BELIEVE   ME   NOT!    - -     A   SKEPTICs   GUIDE  

Atwood's Machine:

To illustrate the use of the FBD in nontrivial mechanics problems we can imagine another series of measurements⁹ with a simple device known as Atwood's Machine. The apparatus is pictured in Fig. 2.

  

Figure: Atwood's Machine - one object labelled m₁ is glued to a massless cart with massless wheels that roll without friction on a perfectly horizontal surface. The cart is attached to a massless, unstretchable string that runs over yet another massless, frictionless pulley and is attached at the other end to a second object labelled m₂ that is pulled downward by the force of gravity. [You can see that a real experiment might involve a few corrections!] At the right are pictured the two separate FBD's for m₁ and m₂, showing all the external forces acting on each. Here W₁ is the weight of m₁ and N is the normal force exerted on m₁ by the horizontal surface (through the cart) to keep it from falling. Since it does not fall, N must exactly balance W₁. The only unbalanced force on m₁ is the tension T in the string, which accelerates it to the right. The tension in a string is the same everywhere, so the same T pulls up on m₂, partly counteracting its weight W₂.

It is easy to see that the two vertical forces (W₁ and N) acting on m₁ must cancel. The rest is less trivial. The weight of m₂ is given by $W_2 = m_2 \; g$; thus for m₁ and m₂, respectively, we have the ``equations of motion''

$$a_1 \; = \; {T \over m_1} \qquad \hbox{\rm (to the right)}$$

and

$$a_2 \; = \; {m_2 \; g - T \over m_2} \qquad \hbox{\rm (downward)}.$$

But we have here three unknowns (a₁, a₂ and T) and only two equations. The rules of linear algebra say that we need at least as many equations as unknowns to find a solution! Our salvation lies in recognition of the constraints of the system: Because the string does not stretch or go limp, both masses are constrained to move exactly the same distance (though in different directions) and therefore both experience the same magnitude of acceleration a. Thus our third equation is a₁ = a₂ = a and we can equate the right sides of the two previous equations to get

$${T \over m_1} = {m_2 \; g - T \over m_2}$$

which we multiply through by $m_1 \; m_2$ to get

$$m_2 \; T \; \; = m_1 \; m_2 \; g - m_1 \; T$$

$$\hbox{\rm or} \qquad T \; [m_1 + m_2] = m_1 \; m_2 \; g$$

Plugging this back into our first equation gives

$$a \; = \; g \; {m_1 \over m_1 + m_2}.$$

A quicker, simpler, more intuitive (and thus riskier) way of seeing this is to picture the pair of constrained masses as a unit. Let's use this approach to replace the distinction bewteen gravitational and inertial mass, just to see how it looks. The accelerating force is provided by the weight W₂ of m₂ which is given by $W_2 = g\, m_{2_G}$, where m_2G is the gravitational mass of m₂. However, this force must accelerate both objects at the same rate because the string constrains both to move together (though in different directions). Thus the net inertia to be overcome by W₂ is the sum of the inertial masses of m₁ and m₂, so the acceleration is given by

$$a = {W_2 \over m_{1_I} + m_{2_I} } \; = \; g \; {m_{2_G} \over m_{1_I} + m_{2_I} }$$

$$\hbox{\rm or} \qquad \qquad {a \over g} \; = \; {m_{2_G} \over m_{1_I} + m_{2_I} } . \qquad \qquad$$

The latter form expresses the acceleration explicity in units of g, the acceleration of gravity, which is often called ``one gee.''

Suppose we have three identical objects, each of which has the same inertial mass $m_{\scriptscriptstyle I}$ and the same gravitational mass $m_{\scriptscriptstyle G}$. [This can easily be checked using a balance and a standard force like a spring.] Then we use two of them for m₁ and m₂, set the apparatus in motion and measure the acceleration in ``gees.'' The result will be $a/g = m_{\scriptscriptstyle G}/2m_{\scriptscriptstyle I}$. Next we put two of the objects on the cart and leave the third hanging. This time we should get $a/g = m_{\scriptscriptstyle G}/3m_{\scriptscriptstyle I}$. Finally we hand two and leave one on the cart, for $a/g = 2m_{\scriptscriptstyle G}/3m_{\scriptscriptstyle I}$. If the measured accelerations are actually in the ratios of ${1\over2}:{1\over3}:{2\over3}$ then it must be true that $m_{\scriptscriptstyle G}/m_{\scriptscriptstyle I}$ is constant - i.e. that $m_{\scriptscriptstyle G}$ is proportional to $m_{\scriptscriptstyle I}$ or that in fact they are really basically the same thing (in this case)! Unfortunately we have only confirmed this for these three identical objects. In fact all we have really demonstrated is that our original postulates are not trivially wrong. To go further we need to repeat the Eötvös experiment.