a dot product: A·B = ABcosθ if we want to find the angle θ, we have to realize that there are two vector quantities present (Force and displacement) AND THE ANGLE BETWEEN THOSE TWO VECTORS is θ so... now let's do some basic rearranging: A·B/AB = cosθ Cos-1(A·B/AB) = θ so now we calculate the 'dot' product of A and B and then divide that quantity by the product of the magnitude of A and B *YIKES* Now do 7.3--- oh and add in a part where you have to find the angle that the force makes to the displacement (OUCH) So...
═══════════════════════════ HOMEWORK
|