Suppose I have an Atwood machine, that is, two different masses connected with an inextensible, massless rope over a pulley. Assuming no friction between the rope and the pulley, the heavier mass will accelerate towards the ground, the lighter mass will accelerate towards the pulley, and the rope will accelerate towards the heavier mass. These three accelerations will be equal in magnitude. But this makes no sense to me.
Force causes acceleration. But there is no force acting on the rope. And even if there was, the acceleration of the rope would be infinite because its mass is 0. So why does the rope accelerate? And how can the magnitude of this acceleration be finite?
[Physics] Can a massless rope accelerate
accelerationforcesfree-body-diagramnewtonian-mechanicsstring
Related Solutions
On the pulley on the left, there are 4 forces applied, $T_1'$, $T_2'$, the gravitational acceleration on the pulley (its weight) $m' g$ (directed downwards), and the tension of the rope at the center of the pulley $T$, which is the one that you draw, but directed upwards. Now, the tension $T$ balances the weight $mg$ and the other two tensions $T_1'$ and $T_2'$, and the pulley don't move. However, the toques of the tensions $T_1'$ and $T_2'$ may not balance, and may result in a rotation of the pulley. In fact, if $L=I\omega$ is the angular momentum of the pulley, $I$ the momentum of inertia, and $\omega$ the angular velocity, one has $$ \frac{d L}{dt} =I\frac{d \omega}{dt}=r T_1'-r T_2' $$ where $r$ is the radius of the pulley and the terms at the right side of the equations are the torques of the tension forces applied to the pulley.
If your problem is just to determine the static equilibrium of the system, and not its dynamics, you may want to assume $\frac{d L}{dt}=0$ and therefore balance the two torques $r T_1'=r T'_2$, that is, $T_1'=T_2'$.
It is best to draw free body diagrams for the two masses.
$F$ is the applied force and $T$ the tension in the massless and inextensible rope joining the two masses.
There is no friction and both masses have the same acceleration $a$.
Applying Newton's second law for each of the masses:
$T = m_1\;a$ and $F-T= m_2\; a \Rightarrow F = (m_1+m_2)\;a$ so $F>T$
You can think of it as the force $F$ is accelerating both masses whereas the force $T$ only has to accelerate mass $m_2$.
Best Answer
When the (inertial) mass is zero, then the acceleration can be non-zero for zero force.
This is similar, conceptually, to what has been discussed recently regarding an ideal conductor.
Consider Ohm's Law:
$$V = IR$$
Now, what if $R = 0$ as is the case with an ideal conductor?
Clearly, the voltage must be zero for any current. The current through the conductor, then, is determined by constraints external to the ideal wire, i.e., by whatever the ideal wire is connected to.
Consider Newton's 2nd Law:
$$F = ma$$
Now, what if $m= 0$ as is the case with the massless rope?
Clearly, the force must be zero for any acceleration. The acceleration, then, is determined by constraints external to the massless rope, e.g., the attached masses.
Yes, the massless rope is ideal and, thus, not physical but, there can be effectively massless ropes just as there can be effectively ideal conductors. Which is to say that, to the precision one is working to, the rope has zero mass and zero force acting on it but non-zero acceleration.