6
$\begingroup$

Consider the following simple Atwood machine with an ideal pulley and an ideal string

enter image description here

According to my textbook, the tension on the clamp that holds the machine to the wall equals $2T$. I don't understand why that is. The tension in $T$ in the string is equal in magnitude to $m_1g + m_1a = m_2g - m_2a$, assuming that $m_1$ is accelerating upwards.

Also, the acceleration of masses in an atwood machine is given by

$$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$$

Substituting this in, we get the tension equal to

$$T = m_1g + m_1\frac{(m_2 - m_1)g}{m_1 + m_2} = m_1g\left(1 + \frac{m_2 - m_1}{m_2 + m_1}\right) = \frac{2m_2m_1g}{m_1 + m_2}$$

So, according to my textbook, the tension on the pulley clamp should be:

$$2T = \frac{4m_1m_2g}{m_1 + m_2}$$

But, aren't all these forces internal forces? If we consider the whole atwood machine as the system (excluding the clamp), the only forces acting on it are the force of gravity, $(m_1 + m_2)g$ and the tension in the clamp, $T_c$. Since the system is at rest

$$T_c = (m_1 + m_2)g$$

Am I right, or is there a flaw in my argument?

$\endgroup$
  • 1
    $\begingroup$ You found $T$, and the text book has that same equation multiplied by a factor of 2. There is no problem here. $\endgroup$ – Ruben Feb 7 '14 at 13:41
  • 4
    $\begingroup$ Hint: The system is not at rest. $\endgroup$ – DR10 Feb 7 '14 at 13:51
  • 3
    $\begingroup$ Nick's answer is complete but I liked your question because it shows the effort to understand the PRINCIPLE under calculations. So it's important in my opinion to understand why the system isn't at rest. $\endgroup$ – DR10 Feb 7 '14 at 14:06
  • 3
    $\begingroup$ True, every calculation should not only mathematically check out, but the physical interpretation is also a very very very important part! So on the point of the question I'd say good job and keep up the good work! $\endgroup$ – Nick Feb 7 '14 at 14:10
  • 1
    $\begingroup$ If it helps, you can show that the center o fmass of the two masses $m_{1}$ and $m_{2}$ is accelerating downward, and though it looks like the support is holding the wheel steady, it is actually letting the wheel/mass system accelerate downward because of this. $\endgroup$ – Jerry Schirmer Feb 7 '14 at 15:01
3
$\begingroup$

Your result holds when the two masses are the same, in that case $a=0$ and you'd have that:

$T = m_1 g = m_2 g$.

Or:

$2T = 2m_1 g=2m_2g=(m_1+m_2)g$.

In the case that the masses are not the same, then both masses are accelerating, which in its turn apply yields a lower force on the pulley-system (and on the clamp).

This can be easilly checked with your formula of the tension!

$T = \frac{2m_1m_2g}{m_1+m_2},$

If I were to define the total mass as: $M=m_1+m_2$, then I could express $T$ as:

$T=\frac{2m_1(M-m_1)g}{M}=\frac{2g}{M}(m_1(M-m_1)).$

You can check if you'd plot $T$ as a function of $m_1$, that it reaches a maximum in $m_1=M/2$, which means that the tension becomes maximal if the two masses are equal, the tension then becomes:

$T=\frac{Mg}{2}=\frac{(m_1+m_2)g}{2}$,

or as you were thinking:

$2T=(m_1+m_2)g$

For completeness the plot of the tension in function of the mass $m_1$ in terms of dimensionless quantities.

enter image description here

On this plot you can easilly see that if $m_1=0 \Rightarrow m_2=M$ or $m_1=M \Rightarrow m_2=0$, that there'd be no tension since one of the two masses would be free falling. In the intermediate cases there would be tension since there is a ''pull'' on both sides of the string, the more the masses $m_1$ and $m_2$ equal eachother, the less movement there is and the more pull there is on the string.

$\endgroup$
  • $\begingroup$ So, if my argument was incorrect, it can only mean that the system is not at rest. But how can you say that the system is not at rest? $\endgroup$ – Gerard Feb 7 '14 at 14:36
  • $\begingroup$ In the above case we have a frictionless pulley, with a massless string. The only way the system can be at rest is when the two masses are equal (in your calculation that's the only case when the acceleration equals zero). When this is the case, both masses are pulling with an equal force at both ends of the string. Note that this doesn't necessarilly implies the system is at rest, it can also move with a constant velocity! $\endgroup$ – Nick Feb 7 '14 at 14:42
  • $\begingroup$ @Gerard If you were to add mass to the sting and/or friction to the pully, then there might be other situations in which the system is/becomes at rest. $\endgroup$ – Nick Feb 7 '14 at 14:43
  • 2
    $\begingroup$ It's not true that it remains in the same place. Its center of mass is accelerating because even if m_1 is going upward and m_2 downward the masses are different so they have a different "weight" in the global motion. So if m_2>m_1 and m_2 is accelerating downward, then the center of mass is going downward. $\endgroup$ – DR10 Feb 7 '14 at 15:40
  • 1
    $\begingroup$ @Gerard: Right, notice that for your purpose (i.e. total force acting on the system) rest or motion is not what you're really looking for. I've been sloppy simply telling you:"the system is not at rest". What's important is total acceleration and in this case it's different from 0. A motion with constant velocity needs no force acting on the system. I'll stop here because we're a little misusing the space for comments. $\endgroup$ – DR10 Feb 7 '14 at 16:34
4
$\begingroup$

The system is not at rest. If you consider the masses and the pulley to be one system, you can understand the behaviour of the system by the behaviour of its centre of mass. Unless the masses are equal, the centre of mass of the system is not at rest.

It might be useful to think of it in this way - Inside the system boundary mass $m_1$ moves down through a distance while mass $m_2$ moves up by the same distance. So, the centre of mass has moved down (or up depending whether $m_1 > m_2$).

So, the tension would be given by the equation:

$$(m_1+m_2)a_{cm} = (m_1+m_2)g - T_c $$

You can further work out that

$a_{cm} = a(m_2-m_1) /(m_1+m_2)$, where a is the value of the acceleration of mass $m_1$ that you have mentioned.

Plug it in the equation and you'll find that:

$T_c=\frac{4m_1m_2}{m_1+m_2}{g}$

$\endgroup$
  • $\begingroup$ This is how I will try to teach this problem. Thank you. $\endgroup$ – commonhare Mar 4 at 18:04
  • $\begingroup$ Any chance you or @Nick could comment on the solution taking the form 4g*mu? I know it may be beyond the scope of the problem, but when I see connections like that, I try to understand them. $\endgroup$ – commonhare Mar 4 at 18:13
0
$\begingroup$

There is indeed a flaw in your argument. In short, the tension on the pulley clasp is only required to cancel the total gravitational force on the system when everything is in equilibrium and there is no acceleration. However, if the masses are unbalanced then one of them will fall and the other will rise, and it is not clear that this will keep the total force at the same value as the balanced case.

In fact, you can check that when the two masses are equal then the answers coincide: the correct tension on the pulley clasp is $$ T_\text{clasp}=2T=\frac{4m^2}{m+m}g=2mg=(m+m)g. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.