But as always, physics calls.
While we were going over homework, Coats-Haan was speaking her mind like she usually does (you understand), and for some reason, she was staring at Jeff and said, "You know why I think fourth period is stupider? It's because I don't have any gingers in it." I suppose she has a point––fourth period did have a class average of 73.10% on the Newton's Laws Test compared to everyone else's 80% and up averages.
As for the homework, someone asked to go over #43, which if you recall is about to a 25 kg block sliding on a frictionless surface with a 4 kg block on its side, and you had to find the push force necessary to put on the 25 kg block to prevent the 4 kg block from falling down. In this problem, friction equals gravity, and you have to use the f=uN equation to help solve this problem. The answer ends up being a 40 N push. If I were you, I would ask Coats-Haan for the details, because there are several forces working in this problem and they are easy to mix up on a blog.
We put away our homework as Coats-Haan passed out a worksheet, which is our homework for the weekend. It's seven problems about overcoming friction, and it doesn't look too thrilling. But as Coats-Haan said today, "I'm sure you had f=uN on last night's homework."
Jeff shook his head. Kelly smiled. Onur stared at her like usual.
After that, Coats-Haan told us to get out our example sheet (which I really insist on calling a packet) to do #4, which asks what force is needed to push a 100 N cartoon [of gumm[y] worms] up a 30 degree incline if the coefficient of static friction is 0.400 and the coefficient of kinetic friction is 0.350.
Now I'm going to warn you, I don't feel qualified to explain this over a blog because I don't have the skill to put detailed visual diagrams in your head. Plus, Coats-Haan even told us in class that grasping the concept may be a little challenging.
Basically, because the problem is set on an incline, we have to turn our x and y axes so that x is parallel to the ground surface and y is perpendicular to the ground surface. By doing this, the normal force, push force, and kinetic friction force are along the axis, and the only force which we need to find the components of is mg.
Next, we made a triangle using mg as the hypotenuse and it's x and y components as its other sides. Using geometry, we calculated that the angle opposite from the x component side is always 90 degree – theta, which is equal to the incline given to us in the problem (30 degrees). The x component = mgsintheta, and the y component = mgcostheta. Now you may be wondering why x is proportional to sin rather than cos like it usually is, but I have made a more detailed expression of why this is true in the QotD below.
Furthermore, there are three steps to solve these incline problems:
1. Sigma Fy = 0
2. f = uN
3. Sigma Fx = ma
The below info is more detailed for this problem.
1. Sigma Fy = 0; N = mgcostheta (This and step 2 are not really steps for Honors Physics because all of our incline problems will always be set up with these conditions. AP Physics may be different, however.)
2. f = uN = umgcostheta
3. Sigma Fx = ma; (which in this problem is P = umgcostheta + mgsintheta = (0.35)(100N)cos(30) + (100N)sin(30) = 80.3 N)
On the examples packet, #5 is similar to #4 because it is on an incline, only you need to use step 3 to solve for u. Also, friction moves up the incline instead of downwards because as you walk, you push back on the ground, and the ground pushes you forward.
For the pair check we got today, Coats-Haan only wanted us to do #1-4 for Monday. She told us that #3, in which we are wearing glass slippers on top of a glass hill, is more like #5 on the examples packet than #4. Onur and I breezed through these questions, but Jeff refused to believe in the possibility of him wearing glass slippers.
THE QUICKIE DETAILS:
Activities:
Checked our homework (see below)
Learned about two-dimensional physics, specifically on an incline.
Completed #4-5 on our examples packet
Worked on Two-Dimensional Friction Pair Check
Returned to us:
1. Pair Check on One-Dimensional Friction
2. Hewitt Reading on Friction (a worksheet)
Checked Homework:
Textbook homework problems, p.128 #34, 35, 37, 39-44 (Turned nothing in)
Homework for the weekend:
Forces and the Laws of Motion: Overcoming Friction, #1-7 (a worksheet)
Pair Check Problems #1-4 (The rest of the pair check will be finished Monday)
Pendulum Lab Report (due Tuesday)
Question of the Day:
Why is the x component not always proportional to cos theta?
Answer:
Cos theta is defined as (adjacent/hypotenuse), which is y/mg if theta is opposite of x like in problem #4 on our examples packet. In this case, y is proportional to cos theta. The x component is instead proportional to sin theta, which is (opposite/hypotenuse) or x/mg.
Therefore in these problems:
sin theta=x/mg, or x=mgsintheta
cos theta=y/mg, or y=mgcostheta (Coats-Haan asked Onur if he was fine with this being true. Onur's fine with it.)
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