Chapter 4 (or 5) Tutorial #2
Given the pulley system below, there IS friction between any pair of surfaces in contact. The acceleration of the mass m is 2 m/s2. The acceleration of the mass M is 1 m/s2. There is no slipping or tangling of the string. Solve for the following:
· f1, the friction between mass m and mass M
· f2, the friction between mass M and the tabletop
m
M
400 g
M = 600 g, m = 100 g
The purpose is to walk you through logic in setting up a problem. Work through all of the steps below as a review of what we did in class last time.
Make sure you know the definition of victim of agent:
Victim: the matter that is feeling the forces caused by something else. All FBD’s show forces that act on the victim being diagramed. People who are not clear on this will get nothing right. It will be impossible to continue without understanding this.
Agent: the matter that is exerting the force on any victim. When you look at a FBD, you never see a picture of the Agent. However, each force has an agent, and you must be able to picture each of these agents in your mind.
Methods:
A. Make three good FBD’s as usual. Define a coordinate system for each.
B. FBD’s require mad skills. So evaluate every force in each of your FBD’s:
· For a given FBD, only ONE force can be gravitational. For the FBD you are looking at, identify the gravitational force. Now, besides this one, have you named any of the others with a name that looks or sounds anything like gravitational. If so, it is wrong. Erase that label now.
· All forces that are not gravitational are electrical. And we know them as contact forces. For the FBD you are looking at, is all contact represented:
o If the flat surface of the victim is in contact with a flat surface of another object, have you drawn a force vector for perpendicular contact across that surface? Have you called such a force a normal force?
o If the victim is experiencing normals from two different things touching it, have you labeled these normals differently, not with the same letter n? Have you used subscripts such as nTable, n1, etc.
o If the flat surface of the victim is in contact with a flat surface of another object, have you drawn a force vector for parallel contact along that surface? Have you called such a force a friction force?
o If the victim is experiencing frictions from two different things rubbing against it, have you labeled these frictions differently, not with the same letter f? Have you used subscripts such as fTable, f1, f2, etc.
o Have you used a vector to represent any other contact that is influencing your victim object. Is all contact represented now? For example, for any string pulling, have you drawn a vector?
· For every force you have drawn, can you easily and quickly name the AGENT who is causing that force? If not, then it is not a real force and should be erased.
· Labels: All forces are interactions. If you have FBD’s for objects in contact with each other, some of the forces drawn represent these two objects interacting with each other. Do the labels for these interaction forces obey Newton’s Third Law? They must. So read Newton’s Third Law and decide if your interaction forces are labeled correctly. Warren’s Version of Newton’s Third Law: “Mutual interaction forces are identical.”
C. Once the FBD is right, set up åFy = May AND åFx = Max (just like Tutorial 1 and Tutorial 2 advised) and manipulate algebraically until you have a solution.
Sample Beginning Student Solution:
Evaluate the following FBDs for masses M and m as the victims.
n mg n
T
T
f Mg f mg
Find and specifically identify ALL the mistakes in those two FBD’s, and after you have done so, go to the next page to find out what all the mistakes are. The T shown there is a common symbol that people use for string tensions. But again, the letter you use isn’t so important as long as you are consistent.
Mistakes:
Diagram with mass M as victim:
· Gravity is represented by the force Mg. No other forces are gravitational. The force labeled “mg” acting on mass M is entirely inappropriate. Label must be changed.
· Not all normals (i.e. basic perpendicular contact) are represented. It is assumed that the “n” shown is caused by the table’s atoms. Student has forgotten the normal due to mass m’s atoms.
· Not all friction (i.e. basic parallel contact) is represented. It is assumed that the “f” shown is caused by the table’s atoms. Student has forgotten the friction due to rubbing from mass m’s atoms.
· An effort is made to easily and quickly name the AGENT who is causing that T force in the diagram: Can’t be done. “String” you say? Well, the string is not touching mass M, so common sense says erase this force.
Diagram with mass m as victim:
· The basic normal force shown is called “n”. Easily and quickly name the agent of this force: easy, mass M’s atoms are exerting this electrical force. So it’s real force. But the student may not call it “n”. That label was already used to represent the force on M due to the table’s atoms. That is an entirely different interaction. One of the n’s in the two diagrams must change its name.
· The basic friction force shown is called “f”. Easily and quickly name the agent of this force: easy, mass M’s atoms are exerting this electrical force. So it’s real force. But the student may not call it “f”. That label was already used to represent the force on M due to the table’s atoms. That is an entirely different interaction. One of the f’s in the two diagrams must change its name.
That’s it for the mistakes. Now use the corrections just outlined to make the free-body diagrams correct for sure. Only after you have done them yourself and have checked them as I’ve just outlined, go to the next page to see the true FBD’s.
Correct FBD’s: T
nT n n
f
fT f T
Mg mg 4 N
Look and think carefully about why the “n” in the diagram on the left must be identical to the “n” in the diagram in the middle. Similarly, look and think carefully about why the “f” in the diagram on the left must be identical to the “f” in the diagram in the middle. The reason that these must be identical is Newton’s Third Law: All forces are interactions, and mutual interaction forces must be identical. There are many other ways to word this, and I won’t write them all here, since you can read your book. Newton’s Third Law is a vital concept.
You didn’t forget diagramming the third victim of mass 400 g, did you?
If you used different symbols than I did, that is fine, as long as you are consistent with the concepts outlined in this document and as long as you are consistent in the rest of your solution. My fT means friction due to TABLE. My T means string tension. Nobody had to know what the string force symbol T ever meant. It was only ever important to know that the string was making contact, and therefore some symbol had to be a label on a rightward force in the middle diagram and on an upward force in the rightmost diagram.
Equation Set-up:
The rest of this is just like recent Fnet = (mass)a setup.
You have ∑Fx = (mass)ax and ∑Fy = (mass)ay expressions to set-up as they arise from your FBD’s. That’s one equation for each axis of each massive victim. Therefore, for this problem, you have a total of __ equations to write down. Write down these equations, and identify the algebra you can do to solve them. Then do that algebra. Finish it yourself now. The answers will be given below.
Answers: there were five different F=ma expressions to set up. Final answers to the problem were: f1 = f = 3 N, f2 = fT = 2.4 N.
If you got these, congratulations. If you didn’t, try to fix things yourself. Only if you have really tried yourself, turn to the next page for hints that frankly give too much away. Only look at these as a last resort. You need to be doing your best to find your own mistakes now.
Horizontal expression for mass M:
( ) + (- fT ) = M( )
Horizontal expression for mass m:
( ) + (- f ) = m( )
Vertical expression for 400 grams mass:
( ) + ( - T ) = (0.4 kg)( )
Common mistakes and oversights:
· The forces I left blank in the above are meaningful symbols (like f, T, and 4 N). Some are unknowns. Algebra will determine the unknowns if you set up all equations completely. It could be a system of 3 equations/3 unknowns. It could be easier than that. You have to look at what you have and do it all by yourself. If you rely on a friend or tutor to tell you how to solve an algebra system, you won’t develop your own way of doing it. Between the textbook, stuff we do in class, these tutorials, and your opportunity to ask questions in class, you have enough support to finish these problems all by yourself.
· The accelerations I left blank require some thinking. You have to factor in the idea that the string does not slip or tangle. This is a physical interpretation: the mass m and the 400 g mass share the same value of acceleration as a result of it. It is a pure physics concept.
· There were vertical expressions for mass M and for mass m, but they weren’t needed when you did your algebra for this problem. But since you don’t know that will happen for any given problem, be in the habit of writing all expressions down anyway. There will easily be other problems that will require you to set-up and use the vertical equations.
· Back at the leftmost FBD that is correctly done, there is a downward force n that is perfectly done. I sometimes encounter a person who thinks that the label on that force in that diagram can simply be mg. In my effort to talk that person out of the label mg there, the person might respond, “well, n and mg are equal so its fine.” I strongly disagree. The label mg there is flat-out wrong. And there can easily be situations or problems where such a label in the FBD will lead to wrong final answers. The next paragraph will defend my claim.
An Important Detail in the Vertical Expressions from the Middle FBD:
Below is a FBD excerpt. It is incomplete. It only shows forces acting on one of the two masses, and in that one, it fails to show the horizontal forces. So you would need to complete the story for the rest of the problem’s solution by completing the FBD on your own. But what is shown is accurate so far and would lead to the first vertical equation written below. The second vertical expression came from a different FBD.
n
mg
n – mg = m(0)
nT + (-n) + (-Mg) = M(0)
That 0 substituted above is not going to be true for all situations. And that 0 substituted there is the only thing that validates the extrapolation that n = mg. So anyone who walks around saying that n = mg without proving it from the V. I. Method is going to be completely unprepared when having to deal with the more general situation that involves a true vertical acceleration. The problem could take place in an elevator, for example. Or it could be happening on a rocket that’s accelerating upward.
Furthermore, this makes it clear that when two forces such as n and mg are the only forces acting on a thing, and these forces are opposite, the only law that can verify whether or not n and mg are equal is Newton’s ______Law.
What is the reaction force to the “mg” mentioned in the previous discussion?
The following notice is for people who refuse to change their language:
If you filled in Newton’s Third Law at the end of the last page, or were even tempted to state that, then it means that you don’t really believe Newton’s Third Law. The statement where “Newton’s ______Law” was to be filled in on the previous page was talking about the fact that n and mg don’t have to be equal, and it was talking about the conditions under which they could be equal. By now, it’s been seen several times that n and mg don’t have to be equal, so comparing their values would have nothing to do whatsoever with Newton’s Third Law. “Third” is not the answer to the fill-in, and n and mg are certainly not members of an action/reaction pair. Any hint of a temptation to answer “Third” there indicates lack of visualization of agent and victim, and that has to change.
The reason the answer is “Second” is simple: n + (-mg) = m(aVERT) is the 2nd Law statement that comes from the FBD, and it can handle a scenario like, say, the problem taking place on a rocketship that is accelerating vertically with an acceleration equal to aVERT. And n won’t equal mg if aVERT is not zero. So n and mg are conditionally equal, NOT always equal. Meanwhile:
- n does have a reaction force to which it’s always equal, and that reaction force is NOT mg.
- mg does have a reaction force to which it’s always equal, and that reaction force is NOT n.
These last two sentences were NOT the reverse of each other. Sentence 1 should have conjured up the image in your head of Agent Mass M being the victim of n’s reaction force. Sentence 2 should have conjured up the image in your head of Agent Mass Planet Earth being the victim of mg’s reaction force. If these mental pictures are not happing, Free-Body Diagram analysis becomes a waste of time. If agent and victim are visualized, as habit every time, then FBD analysis becomes simple, powerful, and invincible..