# Why can we use Newton's second law for a massless pulley in this problem?

We have the following configuration of pulleys:

The problem states that neither the rope, nor any of the pulleys have mass and they rotate without friction. The task is to find the acceleration of the weights.

In my textbook the following solution is suggested:

$$m_1a_1 = T - m_1g \\ m_2a_2 = 2T - m_2g \\ m_Aa_A=2T - T$$

And since $$m_A = 0$$

$$T = 0\Rightarrow a_1=a_2=g$$

My question is, why is it okay to apply Newton's second law to pulley A, if we assumed its mass is 0? Shouldn't that lead to a contradiction? If, for example, we apply the same logic to pulley C we get:

$$a_Cm_C = 2T - m_2g \\ 0 = 2T - m_2g \\ m_2 = 0$$

which doesn't make sense. Choosing pulley A and applying this equation to it seems completely arbitrary to me. What am I missing?

• $T=0 \Rightarrow a_1=a_2=-g$, not $+g$
– Rol
Nov 13, 2021 at 1:51

The system you have shown is known as a fool's tackle (see: https://www.lockhaven.edu/~dsimanek/TTT-fool/fool.htm)

The funny thing about it is that it is impossible for the pulley to remain in a stable, static configuration for any choice of $$m_1$$ and $$m_2$$. The reasoning is quite simple. The tension along the string, assuming it is ideal, is constant throughout. Hence, if we were to resolve the forces on pulley A, there will always be a net downward force on it, which lead to problems like infinite torque and acceleration.

Of course, we can imagine setting up such a system using real pulleys, strings and masses. Then, the moment we release the system, the whole thing falls apart in a rather complicated fashion. In fact, it may be a great hassle to set up in the first place.

• Interesting article! Thanks! Nov 17, 2021 at 16:13

Massless, frictionless pulleys are an idealization. You never encounter them in real life. But they do simplify this kind of physics problem. And that is a good thing. The problems can be hard enough without further complications.

In this case, the idealization isn't helping. You have $$3$$ moving masses: $$m_1$$, $$m_2+m_C$$, and $$m_A$$. Note that $$m_2+m_C$$ count as one mass because they move together at the same speed and acceleration. $$m_A$$ is separate because it moves at its own speed and acceleration.

So you have to solve $$3$$ equations with $$4$$ unknowns.

$$m_1a_1 = T - m_1g \\ (m_2+m_C)a_2 = 2T - (m_2+m_C)g \\ m_Aa_A=T - 2T - m_Ag$$

The $$2$$nd equation is easy to simplify, because we know $$m_C = 0$$.

The same simplification can be used on the $$3$$rd equation. But as you pointed out, it gets confusing to calculate the force on a mass of $$0$$ g. It leads problems like $$a = F/m =0/0$$. Or if it turns out $$F \neq 0$$, then $$a = \infty$$.

You need to treat the problem as if the mass was not truly $$0$$, but it is small enough that the problem isn't much changed by it. If you felt like being mathematically rigorous about it, you could take the limit as $$m_A \rightarrow 0$$.

The reason for setting the force on A to $$0$$ is a cheat. If it wasn't, then $$a_A = \infty$$. And that isn't right. The point of this kind of problem is to learn about forces and accelerations. It isn't to learn about what happens when you put a force on a $$0$$ mass object. So this problem isn't very helpful.

If you do set $$f_A = 0$$, you get the acceleration of $$m_1$$ and $$m_2$$.

But you can't figure out the acceleration of pulley A from the forces. You have to use another method. It hinges on the length of the string being constant.

You find that $$m_1$$ and $$m_2$$ accelerate downward at the same rate, $$g$$. Suppose that after a certain time, $$m_1$$ and $$m_2$$ have moved downward a distance $$\Delta x$$.

Because pulley C has moved downward, both $$d_{BC}$$ and $$d_{BA} + d_{AC}$$ have grown by $$\Delta x$$.

To keep the string length constant, $$d_{A1}$$ must have shrunk by $$2\Delta x$$.

If $$m_1$$ has moved downward by $$\Delta x$$ and $$d_{A1}$$ has shrunk by $$2\Delta x$$, then pulley A must have moved downward by $$3 \Delta x$$.

With a little thought, you can conclude $$v_A = 3 v_1$$, and therefore $$a_A = 3 a_1 = 3 g$$.

There is nothing mischievous about massless pulleys in this exercise. Here's why:

Let me rewrite the system of equations but this time assuming massive pulleys

$$\begin{cases} m_1 a_1 &= T_1 - m_1 g \\ m_2 a_2 &= T_2 - m_2 g \\ m_A a_A &= -T_1 - m_A g \\ m_C a_C &= 2 T_1 - T_2 - m_C g \\ 0 &= (a_1 - a_A) + (a_C - a_A) + (a_C - a_B) + (a_A - a_B) \\ 0 &= a_2 - a_C \end{cases}$$

The first four equations are from Newton's law, the last two are by constraining the length of the ropes to be constants (long rope tension is $$T_1$$, and the short rope between $$m_C$$ and $$m_2$$ feels $$T_2$$) respectively. Setting $$a_B = 0$$ and solving I get

$$\begin{array} \, a_1&\to \frac{g \left(m_A \left(m_C+m_2\right)-m_1 \left(4 m_A+m_C+m_2\right)\right)}{m_A \left(m_C+m_2\right)+m_1 \left(4 m_A+m_C+m_2\right)}\\ a_2&\to -\frac{g \left(m_A+m_1\right) \left(m_C+m_2\right)}{m_A \left(m_C+m_2\right)+m_1 \left(4 m_A+m_C+m_2\right)}\\ a_A&\to -\frac{g \left(m_A \left(m_C+m_2\right)+m_1 \left(4 m_A+3 m_C+3 m_2\right)\right)}{m_A \left(m_C+m_2\right)+m_1 \left(4 m_A+m_C+m_2\right)}\\ a_C&\to -\frac{g \left(m_A+m_1\right) \left(m_C+m_2\right)}{m_A \left(m_C+m_2\right)+m_1 \left(4 m_A+m_C+m_2\right)}\\ T_1&\to \frac{2 g m_1 m_A \left(m_C+m_2\right)}{m_A \left(m_C+m_2\right)+m_1 \left(4 m_A+m_C+m_2\right)}\\ T_2&\to \frac{4 g m_1 m_2 m_A}{m_A \left(m_C+m_2\right)+m_1 \left(4 m_A+m_C+m_2\right)} \end{array}$$

by setting $$m_A \to 0, m_C \to 0$$ you will get

$$a_1\to -g,a_2\to -g,a_A\to -3 g,a_C\to -g,T_1\to 0,T_2\to 0$$

why is it okay to apply Newton's second law to pulley A, if we assumed its mass is 0?

Because you obtain the same result in both cases: by setting $$m_A \to 0, m_C \to 0$$ in the equations prior to solving them (as you did), or by solving the equations with massive pulleys and then setting $$m_A \to 0, m_C \to 0$$ in the answer at the end (as I did)

If, for example, we apply the same logic to pulley C we get: ... $$m_C = 0$$ which doesn't make sense

For pulley $$C$$, Newton's law becomes $$m_C a_C = 2 T_1 - T_2 - m_C g = 2 T_1 - (m_2 a_2 + m_2 g) - m_C g$$. When $$m_C = 0, T_1 = 0$$ then $$a_2 = -g$$ as expected, no contradiction. Note $$T_2 \neq m_2 g$$ in general

My question is, why is it okay to apply Newton's second law to pulley A, if we assumed its mass is 0? Shouldn't that lead to a contradiction?

Yes it is OK to apply Newton's second law to pulley A. It is OK to apply the law to any object, even if its mass is zero.

If the lengths of the strings are constant (i.e., the strings are inextensible) then it is a constraint on the system that the positions of pulleys A and C must move (accelerate) if the masses 1 and 2 move (accelerate).

Considering pulley A, the string that connects $$m_1$$ to the pulley is the same string that connects pulley A to both pulleys B and C so that the tension throughout the string is the same (assuming the string is both massless and inextensible). Then, if the mass of pulley A is zero, applying Newton's second law means the net force applied, $$2T-T$$ must be zero, ergo $$T$$ must equal zero.

Hope this helps.

This is the free body diagram

for each mass you can write the Newton equation $$~m_i\,a_i=\sum F_k$$

$$m_2\,a_2=-m_2\,g+T_4\\ m_c\,a_c=-T_4+T_1+T_2\\ m_a\,a_a=T_3-T_2+T_5\\ m_1\,a_1=-m_1\,g-T_5$$

$$T_2=T_1\\ T_2=T_5\\ T_1=T_3$$
$$a_2~,a_c~,a_a~,a_1~,T_i\quad i=1..5$$
to solve the problem you need two assumptions which are $$m_c=0\\ m_a=0$$
results $$~T_i=0~$$ and $$~a_1=-g~,a_2=-g$$