Tampilkan postingan dengan label tension. Tampilkan semua postingan
Tampilkan postingan dengan label tension. Tampilkan semua postingan

Kamis, 17 November 2011

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.

Senin, 25 Oktober 2010

Lab idea for general physics: introduction to force components

 I had just introduced the idea of a force acting at an angle in my general physics course.  I needed an experiment for lab day.

In general physics lab, I want as often as possible to be able to make a linear graph, and use the slope or intercept to calculate a verifiable physical quantity.  The data collection process should be as simple as I can make it -- very little messing with computers, with minimal calculation before the linear graph appears.

My first thought was to use the PASCO fan cart.  The fan can blow straight ahead, or can blow at any angle all the way to 90 degrees off of straight ahead.  If we place this cart on a PASCO track --  with grooves keeping the cart from sliding -- a spring scale attached to the front of the cart can measure the force the cart experiences in the direction of the track.  This force would be equal to the force of the fan times the cosine of the fan angle.  A graph of the scale reading vs. the cosine of the fan angle would make a line whose slope is the force of the fan.

Problem is, I have 22 students and 11 lab groups.  I don't even have two fan carts, let alone 11; that many fan carts would bust my budget.

After considerable brainstorming with some AP physics alumni, a new thought occurred to me... what if I could keep the angle of an applied force constant?  Look at the lower picture, the picture of the red PASCO cart.  This is one of the newer plastic "pascar" models.  It has several convenient raised thingamabobbers (that's a technical term) useful for attaching strings. 

One string is attached to the middle of the left-hand side of the cart.  This string is passed over a pulley, and a hanging weight is hanged from the string.  A second string is held across the diagonal of the rectangular pascar, as shown in the picture.  The geometry of the pascar itself determines this angle -- students are instructed to ensure that the rope stays along the diagonal to control the angle.  A spring scale is attached to the angled rope.  The top picture shows the Nachoboy holding the spring scale at the correct angle; the hanging weight is visible above the garbage can.  Component analysis shows that the hanging weight is equal to the tension in the angled rope times the cosine of the string's angle.

Students vary the hanging weight, and measure the tension in the angled rope with the scale.  They graph the hanging weight vs. the tension in the angled rope.  The slope of this graph should be the cosine of the angle of the string!

Each student used the slope of his graph to figure out the angle of the rope.  I measured the angle with a protractor... I got between 21 and 24 degrees, depending on where the rope was anchored.  Those who did the experiment carefully (most of the class) got a slope of 0.90-0.95... giving an angle between about 18 and 25 degrees.  Woo-hoo!

GCJ

Selasa, 05 Januari 2010

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.