Two Masses and a Pulley, and a New Misconception

The badly sketched picture to the right shows a classic mechanics problem.  Two equal masses are connected by a string over a pulley.  In this case, the table is frictionless.

Typically, a student is asked to determine the tension in the rope and the acceleration of the masses.  Great -- that's (mg)/2 and g/2.*  This problem is richer, though, than a mere calculation might suggest.  Take a look at a quiz I gave the other day:

* The quick way to get this is to consider both objects as a single system.  The net force on that system is the weight of the hanging mass, mg; the mass of the system is 2m.  By Newton's second law, a = g/2.

1.       A block of mass m is attached over a pulley to another hanging mass m, as shown above.   The surface is frictionless.  The system is released from rest.

     

(a)    What is the direction of the hanging mass’s acceleration?  Explain.

(b)   Is the acceleration of the hanging mass greater than, less than, or equal to g?  Explain.

(c)    Is the tension in the rope greater than, less than, or equal to mg?  Explain.

2.       A block of mass m is attached over a pulley to another hanging mass m, as shown above.   The surface is frictionless.  This time, the top block is given an initial velocity to the left and released. 

(a)    What is the direction of the hanging mass’s acceleration?  Explain.

(b)   Is the acceleration of the hanging mass greater than, less than, or equal to g?  Explain.

(c)    Is the tension in the rope greater than, less than, or equal to mg?  Explain.

Ideally, 1(a) is answered with a kinematic approach -- the hanging mass is speeding up and moving down, so acceleration is also down.  For 1(b), I've defined "free fall" as the situation in which no forces besides weight are acting.  Since a tension acts upward on the hanging mass, the mass is not in free fall and the acceleration is less than g.*  And in 1(c), acceleration is downward, so net force must also be down.  That means down forces greater than up forces, so the tension is less than the weight.

* Okay, sure, if the upward tension is twice the block's weight, the acceleration could be g, upward.  That's highly unlikely in hanging-block-and-pulley problems.  

Of course, question 2 is identical to question 1!  The hanging mass is moving up but slowing down, so acceleration must still be downward.  (Or, one could argue that the block on the table still experiences only one horizontal force, that of tension, so its acceleration must be to the right; the blocks must move as a unit, so the hanging block has downward acceleration.)  Once it's established that acceleration is still down, questions 2(b) and 2(c) follow as in 1(b) and 1(c).

By far the most common misconception here is that the net force must be in the direction of movement.  A student will commonly get question one reasonably correct, but then say "the block is moving upward, so up forces must be bigger than down forces."  This question is just one more salvo in my arsenal aimed at that piece of nonsense.  

Another typical misconception is that in question 1, since the hanging block is falling near earth, its acceleration must be g.  That's taken care of with a request to state the definition of free fall and a sheepish look from the student.

And, a common mistake is to justify (a) and (c) with circular reasoning:  The acceleration is downward because the weight is greater than the tension; the tension is less than the weight because the acceleration is downward.  This student doesn't earn full credit, but I'm not worried so much about his comprehension. 

I discovered a new misconception today, though.  One of my brighter students said acceleration was equal to g, and he stated the definition of free fall accurately.  He asked, "since the surface is frictionless, the block on the table doesn't require any force to move.  So why won't the rope will be slack, the tension zero, and gravity the only force on the hanging block?"

At first I was flummoxed.  I set up two carts on my track, and showed him that the string was in fact not slack.  But why on earth would he think that no friction leads to a slack rope?

In further conversation, I discovered that he was referring back to our class's multiple conversations about how no net force is necessary for motion at constant speed in a straight line.  A mass on a frictionless track, once moving, keeps moving, even without any tension to pull it.  My student wasn't processing that this block on this surface was accelerating, not moving at constant speed.  Once I pointed out how the blocks must move together, and therefore accelerate together, he got it.

Multiple Choice quiz: two-body problem in an elevator

Diagram for today's problem, modified from
something in (I think) Serway & Vuille

A couple of nights ago, I assigned a two-body problem in an elevator, from (I think) Serway & Vuille.  Two blocks were hanging from an elevator as shown in the picture; the acceleration in the original problem was upward.  On the homework, I asked (among other things):

• Draw a free body diagram for each object.
• Is the tension in the lower rope greater than, less than, or equal to 35 N?
• Calculate the tension in each rope.
• The ropes have a breaking tension of 85 N.  Calculate the maximum acceleration that will cause a rope to break.
• When a rope is observed to break, explain how the elevator was moving.

This problem is one of the best at separating those who are following an appropriate physics problem solving procedure from those who are just trying to plug numbers into some random equation.  The students who used the free body diagrams to write (up forces) - (down forces) = ma got the right answers, and got them quickly. 

On the other hand, the students who didn't carefully write the equations were confused for most of an hour, got the final answers correct because they asked friends for help, but usually earned little credit -- if after collaboration they just wrote "T = ma + 35 N, so T = 40 N" I marked the answer wrong.  Why?  Because I saw no evidence of how they got to that equation, other than listening to a friend without understanding.  Would an English teacher give credit for a one-sentence essay, even if the one sentence is spot-on in its conclusion?  Of course not.  So why on homework should I reward the correct numerical answer when it was essentially derived through magic?

I invited in for extra help the students who didn't follow the correct method.  They now feel much more confident about two-body problems, because they see that all they have to do is write the correct Newton's Second Law equations from the free body diagrams.  But it's still worth a follow up quiz -- either I build significant confidence, or I discover further misconceptions.

Below is today's three-question quiz that I'll give at the opening of class.  (It refers to the diagram above, in which the acceleration is DOWNWARD.  Yeah, I switched the direction of acceleration for the quiz.)  The "distractor" answers in the second question quote some students verbatim.  

Two 3.5 kg blocks hang from ropes in an elevator, as shown above.  The acceleration of the elevator is 1.6 m/s², downward.  While the elevator has this acceleration, the tension in the bottom rope is 29 N.

19. Which of the following best describes how the elevator’s speed is changing?

(A) The elevator is speeding up.

(B)  The elevator is slowing down.

(C)  The elevator is moving at constant speed.

(D) Whether the elevator is speeding up or slowing down cannot be determined.

  

20. Which of the following describes the meaning of an acceleration of 1.60 m/s²?

(A) The elevator gains or loses 1.6 meters per second of speed each second

(B)  The elevator gains or loses 1.6 meters each second

(C)  The elevator travels 1.6 more or fewer meters each second

(D) The elevator travels 1.6 m/s² more or less each second

(E)  The elevator is either speeding up or slowing down by 1.6 meters for every second squared.

  

21.  Now the magnitude of the elevator’s acceleration is doubled to 3.2 m/s², still directed downward.  What is the tension in the bottom rope now?

(A) 41 N

(B)  35 N

(C)  32 N

(D) 24 N

(E)  0 N (i.e. the rope goes slack)

Two-body problem quantitative demonstration

Consider the classic two-body problem, in which a cart on a horizontal low-friction track is attached via rope-and-pulley to a hanging mass.  The cart has mass 340 g, and the hanginging mass is 100 g.  I release the cart, which speeds up.

Is the tension in the rope...
(A) greater than 1 N
(B) less than 1 N
(C) equal to 1 N?

"Equal," say a majority of the students -- because the rope is attached to the 100 g hanging mass, which has a weight of 1 N.

"Less than 1 N," the cleverer ones respond.  The hanging mass has a downward acceleration, because it is moving down and speeding up.  So the down forces must be GREATER THAN the up forces, meaning the tension is less than the 1 N weight.

"Let's see," I say.  The picture shows my cart with a Vernier force probe taped inelegantly on top.  The string is attached to the force probe, is run over a low-friction Pasco pulley, and connected to a 100 g hanging mass.  I tell Vernier's Logger Pro software to collect force probe data.  Before I let go of the cart, the probe reads 1.0 N.  As soon as I let go, the force probe's reading very obviously dips, to something like 0.7 - 0.8 N.  Looks like the cleverer ones were right.

Next, I use free body diagrams and Newton's second law to predict the acceleration of the cart and the tension in the rope -- I get 2.3 m/s² and 0.77 N.  Sure enough, that's just about what the force probe and a motion detector read.  (I get the acceration from the slope of a velocity-time graph, which I make with the vernier motion detector.)

Next question:  Instead of letting the cart go from rest, I give the cart a shove to the right, away from the rope. 

After I let go but as the cart is still moving to the right, is the tension...
(A) greater than 1 N
(B) less than 1 N
(C) equal to 1 N

"Greater than 1 N," say the majority.  "The hanging mass is now moving upward, so the up forces must be greater than the down forces."

"Nonsense," I say.  The mass is slowing down while it moves upward.  Slowing down means that acceleration and velocity are in opposite directions.  Thus, the acceleration must still be in the downward direction, and the tension must still be less than the weight.

In fact, the entire set of free bodies is unchanged from the previous problem -- the force that I pushed with doesn't act once I let go!  So, all calculations are the same, and the acceleration and tension should be unchanged.

Of course, I finish the class by doing the experiment -- sure enough, the tension and acceleration readings are the same as before.

GCJ

Email discussion group and an excellent conversation

I set up an internal email conference accessible only to my classes.  Each class has a separate folder, to which I post assignments and commentary.  Students are also encouraged to ask specific questions of each other and of me on the folder.

Getting the students to use this folder properly and usefully is difficult.  Most are too timid to post anything.  Then when someone does post, they say "I can't do #4.  Help."  I have to tell such a student to try again with a SPECIFIC question:  say, "On #4, I drew the free body diagram, but I can't figure out how to get the components written."  Specific questions can provoke excellent e-discussions.

I also use this conference myself to hammer home points that I don't want to make in class; or, points that I have made in class, but need reinforcement.  For example, yesterday when I graded homework I saw some ridiculous free body diagrams; posted the picture above with the subject line "What are the two things wrong with this free body for the boat?"  The body of the message included the statement, "First to post the answer gets candy!"

A few minutes later, I found a student's homework contradicting Newton's second law.  Instead of screaming, I merely posted to the folder, offering another piece of candy:  "What's wrong with the statement 'Since the resistive force exerted by the water is less than the forces exerted by the ropes, the boat can move in the direction of the ropes.' "

I got excellent answers.  To the latter question, student Peter Chen replied: 

"You don't necessarily need to exert greater forces than the resistive force to make it move. What you really need is the same quantity of forces on the opposite direction to the resistive force. If you don't fulfill this requirement, no matter how great the forces are, you cannot move the boat.  eg: Even if you exert 1 billion Newton of forces to both sides of the boat, which are perpendicular to the direction of the resistive force, you will never move the boat. Because the quantity of forces on the opposite direction to the resistive force is 0.

To the question about the free body diagram, I reposted James Moyler's answer with commentary:

1.You shouldn't write 600N on the free body diagram. It is a rope so it should say T1 and T2. You don't put quantitative values on a FBD.

Correct.

2. F isn't enough. F of what? I believe it should be Fr for Force of Resistance.

In this particular problem, the resistive force was DEFINED as "F." So the plain old label "F" is okay here. But generally, Mr. Moyler is correct -- put a subscript or words on the label on a free body.

Another student, Frederic Lamontagne, followed up:

Firstly, there is no normal force when the object is on water. The force on water is Buoyant Force.  Secondly, [this force] would not be acting to the left (as it is indicated in the free body diagram)...it would act perpendicular to the water.

Absolutely correct.

Thing is, everyone in the class read that exchange.  Everyone in the class had done those problems just the day before, so the information was current and relevant.  The five minutes of my time it took to post and reply were worth 20 minutes of lecture time.

Pushing a truck and Newton's Second Law

At my Manhattan College Summer Institute, participants can earn graduate credit by submitting a "final project."  As one focus of the course is the development and use of quantitative demonstrations, an option for the final project is to create a couple quantitative demonstrations for use in participants' classes.

Alex Tisch, a colleague of mine starting this year, took this institute and submitted a clever but involved demonstration he had brainstormed.  Before I give the details, I ask you to BE CAREFUL with this one -- it involves a live, full-sized, moving pickup truck.  I generally scoff at the dire warnings that textbooks present with their lab ideas ("Drop a tennis ball and time its descent.  But WEAR GOGGLES!!!  And a cup!")... but don't do this particular experiment unless you're sure you can accomplish it safely.  If you're worried, have the class watch while a few faculty members carry it out.

Alex envisions his pickup truck in neutral on a level surface.  One person is in the driver's seat, just to keep the truck going straight.  One person sits in the back with a turkey baster full of paint; maybe another person is back there with a stopwatch.

A big strong person pushes the resting truck, trying to apply a relatively constant force.  Once he starts pushing, the painter drops one dollop of paint off of truck every second.

The class can now measure the distance from the starting point to each dollop of paint on the road.  From this data, a position-time graph can be made; analysis of this graph can lead to a calculation of the acceleration of the truck.  Knowing the mass of the truck + occupants, the average force applied by the pusher can be calculated using Newton's second law.

Neat, but not awesome yet.  Alex's big brainstorm was to MEASURE the force of the pusher while he's pushing.  He envisioned a bathroom scale with a nearby observer to record the force every second.  I suggested a force plate.  (A force plate is essentially a bathroom scale that connects to Vernier or Pasco data collection softward, so can make a live graph of force vs. time.)  Vernier's force plate even comes with handled attachments that make pushing pretty easy.

The force calculated from Fnet = ma can be compared to the force measured directly from the force plate.  Awesome.

(Picture from Leatherneck magazine.)

Three masses connected over a pulley

In my 5 Steps to a 5 AP Physics prep book, page 70 is a review of tension problems, also known as many-body problems.  I give eight different situations in which blocks are connected by ropes.  The goal of each problem is to find the tension(s) in the rope(s), and the acceleration of the system.

The approach that I advocate is to draw a separate free body diagram for each block, then write Newton's second law separately for each block.  The acceleration and tension(s) are solved for by adding the Newton's second law equations together.

Ruth Mickle, of Atlanta, noted yesterday that she gets a different answer to problem 6 than is printed in the book.  I agree -- for whatever reason, the answer in my book is wrong.  Below I give a thorough solution. 

The problem shows three blocks connected by strings over a pulley, as shown at the top of the post.  Given that m is 1.0 kg, the question asks for the tensions and acceleration.

Start by drawing three free body diagrams.  Note that the two ropes will have two tensions; I'll label these T1 and T2.

The acceleration will be toward the heavier blocks.  Thus, the mass m will accelerate upward, and the other masses will accelerate downward.  So when we write Newton's second law, for mass m we'll write "up forces - down forces = ma."  For the other masses, we'll write "down forces - up forces = ma."  Always start Newton's second law in the direction of the acceleration.

T2 - mg = ma          2mg + T1 - T2 = 2ma                       4mg - T1 = 4ma

Now, add 'em up.  Note that the T1s and the T2s will cancel in the addition:

-mg + 2mg + 4mg = 7ma

Solving for a, we get a = (5/7)g, or 7.1 m/s².

Now just plug back into the equations above to find that T2 = 17 N, and T1 = 11 N.

Clicker activity -- basics of Newton's Second Law

My wife has been gone for four days. She's hiking with the sophomores on their Outward Bound trip to North Carolina. That means, however, that I'm in charge of six-year-old Milo all by myself. I quite enjoy occasional solo time with the boy. But what to do on Saturday morning, when I had to teach a physics class some more about Newton's Second Law?

I prepared a clicker exercise. Milo loves using the clickers, and he loves being part of a class with juniors and seniors. In turn, the juniors and seniors are welcoming and friendly to Milo -- even moreso due to the "Milo Questions."

I had Milo write a set of multiple choice questions about himself. I promised to use these as part of the in-class activity. Thus, the overall set of questions for the day's clicker exercise consisted of two Newton's Second Law questions, followed by one Milo Question, followed by two more second law questions, etc. The class (including Milo!) was divided randomly into teams; the team that got Milo was excited, because they knew that they had the Milo questions in the bag. As always, each team could collaborate and submit a single answer to each question. They earned one point for a correct answer, and one bonus point for each group who did NOT get the correct answer.

The actual set of questions is below. Feel free to use them. They may sound really easy, but remember how difficult it is to remember and assimilate even the most basic facts about the second law. It takes an amazingly huge number of repetitions before we can break down the most common misconceptions like "motion requires a force" and "acceleration tells which direction something is moving."

(I gave a "fundamentals quiz" about some of these same ideas a few days later. I'll try to post that soon.)

GCJ

1. A bucket whose mass is 10 kg hangs by a rope in which there is 63 N of tension. What is the weight of the bucket?
(A) 100 N
(B) 10 N
(C) 10 kg
(D) 100 kg
(E) 63 N
(F) 63 kg
(G) 73 N
(H) 73 kg
(I) 163 N
(J) 37 N


2. A bucket whose mass is 10 kg hangs by a rope in which there is 63 N of tension. What is the net force on the bucket?
(A) 37 N
(B) 163 N
(C) 100 N
(D) 63 N
(E) The answer depends on which way the bucket is moving.


3. What color is Milo’s house?
(A) Green
(B) Blue
(C) Purple
(D) Yellow
(E) Grey
(F) Brown
(G) White


4. A bucket whose mass is 10 kg hangs by a rope in which there is 63 N of tension. What is the magnitude [i.e. the amount] of the bucket’s acceleration?
(A) 6.3 m/s2
(B) 0.63 m/s2
(C) 3.7 m/s2
(D) 0.37 m/s2
(E) 10 m/s2
(F) 1.0 m/s2

5. A bucket whose mass is 10 kg hangs by a rope in which there is 63 N of tension. What is the direction of the bucket’s acceleration?
(A) Up
(B) Down
(C) The direction of acceleration is unknown


6. What does Milo do after seated meal?
(A) Go out back
(B) Go home
(C) Come here
(D) Go to bed


7. A bucket whose mass is 10 kg hangs by a rope in which there is 63 N of tension. What is the direction of the bucket’s velocity?
(A) Up
(B) Down
(C) The direction of velocity is unknown


8. So how could it possible for the bucket to move upward, then?
(A) The bucket must be slowing down
(B) The bucket must be moving at constant speed
(C) The bucket must be speeding up
(D) The tension has to increase to more than 100 N


9. How many bunnies does Milo have?
(A) 0
(B) 7
(C) 2
(D) 1

Mailbag: Forces on a Table

From Michael Herrin, from Chestatee High School in Gainesville, GA:

I’m having trouble working a problem. It’s from the Cutnell and Johnson’s book chapter 4, and I am curious if you could help point me in the right direction.

Mr. Herrin actually pointed me to a problem in the 5th edition, but I found a slightly different version in the newer edition. In the diagram shown, all pulleys are massless and the surfaces are frictionless. Find the tension in the rope and the acceleration of the masses.

The problem solving process that I teach for Newton’s second law problems is:

1. Draw a free body diagram.
2. Break angled forces into components, if necessary.
3. Write (up-down=ma) and (left-right=ma).[1]

This is a two-body problem; therefore, we start with not one but TWO free body diagrams: Look here -->

Note the tricksiness of the diagram for the 3 kg mass. The rope is pulling up TWICE on this mass: once from the left side of the pulley, once from the right side. So we put two tensions on the diagram.

Jacobs’ law of tensions says “One Rope Equals One Tension.” That’s why I didn’t label the tensions T₁ and T₂, but just T. The tension will be the same throughout.

Now, write Newton’s second law twice, once for each block. The direction of acceleration will be to the right for the 10 kg block, and down for the 3 kg block. (We can see that by imagining releasing the masses from rest, and seeing which way the blocks speed up.)

T = (10 kg)a and (3 kg)g - (T+T) = (3 kg)a

Now, the PHYSICS IS DONE: I have two equations and two unknowns, a and T. Everything else is a known value. All that’s remaining is mathematics.

I think it’s easiest to solve by addition – multiply the first equation by 2, and add the equations together. This cancels out the tension terms; solve to get a = 1.3 m/s²? . Plugging back into either equation, I get T = 13 N. That’s reasonable: the tension is less than the weight of either mass, but is on the same order of magnitude.

Now, ask your students: what would happen to the acceleration if we were to give the 10 kg mass an initial shove to the left. Would the acceleration be greater than, less than, or equal to 1.3 m/s²? Post YOUR answer in the comment section.

GCJ



[1] Well, okay, sometimes we write (right-left=ma), not (left-right=ma). How to know which is which? Determine the direction of acceleration, and start with that direction.