It's official, Mechanics is hard!

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  • #51
kuruman
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Well I didn't like the A-Level books I looked at. They just all looked rubbish to me. Too much waffle and decoration. Then I found the Exam Solutions website. I generally like the no frills approach. The other option is the Khan Academy (again online videos).

Perhaps someone else knows a good textbook for High School level?

This OpenStax textbook claims to be a connection for AP courses.
https://cnx.org/contents/jQSmhtXo@16.1:6pB5TgBD@4/Connection-for-AP-Courses
Khan Academy is good as a first pass to the subject, but difficult to search for something specific that one might be looking for.
 
  • #52
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I've been looking at the OpenStax since you pointed it out. Looks really, really impressive. Very much appreciated!
 
  • #53
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I should add, that I've used "ilectureonline" - Michael VanBizen, and DrPhysicsA (British) also Khan Academy. All freely available. We live in a "golden age" of learning, for those who wish to advance their knowledge and understanding.
 
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  • #54
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As @ZapperZ suggests, but before any quantitative work or comparison, throw a piece of chalk straight up in the air and ask
1. What is the acceleration of the chalk at the top of its travel?
Preconception: Zero, because an object has to be moving in order to have acceleration.
2. Other than air resistance, how many forces act on the chalk while it is on its way up?
Preconception: Two, gravity and "the force of the hand".
3. Other than air resistance, how many forces act on the chalk while it's on its way down?
Preconception: One, gravity because "the force of the hand" was depleted by the time the chalk reached maximum height.
4. Ask them to draw the net force on the chalk at different points if thrown on a parabolic trajectory.
Preconception: The force is tangent to the path.
I'd like to make study cards out of the above questions. I'd like to have the correct answers however. Would you like me to try to answer them, or would you prefer to spare yourself the agony and just give me the correct answers? I'm very grateful to you for your help.
 
  • #56
kuruman
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I'd like to make study cards out of the above questions. I'd like to have the correct answers however. Would you like me to try to answer them, or would you prefer to spare yourself the agony and just give me the correct answers? I'm very grateful to you for your help.
Actually, teaching is a pleasure, not an agony. Do the best you can and I or someone else will guide you through the pitfalls and dead ends. It is important to get to the answer by yourself. It is also important to realize is that you have to ask and answer intermediate questions in order to reach your goal. We at PF specialize at asking intermediate questions and out hope is that eventually you will learn how to do it on your own and no longer rely on us as much.
That apparently was @kuruman's intention, and aligns with our policy here at PF.
Yes.
 
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  • #57
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1. What is the acceleration of the object at the height of its trajectory?

My co-ordinate system is set up such that + velocity is upward. Acceleration due to gravity is directed downwards. As object climbs, it's slowing down because velocity and acceleration are anti-parallel. Object reaches crest, but it's not as if suddenly gravity stops working, so there's always an acceleration due to gravity. So, even at the height of it's trajectory, there is a non-zero acceleration - which is due to gravity.

2. How many forces operate on the object on its way up?
My Aristotelian/medieval urge is, obviously, to say two. But, I'm trying to convince myself that once that object has left my hand, it no longer has the force of the initial impetus behind it. It's slowing down as it climbs due to the force that's acting on it, gravity - the only force acting on it.

3. How many forces acting on it as it descends?
One. Gravity. It doesn't go away.

4. What are the net forces operating on an object that has a parabolic trajectory?
I'm going to say that the net forces on the object operating in the vertical never alter throughout its path. It's always gravity, usually denoted as negative. There is no force I'm aware of operating in the X direction that causing the object to either speed up, slow down, or turn. So, acceleration in the X direction is zero???

If I'm wrong, please don't ask me to go stand in the corner with the Dunce Hat on. Thanks again for your help.
 
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  • #58
kuruman
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Your answers are correct and even if they weren't, nobody will ask you to stand in the corner. We understand that you were not born knowing this stuff and that if you were, you would not be posting here. I have a few remarks that will help you perceive and think with more clarity.

Question 2. If you throw a ball up in the air and the ball is no longer in contact with your hand, the hand cannot possibly exert a force on the ball. If an alien vaporizes your hand as soon as the ball leaves it, would that affect the ball's motion? Impetus is no longer used in physics and the concept behind it has been replaced by two different concepts, momentum and impulse which are related but not identical ideas. Once you learn about them, kinematics ideas will be sharper. That's what I meant when I suggested that you move forward a bit and then look back.
Question 4. Net force is singular as there is only one. It means "the sum of all the forces" and is a single entity. It's like your "net worth"; there is only one. I am not nitpicking about grammar here. It's important to thing of net force as a single entity to avoid confusion later when you start using free body diagrams. You have put three question marks when asking if the acceleration is zero. The answer is yes and I will give you a rule, but you have to explain why this rule is true. Hint: Use the definition of acceleration.
Rule: If you release a mass from rest, it will always move in the direction of the acceleration.

Other than that, you've done well. I was pleased to see that you understand something that a lot of students miss at first, namely that when the acceleration and the velocity are in the same direction, the speed increases and when the two are in opposite directions, the speed decreases. Good job.
 
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Good teaching is often characterized by good questions. The skill of questioning is a vital tool in the teacher's tool-box and your questions are/were spot on.

So, why will a body released from rest always move in the direction of acceleration?
Forces cause acceleration and acceleration is change in speed and/or change in direction as a function of time. Going back to the object in parabolic motion, there's no force acting on it in the X direction, thus nothing to cause it speed up, slow down or turn in that direction. The only direction of acceleration is the one due to gravity, so the object will be directed only in that direction.
 
  • #60
kuruman
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Good teaching is often characterized by good questions. The skill of questioning is a vital tool in the teacher's tool-box and your questions are/were spot on.

So, why will a body released from rest always move in the direction of acceleration?
Forces cause acceleration and acceleration is change in speed and/or change in direction as a function of time. Going back to the object in parabolic motion, there's no force acting on it in the X direction, thus nothing to cause it speed up, slow down or turn in that direction. The only direction of acceleration is the one due to gravity, so the object will be directed only in that direction.
Please read the rule carefully. It says nothing about gravity or forces. You used a specific instant (gravity) to prove something more general. This rule is applicable to a body near the surface of the Earth (your case), to a body in free space, to a charged particle in an electric field, to a magnet near another magnet and so on. I gave you a hint, use the definition of acceleration and by this I meant the kinematic definition without recourse to forces. You rattled off what you have internalized, "acceleration is change in speed and/or change in direction as a function of time" without stopping to think what it says or how it is relevant to the rule. Memorizing a definition and an equation does not necessarily imply knowing how to use them to gain insight. The definition of the acceleration is $$\vec a = \lim_{\Delta t \rightarrow 0}\frac{\Delta \vec v}{\Delta t}.$$ That's all you are allowed to use and first you have to interpret what it's saying to you. By this I don't mean "The acceleration is equal to the limit of the change in velocity over time as ##\Delta t## goes to zero." That's paraphrasing, not interpreting. You need to put an order into how you think about things and how you understand them.
 
  • #61
ZapperZ
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So, why will a body released from rest always move in the direction of acceleration?

Let's clarify this a bit. Are you asking why F = ma, and why a = dv/dt?

I remember way in the beginning, when the issue of Newton's laws were brought up, you specifically clarified that you "understood" what they are, etc. So now it appears that we are back to actually discussing not gravity in particular, but the understanding of Newton's basic principles. This makes your earlier claim to not be true.

If this is so, then the issue isn't really gravity, but the understanding of the meaning of Newton's laws. Asking question is important, but so it trying to diagnose and pin-point the exact problem or issue here. Otherwise, we will be running around in circles, like what have been done so far in this thread.

Zz.
 
  • #62
kuruman
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Let's clarify this a bit. Are you asking why F = ma, and why a = dv/dt?
If this is so, then the issue isn't really gravity, but the understanding of the meaning of Newton's laws.

Zz.
I think OP asked a rhetorical question as a segue to his attempt to answer it in italics immediately below.
If this is so, then the issue isn't really gravity, but the understanding of the meaning of Newton's laws. Asking question is important, but so it trying to diagnose and pin-point the exact problem or issue here. Otherwise, we will be running around in circles, like what have been done so far in this thread.
One way to stop running around in circles is ascertaining that OP understands acceleration first and that's what I am attempting to do.
 
  • #63
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The definition of the acceleration is $$\vec a = \lim_{\Delta t \rightarrow 0}\frac{\Delta \vec v}{\Delta t}.$$ That's all you are allowed to use and first you have to interpret what it's saying to you.

The definition you've provided is one I'm vaguely familiar with. If I had to guess, I'd say that it's Instantaneous Acceleration. So, what's that? I'm guessing the rate of acceleration for a time interval that's infinitesimally small.

But,back to the question: Why will a body released from rest always move in the direction of acceleration? I noticed that in defining the acceleration we now have acceleration clearly notated as a vector; the direction is preordained. Consequently, on the right hand side of the equation, we must also have a vector quantity, which here, is the velocity. So, the change in position - the move in a particular direction - is defined, is determined by the acceleration vector. Motion must be in the direction of the acceleration.
 
  • #64
kuruman
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The definition you've provided is one I'm vaguely familiar with. If I had to guess, I'd say that it's Instantaneous Acceleration. So, what's that? I'm guessing the rate of acceleration for a time interval that's infinitesimally small.

But,back to the question: Why will a body released from rest always move in the direction of acceleration? I noticed that in defining the acceleration we now have acceleration clearly notated as a vector; the direction is preordained. Consequently, on the right hand side of the equation, we must also have a vector quantity, which here, is the velocity. So, the change in position - the move in a particular direction - is defined, is determined by the acceleration vector. Motion must be in the direction of the acceleration.
Your answer is a good example of what pointed out earlier, namely that novices tend to rely on preconceptions and primitive but not well-formed notions about the physical world that put up a barrier to understanding it. It may look like I am criticizing you in what I will say below, but I that's not my intention. I just want to bring you face to face with the problem so that you can do something about it. Fixing it has to come from within, I cannot do it for you.

I gave you a definition for the acceleration and I asked you to interpret it. You responded that you are vaguely familiar with it, which probably means that you have seen it but ignored it. That equation is one of the first equations one sees in an intro physics text and is mostly ignored by most students because they deem that it's not useful because "it does not allow one to calculate anything" like the SUVAT equations. That's a normal reaction as most students believe that they are the sole arbiters of what they should know. Now look at your response when I asked you to interpret it and then used it. You did neither probably because you couldn't do it either. Instead you invoked "preordainment" and begged the question by asserting what you had to show. Aristotle asserted that masses fall because it's in their nature to fall. So what could you have done instead? I will spell it out.

(a) Identify the symbols in the equation.
Left side is the acceleration vector. Right side is the ratio of the change in velocity (final minus initial) over an infinitesimally small time interval.
(b) What is is the equation saying in plain English?
The acceleration is the same as the change in velocity over a time interval in the limit that this time interval is made very small. In other words, at any instant in time you can get the velocity at the next instant, by adding to the existing velocity vector a vector equal to the acceleration vector multiplied by the time difference between instants. That comes from ##\Delta \vec v =\vec {v}_f-\vec {v}_i= \vec a \Delta t## which gives ##\vec {v}_f=\vec {v}_i+\vec{a}\Delta t.##
(c) Special case, mass instantaneously at rest and it is understood that ##\Delta t## is very small.
When a mass is instantaneously at rest, ##\vec {v}_i=0.## Then the equation becomes ##\vec {v}_f=\vec{a}\Delta t.## The vector on the left is the same as the vector on the right. Two vectors are the same when they point in the same direction. Thus, this equation says that the instant after a mass is instantaneously at rest its velocity will be in the same direction as the acceleration vector. Furthermore, if the acceleration itself is zero, the mass will remain at rest.

There is quite a bit in that simple definition for the acceleration. I would strongly recommend that you learn how to interpret equations because they are shorthand notation of how we think the physical world is put together.
 
  • #65
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I've just arrived at work and skimmed over your reply. Just wanted you to know:

1. I'm extremely grateful to you and not at all put off by your frankness.
2. I will have to mull over this, do some research, and within a few days I hope to get back to you.

Very much appreciate your patience and understanding,
Beanyboy
 
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  • #66
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I think we've covered most of what the OP wanted to discuss here and that the topic is getting entangled with technical and non-technical stuff I think its time to close it.

@Beanyboy if you have questions about the technical stuff from @kuruman you can open a new thread.

Thank you all for participating.

Jedi
 
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