Physics: Solving a TwoBody Pulley Problem

1
Physics Solving a Two-Body Pulley Problem

• By David M. Scrofani
• MD 400A, Spring 2005

Objectives
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Objective

• The purpose of this tutorial is to teach the
  viewer how to solve for the acceleration of each
  mass in a standard two-body pulley system (Atwood
  machine). Such problems are typical in
  introductory level physics courses and are used
  as an example of the application of Newtons
  Laws.
• This tutorial assumes basic knowledge of
  kinematics and dynamics as would be typical in
  the first few months of a high school physics
  course. Familiarity with solving algebraic
  systems of equations is also assumed.
• Click on the next slide button to see
  instructions and an animated example of such a
  pulley system.

Next
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Quick Review Newtons Laws

• (1) A body at rest tends to remain at rest a
  body in motion tends to remain in motion.
• (2) SF m a
• (3) For every action force, there is an equal and
  opposite reaction force.

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Example of a Two-Body Pulley Problem

• In a standard two-body pulley problem, the
  student is given the mass of each of the two
  masses that hang from a pulley and asked to
  calculate the acceleration. Usually, the mass of
  the pulley, itself, is neglected and the system
  is assumed to be frictionless.
• Clicking on the link below will take you to an
  applet on the web which simulates such a two-body
  pulley system. You may vary the mass of each of
  the hanging masses. But you should choose zero
  for the mass of the pulley (mass 3) and also
  make the system frictionless.
• Try some different values for the different
  masses to get a feel for how the system behaves.

Applet
Next
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Structure and Navigation of this Tutorial

• In order to learn how to calculate the
  acceleration of the masses in the pulley system,
  you will follow a series of steps (six steps
  total) that will be taught to you on a series of
  slides. These slides ask you to proceed in a
  specified order through the steps. At any time
  you may go back to the previous slide or jump
  back to the Six Steps Summary Slide
• After viewing the last of the six step slides,
  you will have a chance to try an example
  calculation for yourself to see if you can apply
  the six steps.

6-Step Summary
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Six Steps Summary Slide

• Step 1 Draw a Picture of the Situation
• Step 2 Draw and Label all Force Vectors
  According to Standard Convention
• Step 3 Decide on a Direction for the
  Acceleration of Each Mass
• Step 4 Write Newtons Law for Each Mass
• Step 5 Solve the System of Simultaneous
  Equations that Results from Step 4
• Step 6 Examine the Answer to See that it
  Makes Sense Physically

Step 1
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Step 1 Drawing a Picture

• Lets assume that you have been given the
  following problem to solve A 2kg mass and a 4kg
  mass hang on opposite sides of a mass-less,
  frictionless pulley. Determine the acceleration
  of the masses.
• The first thing you should do is to draw a
  picture of the situation. Like this . . .

• Notice that we have drawn the less massive block
  a little smaller than the more massive one. This
  is not essential but it is good practice.
  Anything you can do to represent things
  realistically will help you understand what is
  happening in the long run.

6-Step Summary
Step 2
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Step 2 Draw and Label Force Vectors According to
Standard Convention

• We have drawn the two vectors for the force of
  gravity on each mass (their weights). Since we
  already know that the force of gravity is the
  mass times the gravitational field strength at
  the Earths surface (9.8m/s2 which we will round
  to 10m/s2 here for simplicity) we write those
  values right on the diagram. W mg and we get
  the two values 40 Newtons and 20 Newtons.
• Note that we have tried to draw the 20 N force
  vector about half as long as the 40 N one.

• The two upward force vectors represent the force
  of string tension on each mass. The force is
  equal on each mass thats from Newtons third
  law for equal and opposite forces. Since we dont
  know the value for this tension force we just
  label it T.

All of these force vectors have been drawn to
standard convention. Click on this button to see
an example and explanation of an incorrect
drawing.
Example
6-Step Summary
Step 3
Step 1
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Incorrect Drawing of Force Vectors

• Here we have just been sloppy with our force
  vectors.
• We have not drawn the forces of gravity (the
  weights) as originating from the objects centers
  of mass
• We have drawn the weight of the 2 kg mass as long
  as (if not longer than) the weight of the 4 kg
  mass.
• The tension force vectors are off center and not
  originating from the point of contact.

• Such a poor drawing may not cause you to obtain
  an incorrect answer for this problem. But it will
  make it harder for someone else looking at your
  work to understand your thinking. And perhaps,
  worse, it allows bad habits to develop which will
  ultimately make things harder for you in the long
  run.

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Step 3 Decide on the Direction of Each Mass
Acceleration

• This is an important step. By examining the
  situation you can tell that the 4 kg mass will
  accelerate down and the 2 kg mass will accelerate
  up.
• We put arrows alongside the masses to indicate
  this.
• The direction of acceleration is taken to be the
  positive direction for forces acting on that mass
  when we write Newtons Law in the next step. It
  is very important to observe this convention or
  your answer will be incorrect.

a
a
6-Step Summary
Step 4
Step 2
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Step 4 Write Newtons Law for Each Mass

• For the mass on the left
• Since the acceleration is downward (see last
  slide) the downward force is positive and the
  upward tension force is negative. Filling in
  Newtons Law we get

• For the mass on the right
• Since the acceleration is upward (see last slide)
  the upward tension force is positive and the
  downward force of gravity is negative. Filling in
  Newtons Law we get

T
2kg
6-Step Summary
Step 5
Step 3
20 N
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Step 5 Solve Equations for Two Variables (a and
T)
Here we are most concerned with the value of the
acceleration and not the value for the tension
force since it was not asked for in the original
problem statement. On the next slide well take a
moment to consider whether or not these answers
make any physical sense to help us determine
whether or not we have made any mistakes. The
solution on the left is simply presented to you
assuming that you have some familiarity with
solving a system of two equations with two
variables. If you would like to see the algebraic
steps used in this solution click the button to
the left below.

• Equation 1
• Equation 2
• Solution

Algebra
6-Step Summary
Step 6
Step 4
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Algebra for the Equations Solution
Starting with these two equations we solve the
second one for T
Now we plug this value for T into the top equation
Now we solve this equation for a
We solve this for a which gives us a 3.33
m/s2. Plugging this value for a back into either
original equation gives a value for T.
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Step 6 Examining the Answer

• This step which is often neglected by beginning
  problem solvers is very important. Often, if you
  have made a mistake in the calculations you will
  know it right away if your answer does not make
  any physical sense. Lets check our answers this
  way.
• Recall the value of the acceleration, a 3.33
  m/s2. Is this number reasonable? The largest it
  could ever be is 9.8 m/s2 (or 10m/s2) since that
  is the acceleration caused by gravity in free
  fall. And the smallest it should ever be is 0
  which would be the case if both masses were the
  same mass. Our value of 3.33 m/s2 is somewhere in
  the middle which is a good thing.
• Now recall the value for the tension force T
  26.67 N . This value is between the two values of
  weight for the two masses. Think of it this way
  26.67 N is less than 40 N and so the mass on the
  left accelerates down. And 26.67 N is more than
  20 N so the mass on the right accelerates up. Now
  that makes sense!
• We conclude that the solution is probably correct
  since our answers fall within an expected range.
  If we had gotten an acceleration greater than 9.8
  m/s2 or a tension force greater then 40N (or less
  than 20N) we would know that we had made a
  mistake somewhere.

6-Step Summary
Practice Problem
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Practice Problem

• Now its time for you to try to apply these
  steps. Try this problem.
• A 12 kg mass and a 3 kg mass hang on opposite
  sides of a pulley. Determine the acceleration of
  the system and the force of tension in the cable
  connecting the masses.
• Remember the six steps are
• Step 1 Draw a Picture of the Situation
• Step 2 Draw and Label all Force Vectors
• Step 3 Decide on a Direction for the
  Acceleration of Each Mass
• Step 4 Write Newtons Law for Each Mass
• Step 5 Solve the System of Equations that
  Results from Step 4
• Step 6 Examine the Answer to See that it Makes
  Sense Physically

Click for a correct drawing
Click for correct equations
Click for the solution
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Drawing With Force Vectors
a
T
T
30 N
a
120 N
Back
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Equations
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Final Solution

• Instead of providing the solution here I have
  provided another link to the applet web site that
  you visited at the beginning of this tutorial. Go
  there and enter the settings for this problem (be
  sure to make mass 3 the pulley mass zero,
  and make the system frictionless). Check your
  answer by observing the simulation.

Applet
Conclusion
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Conclusion

• Congratulations on successfully completing this
  tutorial. At this point if you would like, you
  may go back to the applet on the web site and
  experiment further by allowing the pulley to have
  mass and by allowing the system to have friction.
  Such complications are beyond the scope of what
  we will be covering in this course but why not
  experiment?
• Thank you for viewing this lesson.

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