12/5 Tuazon

My birthday is now 22 days away. I should just take everyone's blogs until break so I can continue this countdown. 


I am going to be honest--I didn't remember that I had to write the blog again until the very end of class, but rest assured I will make sure you are well-informed about what took place today. 


While Coats-Haan passed out a packet to do tonight's homework on p. 97 of our lab manuel, we checked our homework from over the weekend, which was the Forces and the Laws of Motion: Overcoming Friction worksheet. No one seemed to have any questions about the homework, so we went straight on to our next order of business, which was finishing #6 on our examples sheet.

Before we began, Coats-Haan told us a little fun fact. She said it made her "conceptually happy" to draw triangles with the slope going down to the right (see figure 1) when something is going downhill, as opposed to drawing the slope going up to the right (see figure 2) when something is going uphill . By doing this, the motion of the object will always go in the positive x direction. Coats-Haan says that if you're better at spatial manipulation than she is, however, you can always draw the triangle going a single direction. Either way you choose to go, she won't take off points from your test for the way you draw the triangle. 

FIGURE 1

FIGURE 2 

(I apologize that one triangle is labeled and the other is not. I'm also sorry for the big gap after this figure. I don't know how to fix it.)










Now for #6 on our examples sheet. Coats-Haan began drawing the triangle in the same direction as figure 1. The three forces acting on the skier are a normal force, a gravitational force, and a frictional force going up the slope.


First we start with the three steps:


1. Sigma Fy = 0; N = mgcostheta
2. f = uN = umgcostheta
3. Sigma Fx = ma (Because the skier is not at rest or going at a constant velocity, we cannot set this equal to zero)


Note that we neglect air resistance. In #7 on our pair check, we have to include a 50.0N air resistance in our calculations, yet doing so is not much different than simply adding another force in the x direction. 


The sum of the forces in the x direction (or ma) is the x component of mg minus friction, which ends up being mgsintheta – umgcostheta = ma. All the "m's" cancel out. 


From this we get a = g(sintheta – ucostheta) = 2.76 m/s. Using the fourth kinematics equation, we solve for Vf to get 27 m/s. 


To find time, Coats-Haan says we can use either the third kinematics equation or the first kinematics equation. Again, either way you choose to go, she will not take points off on the test. When you calculate t, you get 8.4 s.


---


While Coats-Haan was showing us how to do the above example, she dropped her SMART Board eraser and had to chase it around behind her desk while mumbling incoherent things to herself. 


Soon afterwards, she dusted herself off and went to the back of the room to explain the Coefficient of Friction lab, for which we have to complete a lab report, due next Tuesday. (Coats-Haan says that there is only one more lab report to do after this one for the rest of the school year. We'll hold her accountable for that.) In a bunch of bins at the back counter there were a variety of wooden blocks, a protractor, a weight, a ramp, etc. We were to use these things to measure the angle at which various items begin to slide down the ramp. 


The remainder of class was spent finishing up our Two Dimension Friction Pair Checks, which we turned in. I was pretty thrilled by working through it. I tried explaining things on the pair check that I myself was trying to understand, but somehow, explaining what I kind-of knew to them helped me to fully understand it now. 


It was a pretty typical day in second period physics. 

THE QUICKIE DETAILS:


Activities:
Checked our homework (Forces and the Laws of Motion: Overcoming Friction, #1-7)
Completed #6 on our example sheet
Finished #5-8 on the Two Dimensional Friction Pair Check (which we turned in)
Started working on the Coefficient of Friction lab, which we will finish up tomorrow

Returned to us:
Nothing (Except for me. I got my blog grade sheet for last Friday's blog returned.)

Checked Homework: 
Forces and the Laws of Motion: Overcoming Friction, #1-7 (a worksheet)

Homework:
Pendulum Lab Report (due tomorrow)
Hewitt Center of Mass questions (p. 97 of lab manuel) based on a packet that Coats-Haan passed out
Coefficient of Friction Lab Report due next Tuesday, 12/13

Question of the Day: 
What are the two situations where the sum of the forces are zero in the x direction for inclined plane friction problems?

Answer: 
Before I answer, I'd just like to say that I wrote this down in the margin of my example sheet with a gut feeling that I would have to know it later.


It is later. 


Two situations where the sum of the forces are zero in the x direction for inclined plan friction problems are when the object is:


1. at rest.
2. moving at a constant velocity. 


In fact, we had one of these constant velocity situations on our Two Dimensional Friction Pair Check in #6-7. Onur and I got the correct answers (without looking at the board), which end up being 675.58 N and  69.12 N, respectively.