I'd encourage you to leave your explanations to quiz questions in the form of comments to this message (and similar other future messages). I would like to see the simplest scientific explanation within the principles that we covered in class. Explanations based on everyday experience are OK, but only as far as the connections between them and the physics principles of our lectures are clearly made.
If a good final solution emerges from someone's comment, that person will be inducted into "the hall of fame of the best commentators," and she/he will be given the same bonus as a person who spoke out in class with a good explanation. As in class, I will also consider giving reasonable partial bonus points, for reasonable but incomplete comments. [Note: if you already got bonus points for the same quiz question or the same quiz session, please refrain from speaking out here.]
At this time, I have 5 quiz questions for discussions.
Here is Q1:
Here is Q2:
Here is Q3:
Here is Q4:
Here is Q5:
10 comments:
This is in response to Q1:
I actually missed this quiz so I hope this answer is right...
I believe the answer is B, the bus is accelerating to the left.
Its easy to visualize this scenario and come to this conclusion from experience, but we can also use physics. In order for the handle to be in the position it is there must be another force acting on it besides gravity. Other wise, the handle would remain at rest extending downward while applying tension to the rope. Since the motion of the handle gives of the clue that there is a force acting on it we can apply Newtons second law F=ma. According to the equation, in order for there to be force there must exist acceleration, so we can rule out that the bus is traveling at a constant velocity in either direction.
In order to determine the vector of acceleration we shouldn't actually analyze the direction the handle is moving (because nothing is actually pushing it to the right), but actually the direction of the tension of the rope on the handle, because that's what's anchoring it to the bus itself. No matter what coordinate system we use we can conclude that the horizontal direction of the force tension the rope applies to the handle will be the same as the acceleration of the bus because F=ma (and not F=-ma). In this scenario the horizontal tension of the rope on the handle is to the left, so we can conclude that the bus is acceleration to the left.
I hope that was clear =].
This in response to Q2: (which I also missed, so please correct me if I'm wrong.)
I believe the (most) correct answer is A and I'll explain why.
Newtons 3rd law states: If object A exerts a force on object B, then object B exerts an oppositely directed force of equal magnitude on A.
So you and the wall make up a 3rd law pair, connected by the tension of the rope(remember we are assuming the mass of the rope to be negligible, so it won't have any affect on the forces applied to it). So the force acting between you and wall are equal, 1N. The wall and the rope, as well as you and the rope, also form 3rd law pairs meaning the rope will have a tension force of 1N also. The reason that these tension forces don't cancel out is because they are acting between two different pairs. Now I said the best answer is A (1N,1N), but since force is a vector and the wall applies an equal and OPPOSITE force, a better answer would be (1N,-1N).
I hope that made sense =].
Those are cool comments, Abram. Thanks! Other students, please feel free to leave comments/explanations/questions, anonymously or not. Abram's explanations seem really cool in most ways, but you can still express your opinions about these or other problems.
I really like what this guys has to say. Although I think that expressing the normal vector to the tension is a more concrete way to indicate the bus is accelerating to the left.
Maybe also I think that the force the wall exerts is 1 N because direction is arbitrary and it depends on how you define net force. so 1N and 1N is a good guess, I think.
This is in response to the last quiz question I missed on Friday, the "merry go-round" question, so please correct me if I answer it incorrectly. Maybe Sam can add it to the rest of the quiz questions in the original post (As well as future questions).
I'll re-iterate the question:
Instinctively we hold onto the pole of a merry go-round to stay balanced. Which of the following is the BETTER explanation.
A: To fight against the Law of Inertia (Newton's First Law).
B: To fight against the outward force ("centrifugal force").
I believe the correct answer is A, here's why.
I think the most important concept to grasp in order to understand this problem is this "centrifugal force." To be clear, there is no centrifugal (center fleeing) force other than as the reaction pair to the centripetal force you are applying to merry go-round. The centrifugal force is caused (as described by Newton's Third Law) by the equal and opposite force of you pulling yourself towards the center the merry go-round. This centripetal (center seeking) force you apply to merry go-round is exactly was is keeps you traveling in a circle. If you were to let go of the merry go-round, you would travel in a path tangent to the circular path you were traveling in. What cause you to travel in the tangential path? Your inertia, or your mass's "tendency" to stay in motion. So really your fighting against the Law of Inertia.
If you were actually struggling against a centrifugal force, the moment you let go of the merry go-round you would actually fly directly away from the center of the ride, as if an explosion originating from the center pushed you away from the merry go-round. But even a force like this wouldn't be described as a "centrifugal force" in the classical sense.
The discussions are very high standard. However, please think about the following.
Q2.1. Are the wall and the hand exerting forces on each other directly?
Q2.2. Do the wall and the hand form a Newton's 3rd law pair?
Q2.3. If there was a pulley, so that the string is shaped a little differently (bent at the right angle, e.g.), but the hand is still exerting a 1 N force on the string, would the answers to this question change at all?
Q2.4. Does Q2.3 also shed light on the answers to Q2.2? [Think about the directions of tension forces at the two ends and their respective 3rd law pair forces].
Q2.5. Even if the mass of the string is not zero, would the answers to this question change at all (assuming that everything is at rest here)?
Let us call the centrifugal force quiz, Q6. I do not think I need to post an image for it, since the question is already stated clearly in Abram's comment (whose choice of the answer was what I had hoped for, although as I said in class both answers are correct for this one).
Here is something to keep in mind. For those students who were not in the last class, please know that a centrifugal force is not a real force. It is a fictitious force in the sense that only those in a rotating reference frame (thus a non-inertial frame) feels it. If you are an observer A watching another person B on a spinning merry-go-round, then there is absolutely no need for you to dream up a centrifugal force at all. For A, the only horizontal force acting on B is the centripetal force provided by the pole that B is holding on to, and that force (a centripetal force) is equal to ma of Newton's 2nd law, where m is the mass of person B.
Since a centrifugal force is not even real, it definitely cannot form a third law pair with the corresponding centripetal force.
Now, in what sense, then, do we ever need the centrifugal force? Imagine we live in a very large "merry-go-round," like the space station in the homework problem 5.56. In fact, that space station is a perfect example. Imagine no planets are near that station. It is just spinning by itself. The fact is that when we are inside that station, we feel a "downward force" (see picture for 5.56) due to the spinning. This force that we feel when we are on a rotating frame is a centrifugal force. Now, ask yourself these questions, assuming that we spent all our lives inside that space station, oblivious of the outside world or the outside of the space station. Would we think that that the downward force is as real as a gravitational force? Would we ever be aware of the difference between the two? Suppose that the space station is large enough that the local landscape is quite flat. If you did a projectile motion experiment inside the space station and interpret the results using Newton's second law, F=ma, will we have to use a finite F (even though if you ever stepped outside the station, then you will discover that there was no real force!)?
In response to Q3:
The tension on the rope will be (B) equal to the weight of the box.
Mathematically, this ends up being vector addition. we know the acceleration of the box is 0, therefore Fnet=0.
When you draw the free body diagram you get the normal force vector perpendicular to the slope, the vector of gravity pulling directly down, and the vector of the rope pulling horizontal, place those vectors head to tail and you end up with a 45/45/90 triangle with the Normal vector being the hypotenuse.
The other 2 sides are the tension of the rope and the weight of gravity pulling down, and we know via trig thatap those 2 sides/vectors are equal to each other and so we conclude that the tension in the rope is equal to the weight of the box.
Question to the tire pressure problem.
I am organizing this problem to be a ratio of the forces and the pressures. Set force equal to area times Pressure. Area is equal to cancel them.
I am left with acceleration(low pressure)(in terms of x and Vo) times Low Pressure/High Pressure equal to friction coefficient times gravity. then solve to find friction coefficient, using 9.8 for gravity and del x for low pressure tire. I dont know, but I am a bit lost at the algebra step.
Thanks
Nicole, that was a super-clear comment, I thought. You get the double credit bonus for that day. Abram, you, too.
For Q3: If theta becomes (slightly) different from 45 degrees, then the tension is not equal to the weight any more.
For Q2: (1) Please do note that except for gravity, no contact means no force exchange. So, two bodies that are not in contact cannot exert Newton's 3rd law pair forces. (2) Tension force is a vector of course. When one speaks a tension force of a string/rope without specifying which end, however, it means the magnitude, as the tension forces at the two ends usually point in different ways (following the shape of the string), but they have the same magnitude (under ordinary circumstances). (3) In this quiz question, it is not necessary to assume mass = 0 for the string.
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