Atwood Machines

    
Introduction

Atwood Machines have a frictionless, torque less, pulley with masses hanging from both sides.

1.	The tension on the rope is the same quantity on both sides.  Tension is always an upward force on the masses.
2.	The side with the greatest total mass will accelerate downward, and the other side will accelerate upward.
3.	Fill in the equation  F = ma  for each mass.
4.	Solve each equation for T.
5.	T = T  and other sides of the equations are also equal.
6.	Solve for acceleration.
7.	Use either equation in step #4 to calculate T.
3.	If there is more than one mass on a side
	a.	First calculate the acceleration for each side using the total masses.
	b.	Use the acceleration from the first step to calculate the tension on the rope connecting the two sides.
	c.	Fill in the equation  F = ma  for the other masses to calculate the tensions on the ropes connecting masses on the same side of the pulley.
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Problem #1

The red mass is 20.0 kg and the blue mass is 10.0 kg.  Calculate the acceleration of each mass and the tension on the rope.

Fill in the equation  F = ma  for each mass.

Down is positive and up is negative.  Since the red mass is heavier, it will accelerate downward; and the blue mass will accelerate upward.

Force _gravity is always downward.

mass _red g  - T = mass _red a                          Red mass accelerates downward

mass _blue g - T = - mass _blue a                      Blue mass accelerates upward

Solve each equation for T. 

T = mass _red g  - mass _red a

T = mass _blue g + mass _blue a

T = T  and    mass _red g  - mass _red a = mass _blue g + mass _blue a

Solve for a

mass _red a +  mass _blue a  =  mass _red g - mass _blue g
    

        ( mass _red - mass _blue )g           (20.0 kg - 10.0 kg) 9.80 m s ^-2 
a = ------------------------------------ =  ----------------------------------------- =   3.27 m s ^-2    
       mass _red  +  mass _blue                         20.0 kg + 10.0 kg

Substitute into either equation to calculate the tension.

 T = mass _red g  - mass _red a = 20.0 kg (9.80 m s ^-2  -  3.27 m s ^-2 )  =  131 N

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Problem #2

The red mass is 20.0 kg, the blue mass is 10.0 kg, and the green mass is 5.00 kg.  Calculate the acceleration of each mass and the tension on the rope.

Fill in the equation  F = ma  for each side.  The red mass and green mass will have the same acceleration and can be combined to find the accelerations and tension on the black rope.

Down is positive and up is negative.  Since the red mass + green mass is heavier, they will accelerate downward; and the blue mass will accelerate upward.

Force _gravity is always downward.

mass _{red + green} g - T =  mass _{red + green} a

mass _blue g - T = - mass _blue a 

    
Solve each equation for T

T = mass _{red + green} g - mass _{red + green} a

T = mass _blue g + mass _blue a 

T = T  and  mass _{red + green} g - mass _{red + green} a  =  mass _blue g + mass _blue a 

Solve for a

mass _blue a  + mass _{red + green} a = mass _{red + green} g - mass _blue g
   

       (mass _{red + green}  - mass _blue )g        (25.0 kg - 10.0 kg) 9.80 m s ^-2 
a = ------------------------------------------ = ------------------------------------------- = 4.20 m s ^-2  
       mass _blue   + mass _{red + green}               25.0 kg  +  10.0 kg

The tension in the black rope can be calculated using either equation.

mass _{red + green} g - mass _{red + green} a = 25.0 kg (9.80 m s ^-2 - 4.20 m s ^-2) = 140. N

The tension on the dark red rope between the red and the green masses can be calculated by using 

F = ma for the green mass.

mass _green g  -  T = mass _green a                                Solve for T.

T = mass _green g  -  mass _green a  =  5.00 kg (9.80 m s ^-2  - 4.20 m s ^-2) = 28.0 N

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