# How does a string "distribute" tension?

This question deals with a massless, inextensible string. The problem arises with the question below. We know by Newton's third law that the force a body will apply on the string will be equal to the force the string will apply on the body. This force is called tension.

Now, can a body choose how much force it decides to apply to the string? In the leftmost figure, if we solve like how we're supposed to solve:

For block m:$$\quad T-mg=ma$$

For block 2m: $$\quad 2mg-T=2ma$$

Solving, we get $$T=\frac{4mg} 3$$. But $$\frac{4mg} 3$$ is greater than $$mg$$. So, the block m applied a force greater than it could on the string? How is that possible?

Now for the rightmost figure. A force $$F=2mg$$ acts on the right. So, will tension still act over there? There's no mass present, so what exactly applied that force on the string? And will the string apply a reaction force? If yes, on what?

"But 4𝑚𝑔/3 is greater than 𝑚𝑔. So, the block m applied a force greater than it could on the string?"

No, $$mg$$ is the pull of the Earth on the mass $$m$$; it's wrong to assume that it is equal to the force that the mass exerts on the string. Consider, for example, pulling a body of mass $$m$$ sharply upwards using a string. The tension force from the string will need to be greater than $$mg$$ in order to give the body an upward acceleration. The situation is very similar for mass $$m$$ in the left hand diagram.

"so what exactly applied that force on the string [in the right hand diagram]?" A hand, perhaps? We can only guess, but there must be something that is applying the force. And the string will apply an equal and opposite force to the hand (or whatever it is).

• But will the string then apply a tension at the point where the force $F=2mg$ is applied? My intuition says no. What would that Free body diagram look like? Jul 31 at 16:57
• I'm afraid that I don't understand these questions. (a) For one thing, a tension in a string isn't a force, but a combination of two equal and opposite pulling forces exerted on a segment of string by neighbouring segments. We can, though, talk about these forces as 'tension forces'. Sorry if this seems pedantic, but you are clearly aiming for a proper understanding. (b) the string applies an (upward) tension force $2mg$ to the hand and the hand applies a downward force $2mg$ to the rope. (c) For what body do you want the free body diagram? Jul 31 at 17:22
• (a) What exactly do you mean by "equal and opposite pulling forces"? I was taught that this was only for a string with mass, but in that case, the pulling forces wouldn't be equal. Or am I missing something? Also, if we assume that to be true, then drawing a free body diagram of a segment of a string means that there will be an upward force $T$ by the string above and a downward force $F$. Hence, the tension at every point in the string would be $F$, right? (c) I was asking it for the string Aug 1 at 10:44
• (a) What exactly do you mean by "equal and opposite pulling forces"? Suppose that we have 3 adjacent segments of string, A, B and C. If A exerts a force (a pull), $\mathbf F$ on B, then C exerts a force $–\mathbf F$ on B (if B's mass is negligible). (c) the free body diagram for B would simply show these two forces. Aug 1 at 18:29

This is only weird because you have slipped into the unphysical limit where the mass of the string is zero.

Let's abstract away the pulley, and imagine a horizontal system where there's a massive rod connecting the two masses:

,-,                     ,---,
| |---------------------|   |
-'                     ---'
m           µ            M


We can arrange for $$m$$ to be pulled to the left by a force $$mg$$ and for $$M=2m$$ to be pulled to the right by $$Mg$$. There is nothing magic about the weights of the objects; we can just pull on them that hard horizontally with our arms or our students or something.

Now our equations of motion are

\begin{align} -mg + T_\text{left} &= ma \tag1 \\ -T_\text{left} + T_\text{right} &= \mu a \tag2 \\ -T_\text{right} + Mg &= Ma \tag3 \end{align}

If we add these three equations, we see that the internal forces cancel and that the whole system is accelerating as a unit in response to the external forces:

\begin{align} -mg + (T_\text{left} - T_\text{left}) + (T_\text{right}-T_\text{right}) + Mg &= (m + \mu + M)a \\ -mg + Mg &= (m + \mu + M)a \tag4 \end{align}

In a typical lab experiment, we might have $$m\approx\rm1\,kg$$, $$M\approx\rm2\,kg$$, and $$\mu\approx\rm1\,g$$. The fourth equation tells us that, with these mass ratios, we'll get pretty much the correct acceleration for the combined system if we approximate $$\frac{\mu}{m+M} \approx 10^{-3}$$ as zero: in fact, neglecting $$\mu$$ introduces a smaller rounding error than disagreeing about whether to use $$g \approx \rm 9.80\,m/s^2$$ versus $$g \approx \rm 9.81\,m/s^2$$.

Likewise your result $$T_\text{left} = T_\text{right} = \frac43 mg$$ is still going to be good to three or four significant figures, even though we snuck in the low-mass string. In fact, our second equation gives the difference between the two tension forces as $$-T_\text{left} + T_\text{right} = \mu g \frac{M-m}{m + \mu + M} ≈ \frac 13 \mu g$$ so the exact result is going to be something like $$T_\text{right} = \frac43 mg + \text{(one-ish)}\times \mu g$$ where we could find the correction factor if we wanted to, but it's clearly still a small correction.

The string doesn't "transmit" the tension. The string is subject to forces on each end, and the nonzero difference between these forces is required to accelerate the string in tandem with the masses. But because the string's mass is so small, only a tiny fraction of the net external force is "expended" in accelerating the string, and pretending the string isn't there at all is a useful simplifying assumption.

So, the block $$m$$ applied a force greater than it could on the string? How is that possible?

Nope. The left block and string apply $$T_\text{left}$$ on each other, and the difference between $$mg$$ and $$T_\text{left}$$ is what makes that block accelerate.

• Because I am a glutton for punishment: \begin{align} T_\text{right} &= mg\cdot\frac43\cdot\frac{ 1 + \mu/2m }{1 + \mu/3m} \\ &\approx mg\cdot\frac43 \cdot\left( 1 + \frac16\cdot\frac\mu m \right) \\ &= mg\cdot\frac43 + \mu g\cdot\frac29 \end{align}
– rob
Jul 31 at 17:02

Imagine you're dangling on a spring balance. Earth is pulling you down with mg so you have to pull yourself up with mg and the spring balance will hence read your weight as mg. Now imagine you are pulling yourself upward. You will have to apply more force than mg on spring balance to have some positive acceleration so you will 'weigh' more. Similarly, the block in first scenario is apply more force than its weight because it have to for upward acceleration.

"so what exactly applied that force on the string." To whatever's pulling it down. If it's a hand, string will pull up on the hand. It there is positive charge on the string which is pulled down by some negative charge, string will pull up on the negative charge. If there is a solar sail attached to it which is pushed by reflecting light, it will apply force, in a way, on whatever emitted the light and on whatever the light will fall.