Monday, June 29, 2015

Episode 050: Just Around The Riverbend...

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So, to conclude our cliffhanger episode last time, here are my notes to accompany the podcast and address the questions how far down the river did our rower go, as well as how long did he traverse? Answers on the inside.

I think that it's obvious that to attack this type of problem you need to understand both vector components and sohcahtoa. If you haven't yet, I encourage you to check out the apps from the app list above. Good luck!

Thursday, June 25, 2015

Episode 049: Furthur Down The River....

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This is the question from Episode #48, which we were able to solve, with relatively sweating. Now, in the coming days, I have a little homework assignment for you. School may be over, but you are not off the hook.

1) If the river is 20m wide, how far down the river does our rower travel?

2) How long does his traverse take?

Answers forthcoming. In the meantime, click below for the not so subtle musical clip of the week!

Tuesday, June 23, 2015

Episode 048: Vector Components #4- A Boat Down The River!

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Question: A boat is moving across a river with a speed of 3 m/s. The river current is traveling at 4 m/s. What is the magnitude and angle of the boat’s total velocity? This type of problem usually has a lone rower trying to row his boat directly across the river. However, the current moving pushing downstream causes him to move at an angle! Sometimes they ask you to calculate how far downstream he will be based on the velocity and width of the river, but being able to combine these components into the total velocity vector is very important. In the next five steps we’ll be able to see exactly how to do that.

Step 1: Identify the coordinate axis and draw it on the paper. Free to choose, as always, we will align our coordinate axis with the two component vectors in our problem, i.e. the river current and the rower of the boat. Thus, we have our rather standard cartesian coordinate axis with the x-direction pointing to the right and the y-axis pointing upwards.

Step 2: Identify the components. In this problem we’re even given numerical values: the current is traveling along the x-direction at 4 m/s, and the boat is traveling in the positive y-direction at 3 m/s.

Step 3: Use the Pythagorean Theorem to calculate magnitude. Considering the x and y components to be legs of a right triangle, we’re able to calculate the magnitude of our vector, i.e. the resulting hypotenuse, via the Pythagorean Theorem. In the problem we’re conveniently given some perfect squares that calculate nicely to give us a velocity of 5 m/s.

Step 4: Use arctangent to calculate the angle. Since we’re going to be reconstruction the vector, this will always be the hypotenuse of the triangle. Ergo, we will always be using arctangent to calculate the angle, since the components will be the opposite and adjacent sides. In this example, the x-component can be considered the adjacent side, aka the current of the river and y-component or the boat speed can be considered the opposite side. Arctan(.75) is equal to 0.64 rad. Remember, if you see and angle smaller than 5 degrees, do a little mental math and check if you’re in radians or degrees mode. Your answer will look very silly if you wrote 0.64 degrees!

Step 5: Box your answer, you’re done! So, we have a boat which is ultimately traveling 36.86 degrees w/r/t the direction of the river, at a total speed of 5 m/s, with the help of the stream and our crazy boat rower. This mean’s he’s going about 11 mi/hr across the river. Not bad!

Sunday, June 21, 2015

Episode 047: 10 Summer Solstice Facts!

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It's that time of year again. The first day of summer. The Summer Solstice. Many of us hear that phrase thrown around twice a year, for the summer and winter solstices, but what is it really? Here are 10 facts about the summer solstice:

1) How the solstice got its name- The name is derived from the latin- Sol means sun, and sistere means to stand still.

2) Summer Solstice marks the first day of summer.

3) Summer Solstice is when the earth's axis points closest to the sun.- The earths axis goes through the north and south poles. However, that axis does "point" straight up and down with respect to the sun. It is tilted slightly, 23 degrees from "straight up". As the earth moves around the sun, this angle remains the same, so for half of the year the northern hemisphere is directed moreso towards the sun, while for half the year the southern hemisphere is directed closer to the sun.

4) Summer solstice is the longest day of the year- Because the axis points closest to the sun, this is the maximum amount of time the sun remains in the sky, giving the longest day of the year.

5) The summer solstice in the north is the winter solstice in the south- The longest day of the year in the north is the shortest day of the year in the southern hemisphere. This also corresponds to the first day of winter, or winter solstice, for our friends down south.

6) Some places the day is so long it never ends!- In the far north, places like northern Alaska, the earth is directed at the sun such that the sun never sets, or it looks like sunset for a few hours before the sun rises again for weeks or months on end!

7) The solstice occurs somewhere between June 20-22- The solstice most commonly occurs on June 20 or 21, and this year it fell on June 21. The next June 22 solstice will occur in 2203!

8) During the northern hemisphere summer solstice, the earth is at its furthest point from the sun. During the winter, the earth is actually at its closest point, however the north is at this time directed away from the sun. This comes from the fact that Earth's orbit around the sun is not circular, but rather elliptical.

9) Stonehenge aligns with the sun during the solstice- Noone quite knows what stonehenge was designed for, but the symmetry of the monument is clear during the summer solstice- the sun shines directly on the heel stone during the sunrise of the solstice. Other theories regarding Stonehenge's purpose include a calendar, much like a sundial; a predictor of eclipses; an ancient burial ground; and a place of pagan worship.

10) 23,000 people turned out for the stonehenge sunrise this year- Down from 36,000 last year, thousands still go to hang out at stonehenge for the summer solstice and celebrate the longest day of the year.

Monday, June 15, 2015

Episode 046: Vector Components Example #3- Box Pulling A Rope!

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This problem, surprisingly, is very similar to example #2. It’s a force problem, and you’re given a force vector which you need to break into components. The major difference in this problem is that we’re given actual numeric values instead of variables. You will find out that all of these problems are really just the same person with different faces, as we press through the next example. Notice that the steps don’t really change, and that the answer is very easily acquired by simply following the steps.

Step 1: Identify your coordinate axis and draw it on the paper. Because the force that we’re looking for is horizontal component of the pulling force, we want to align our x-axis horizontally. If you’re having trouble determining how to align your axis, this will come easier as you do more problems. Running down false alleys and seeing how the problem may not work out nicely will help give you a better sense of how to orient your coordinate axis in the future. The very nice and important thing to remember is that no matter which axis you choose the final answer has to be the same. It will not change based on your selection in this step. However, the math to achieve it may be messier or cleaner, which will improve based on your expertise.

Step 2: Identify your vector and draw it on the paper. In the problem the rope is being pulled upwards with respect to the x-axis, so we can guestimate the direction that it’s pointed like so.

Step 3: Identify your angle and draw it on the paper. This is given to us in the problem, so the only thing left to do at this point is draw it where it goes.

Step 4: Use sine & cosine relationships for component definition. In this scenario, because of the position of the angle, our x-component will be the adjacent side, and the vector our hypotenuse. This means we’ll be able to use cosine to calculate the horizontal component of the force. This will be the amount that directly opposes the frictional force, when you drag the box across the floor. This is why this force is of interest to us. We’ll also just calculate the y-component for fun. Remember, when you calculate the sin/cos(35), remember that this is 35 DEGREES, or else you’ll have to convert to radians!

Step 5: Box your answer, you’re done! So, a box being pulled 35 degrees w/r/t the horizontal will pull with 8.91 Newtons of force in the x-direction and 5.74 Newtons in the y-direction. So, are you getting the feel for it yet? Do you see how these problems are all basically asking the same question even though the scenarios are all different? If not, go back and refresh the last few examples to see if you see the similarities now.

Monday, June 8, 2015

Episode 045: Vector Components Example #2: Normal Force Calculation!

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Another, and probably by far the most expansive, place that you will see vector components is in Newton’s Second Law, a.k.a. Free Body Diagrams problems. This is where the “identify your angle” step really starts to get tricky, as the forces are usually not as straight forward and in line with the axes as in kinematics problems. This problem gives a general angle theta, and asks for the normal force, N, which is seen pointing upward directly from the incline. The picture gives us some additional information, the weight, W, is pointing directly downwards. Surprisingly, with our 5 steps, this is all the information we’ll need to solve this problem.

Step 1: Identify your coordinate axis and draw it on the paper. Unlike the kinematics problem, this time we’re going to align our coordinate axis a little differently, not just the up and sideways coordinate axis. The alignment we will choose will point the x-axis parallel with the incline, and the y-axis parallel with our eventual normal force. This makes sure that our Normal force will be equal to some trigonometric function times the Weight, and not have to break the Normal force itself into components.

Step 2: Identify the vector and draw it on the paper. Contrary to what the problem asks, for right now we’re after the weight, W. So, as shown above in red, the interest becomes finding the y-component of the weight. This begs the next piece of the puzzle, the angle.

Step 3: Identify the angle and draw it on the paper. Because the interest is in the y-component of the weight, the angle we’re going to need is the one shown above, tucked between the weight and the w-component. Unlucky for us, it’s not the one given in the problem, at least exactly. This is actually the same angle theta, but how is this possible? This is not the same angle theta given in the problem! These are the geometric gymnastics I was hemming and hawing about in the kinematics problem, and why I was so happy they gave us an easy angle back then. On the next page we will look into exactly what is going on with this angle and how exactly this manages to be the same angle.

I think the caption says it all. On to step 4.

Step 4: Use sine & cosine relationships for component definition. Here, given the weight, W as the hypotenuse, with this angle we can say that the y-component, will be identified as the adjacent side of a triangle. We will then be able to define Wy as the weight W times the cosine of the angle theta.

Not really needed for our purposes here, but the Normal force, N, opposes the y-component of the weight. This is where the phrase “equal and opposite” comes in. The weight W is given as the mass of the object times gravity, g. Thus, our normal force is equal to mg*cos(ϴ). If you were bogged down with some of the frills in the problem, i.e. what was exactly going on with the normal force, or the weight, or whether or not you would know how to do this problem without reading the steps, don’t worry. The idea is to see that these same steps can be applied to a variety of different problems, and that the procedure here is roughly the same. This internal skeleton will be the foundation for a much bigger framework as we introduce other physical concepts.

Monday, June 1, 2015

Episode 044: Vector Components Example #1: Object Launch

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An object is launched into the air at an angle of 60 degress with respect to the horizontal. What are the velocities in the x and y directions? When studying kinematic motion, projectile motion, throwing objects up in the air, shooting them from cannons, flinging them from trebuchets, you will constantly be concerned with the velocity in the horizontal or x-direction and the velocity in the vertical or y-direction. This is mostly due to the inclusion of gravity. Since gravity works roughly straight up and down, The motions in the vertical and horizontal directions can be considered independent, and then combined together as a final answer. Usually, the object being launched is not straight up and down. Thus, we’ll have to take this velocity vector, and find out how much of it is directed in the vertical direction and how much in the horizontal direction. We can do this with ease by implementing the 5 steps from last episode.

Step 1: Identify the coordinate axis and draw it on the paper. This is critically important. Usually this step is as simple as drawing the y-direction vertically and the x-direction horizontally. However, there are sometimes cases where it may be easier to make the coordinate axis in some other orientation. So it is very important that everyone understands the coordinate system as the very first item.

Step 2: Identify the vector and draw it on the paper. In this case we have a velocity vector v which is 60 degrees with respect to the horizontal, which has been conveniently aligned with our x-axis. The y-axis, aligned vertically, will work directly opposite gravity. This will be very convenient in problems in force problems and when we dive deeper into kinematics.

Step 3: Identify the angle. In this case it is very straightforward: 60 degrees with respect to the horizontal. Sometimes there will be some more calculations required. Usually the “with respect to the horizontal” is the easiest angle to work with. I pray for this angle to be given in problems. Writing it down on the drawing is critical. It leaves no room for confusion and makes your work complete and precise. It is wise to only place one vector per diagram because with the vectors and angles and labels things can get crowded and messy.

Step 4: Use sine and cosine relationships for component definition. For this very standard case: x-direction aligned with the horizontal, y-direction pointed vertical, and the angle given w/r/t horizontal, you will always be able to use cosine for x and sine for y. This is because everything is set up just as in our unit circle and the coordinates for our radius as we move around the circle is [cos,sin].

Step 5: Box your answer, you’re done! You can verify these calculations in your head if you use the sines and cosines by counting to 4. The final answer is wonderfully generalized so that any vector v at this angle can be plugged in. This is the best type of answer: correct, robust, and specific.