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Atwood Machines with Two Hanging Masses

Atwood Machines with Two Hanging Masses

Solve problems involving Atwood machines with two hanging masses.  Solve for acceleration and tension with variables alone or given masses.

Atwood Machines and Net Forces

Atwood machines are the subject of many AP Physics 1 problems.

An Atwood machine consists of two blocks hanging from a pulley.  The simple pulley with no mechanical advantage changes the direction of the hanging mass force.

The resulting blocks will accelerate if not in equilibrium.

Most problems will assume the pulley and string is massless.

Atwood with Two Hanging Masses

Common Forces in Atwood Machines

It is important to find commonalities so you can create and use mathematical expressions involving multiple objects like mass 1 (m1) and mass 2 (m2) in the animation.

Same Tension force in Atwood Machines

Notice a few facts necessary to understand when solving through Atwood machine problems:

  • Masses are attached to the same string.
  • Though the direction is different, the masses have the same magnitude of acceleration.
  • The side of the string attached to m1, and the other side attached to m2 have the same magnitude of tension (FT).

Determine if the Atwood Machine is in Equilibrium

Masses of an Atwood machine that is in equilibrium (ΣF = 0 N) will be in a state of rest or constant motion.

Masses on an Atwood machine that is not in equilibrium (ΣF ≠ 0 N) will be in a state of acceleration.

Creating an Atwood Machine Equation

Creating Atwood machine Newtons Second Law (Fnet = ma) equations are often used to determine the acceleration of a block or the common tension in the rope between the blocks.

Creating an Atwood Machine Equation in Equilibrium

Creating Atwood machine Newtons Second Law (Fnet = ma) equations are often used to determine the acceleration of a block or the common tension in the rope between the blocks.

An Atwood machine in equilibrium would have equal mass on either side and could be at rest or have the blocks moving at a constant speed.

Atwood Machine in Equilibrium

Newton’s Second Law equation (Fnet= ma) of either mass would be similar.

With up being positive:

ΣF = FT – m1g

The net force on each block would have to be 0 (ΣF = 0 N).

0 = FT – m1g

Therefore, when rearranged:

  • FT = m1g

And since the solution is the similar:

  • FT = m2g

The magnitude of tension in the string is equal to the either mass weight (m1g) or (m2g), only when an Atwood machine with two equal masses hung on each side of a pulley.

Creating an Atwood Machine Equation Not in Equilibrium

When the Atwood machine is not equilibrium (ΣF ≠ 0 N) it is accelerating.

  • The magnitude of each block's acceleration will be the same.
  • The tension (FT) in the string will be the same.
Atwood Machine Not in Equilibrium

Newton’s Second Law equation (Fnet= ma) of block 1.

With up being positive:

ΣF = FT – m1g

When ΣF is substituted in for Fnet, in Newton’s second law Fnet = ma equation and specialized with m1 and a1, the net force equation for block 1 would be:

FT – m1g = m1a1

Atwood Machine Not in Equilibrium m1

Newton’s Second Law equation (Fnet= ma) of block 2:

With up being positive:

ΣF = FT – m2g

When ΣF is substituted in for Fnet, in Newton’s second law Fnet = ma equation and specialized with m2 and a2, the net force equation for block 2 would be:

FT – m2g = m2a2

Atwood Machine Not in Equilibrium m2

Creating Common Variable to use in Substitution

  • FT – m1g = m1a1
  • FT – m2g = m2a2

Commonalities in these equations:

Tension (FT is the same)

With up being positive

  • a1 = +a
  • a2 = -a

Substitution of a creates:

  • FT – m1g = m1a
  • FT – m2g = -m2a

When solving for a

Rearrange the Newton’s Second Law for m1 equation (FT – m1g = m1a) for FT

  • FT = m1a + m1g

Substitute the above FT into the (FT – m2g = -m2a)

Newton’s Second Law equation for m2.

m1a + m1g – m2g = -m2a

 

Rearrange for a:

Step 1: Start by getting a’s on the same side

m1a + m2a = m2g – m1g

 

Step 2: Then factor out common factors a and g

a(m1 + m2) = g(m2 – m1)

Step 3: Rearrange for an isolated a

Acceleration of a Atwood machine with two hanging masses

When solving for Tension (FT)

Rearrange the Newton’s Second Law for m1 equation (FT – m1g = m1a) for a

m1 equation rearranged for a

Substitute the above a into the (FT – m2g = -m2a)

Newton’s Second Law equation for m2.

Net force Atwood equation rearranging step 2

Take steps to get FT on both sides and isolate FT:

Steps to isolate FT

Multiply both sides by m1

More steps to isolate FT

Q1) What is the acceleration of an Atwood machine that has a 10 kg hanging mass on the left and 5 kg hanging mass on the right side of the pulley?

Atwood machine with 10kg and 5kg mass

Look to earlier in the lesson to see how this equation below was derived.

a solution

Since we decided up was positive and:

  • a1 = +a
  • a2 = -a 

Substituting our answer for a above:

  • a1 = +(-3.33)
  • a2 = -(-3.33)

The 10 kg mass will accelerate 3.33 m/s2 down while the 5 kg mass will accelerate 3.33 m/s2 up.

Q2) What is the tension in the string of an Atwood machine that has a 10 kg hanging mass on the left and 5 kg hanging mass on the right side of the pulley?

Atwood machine with 10kg and 5kg mass

FT – m1g = m1a

FT = m1a + m1g

FT = (10)(-3.33) + (10)(10) = 66.7 N

The tension above both masses will be 66.7 N up.

Lets solve the same problem with the derived equation to check our work:

Derived Tension Equation With Two Hanging Masses on and Atwood Machine

The results are hidden until you click below:

checking FT equation

This is the same answer as using a prior equation and checks our work

Links

  • Forward: Modified Atwood Machines
  • Back to Weight Normal Force and Tension (Single Block)
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Unit 1: One Dimensional Motion
Unit 2: 2D Motion
Unit 3: Newton’s Laws and Force
Unit 4: Universal Gravitation and Circular Motion
Unit 5: Work, Power, Mechanical Advantage, and Simple Machines
Unit 6: Momentum, Impulse, and Conservation of Momentum
Unit 7: Electrostatics
Unit 8: Current and Circuits
Unit 9: Magnetism and Electromagnetism
Unit 10: Intro to Waves
Unit 11: Electromagnetic Waves
Unit 12: Nuclear Physics

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