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The standard pulley problem as an example of systems.

Object 1 as a and object 2 as a separate system.

Each system is subject to external influences from the earth (gravity) and the rope (tension).

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diaga1 Diagrammatic Representation

We begin by treating each object separately. We construct free body diagrams:

where the force of tension on each of the objects is exactly the same magnitude. The direction of the force is upwards for both masses because the rope is wrapped around the pulley, and thus hangs downwards at each end (and pulls upward at each end).

matha1 Mathematical Representation

Using the free body diagrams as a guide, we write Newton's 2nd Law for each object. We consider only the y direction, since the x direction simply yields the equation 0 = 0. Because everything is in the y direction, we work with the magnitudes of the forces and explicitly include signs (rather than including lots of tedious "y" subscripts).

To finish this problem, we must use the constraint on the acceleration. The accelerations of objects 1 and 2 are constrained by the fixed length of the rope to be equal in size and opposite in direction. It makes sense to use this fact to eliminate one of the accelerations in the above equations and thus obtain a single expression for the other in terms of the gravitational constant g and the two masses. In our case, we arbitrarily substitute for a₂ using:

giving:

Eliminating the tension from the system of equations then gives:

Since the accelerations are the same size, we can characterize the acceleration of the system as:

Method 2

A second method is to use a "math shortcut" by defining an unphysical system that replicates the mathematical features of the problem.

sysa2 System: sysa2Mathematical construction including the two objects and the rope connecting them.

External forces from gravity on each block.

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This method uses the fact that when a massless, frictionless pulley is involved, the pulley has essentially no role in the system except to change the direction of forces. Thus, the dynamics of Atwood's Machine are completely equivalent to the dynamics of the system sketched here:

Of course, this system is anything but physical. The important point is that the mathematics are identical. This trick of "straightening" the system is often useful when a massless, frictionless pulley is involved. Note that we have chosen to draw the system in the x direction instead of the y direction to remind us that gravity will not pull in the same direction on each mass! Thus, the free body diagram for the system looks like:

and the resulting form of Newton's 2nd Law is:

so that again, the magnitude of the acceleration is:

As in Part A, Method 1.

As in Part A, Method 1.

As in Part A, Method 1.

To find the tension, which is an internal force if we combine the two masses and rope into one system, we must return to the approach of Method 1, treating each object separately. From our work in Part A, we actually have two expressions for the tension:

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In either case, inputting our answer for a_1,y yields:

\begin{large}\[ a_{cm,y} = \frac{m_{1}a_{1,y}+m_{2}a_{2,y}}{m_{1}+m_{2}} = \frac{m_{1}a_{1,y}-m_{2}a_{1,y}}{m_{1}+m_{2}} = \frac{(m_{1}-m_{2})(m_{2}-m_{1})g}{(m_{1}+m_{2})^{2}} \]\end{large}

The two objects, the pulley and the rope.

The system is subject to external forces from gravity and from the axle of the pulley. We will also have to consider the pulley as an independent system acted upon by the rope (tension) and by the axle.

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It is key to realize that although the pulley is massless and frictionless, it is acting to support the system through a force exerted on it by the axle. A simple way to see this is to consider what would happen to the system if the axle was removed.

\begin{large}\[ F_{A} - m_{1}g-m_{2}g = (m_{1}+m_{2})a_{cm,y}\]\end{large}

\begin{large}\[ F_{A} - 2T = 0 \]\end{large}

\begin{large}\[ F_{A} = \frac{4m_{1}m_{2}g}{m_{1}+m_{2}}\]\end{large}

\begin{large}\[ \frac{4m_{2}m_{1}g}{m_{1}+m_{2}} - m_{1}g - m_{2}g = (m_{1}+m_{2})a_{cm,y} \]\end{large}

\begin{large}\[a_{cm}= \frac{(-m_{1}^{2} - m_{2}^{2} + 2m_{1}m_{2})g}{(m_{1}+m_{2})^{2}} = \frac{(m_{1}-m_{2})(m_{2}-m_{1})g}{(m_{1}+m_{2})^{2}} \]\end{large}