bullet

Traditional Atwood Machine

Two masses are connected with a string through a massless and frictionless pulley. The mass m1 is 80 Kg while the mass m2 is 125 Kg (see figure). Find the acceleration of the masses and the tension on the string. Assume all ideal conditions for pulley and masses.
The acceleration of the masses is

 

a)

 1.72 m/s2

 

b)

 9.80 m/s2

 

c)

 4.90 m/s2

N

d)

 2.15 m/s2

 

e)

None of the above.

and the tension on the string is

N

a)

 956 N

 

b)

 1116 N

 

c)

 441 N

 

d)

 2009 N

 

e)

None of the above.

 

Solution:

In this problem, since the masses are different, the masses will accelerate lifting the lighter mass and lowering the heavier mass. Based on the mass definition of the drawing, the pulley will turn counterclockwise. For solving this problem, it is necessary to solve for the acceleration of the two mass and the tension of the string.

It is important to notice that because of the pulley being ideal, no mass or friction, the tension in the two sides of the string is the same, . At the same time, for an ideal string, no stretching under tension, implies that the magnitude of the acceleration for the two masses are the same, . The direction of the accelerations are opposite to each other because the mass m1 accelerates upward while the mass m2 accelerates downward.

Therefore, only two equations are necessary to solve this problem. These two equations will be derived by working the FBDs associated with the individual masses.

FBD Mass m1

The applet at the left shows the development of the free body diagram to its final form for this mass.

The analysis of the diagram lead to the net force (remember that forces pointing up are positives and forces pointing down are negative),

The next relation establishing the net force is derived from Newton's second law,

Equating the two previous relations, it is obtained the first equation for the problem,

     Equation 1

This equation has two unknown; therefore, can not be solved by itself. A second equation is needed for solving this problem.

FBD Mass m2

The applet at the left shows the development of the free body diagram to its final form for this mass.

In this case, the free body diagram provides the following relation for the net force for the second mass (see the last diagram of the applet)

Notice that the tension pulls up both masses (in both cases the string is "holding" the masses). From Newton's second law, the following relation is obtained,

It is very important to notice that the acceleration is negative because this mass is accelerating downward. From these two relations the following equation is derived,

     Equation 2

What about if you can not visualize the situation? There is not problem, it is only necessary to visualize that the two masses accelerate in opposite direction. If you mistakenly chose the wrong direction for the acceleration (clockwise in this case); at the end, the result for the calculated acceleration is a negative number. From there, it is concluded that the actual acceleration is in the opposite direction than the  acceleration originally established.

Now we have two equations and two unknown, such a system of equations can be solved. Subtracting Equation 2 from Equation 1 we obtain for the acceleration,

After solving for the acceleration, this result can be substituted back into Equation 1 or Equation 2 in order to solve for the tension of the string. In the applet below, the tension is solved starting from Equation 1. The final expression for the tension is,

The previous two relations are the algebraic solutions of the problem.

 

The numerical solutions are obtained by substituting the numerical values of the masses and the acceleration of gravity.

Notice that acceleration is positive which means that the direction of the acceleration is the same as the predicted direction of the motion, counterclockwise.

bullet
by Luis F. Sez, Ph. D.    Comments and Suggestions: LSaez@dallaswinwin.com