# Tension in a massless string being pulled at its ends with unequal forces

There is a question in my textbook. If a massless inextensible string is pulled on with a force of $$10 N$$, at both ends, what is the tension in the string?

It’s a very common question. The answer is $$10 N$$, cf. e.g. this & this Phys.SE posts. It can be proved using Newton’s $$2nd$$ and $$3rd$$ law. If we think of the string as a series of links in a chain, for example, or if we think about the adjacent molecules in the string, then we can prove using Newton’s $$2nd$$ and $$3rd$$ law that tension in the string is $$10N$$ at each point along its length.

But what if we pulled on the ends of the string with forces of unequal magnitudes? This question occurred to me and I kind of got confused. My intuition says that the string has a net force acting on it, and hence it would accelerate. But because the string is massless, Newton’s 2nd law did not help me understand this situation. My question is,

If we pull on the ends of a string that is massless and inextensible, with forces of $$60N$$ and $$70N$$ respectively, what would be the tension in the string?

Will it be $$60N$$? Will it be $$70N$$?

I gave it some thought, and I thought this situation is similar to an atwood machine, two masses $$6kg$$ and $$7kg$$ respectively, hanging from a pulley. The pulley is massless and frictionless. The string is massless and inextensible. Because of gravity, one end of the string is being pulled on with $$60N$$, and the other end is being pulled on with $$70N$$, isn’t this situation similar? If I work out the tension in the string using $$T$$ = $$\frac{2m_1m_2g}{m_1 + m_2}$$, it gives $$T$$ = $$64.6N$$.

So can I say that If we pull on the ends of a string that is massless and inextensible, with forces of $$60N$$ and $$70N$$ respectively, tension in the string would be neither $$60N$$, nor $$70N$$, but somewhere in between ($$64.6N$$)?

• Newton's second law tells us what happens when a mass has a net force acting on it – AakashM Dec 5 '19 at 9:32

The arrangement you describe is impossible. The tension of the string will be 70N. Whatever was trying to restrain the end of the string with a force of 60N will be subject to a force of 70N by the string. As a result it will accelerate subject to a net force of 10N. The reaction on the string will be 70N.

• Thanks. Now i get it that the arrangement (Atwood machine) that I have described in my question is similar but not exactly the same as pulling at the ends of the string with 60N and 70N, respectively. Because the point where the string is attached to the 7kg mass is not exactly pulling on the string with 70N. It is pulling on the string with 64.6N, and same goes for the other point where the string is attached to the 6kg mass. I understand it now, thanks for explaining it. – π times e Dec 4 '19 at 16:03

I think you are on the right track. Instead of looking at a net force on the string, which would not be possible as @Marco Ocram points out, consider that the net force is acting on a system consisting of the string and masses connected to each end. Then the system, including the string, can be considered accelerating with no net force on the string itself.

As an example, the diagram below is a system of two masses and a string, in this case without a pulley. The top diagram shows two external forces of 70 N and 60 N acting on the masses. The bottom diagram shows the equivalent system with a net force of 10 N acting on the system.

The acceleration of the system is given by

$$a=\frac {10}{M_{1}+M_{2}}$$

Since both masses have the same acceleration, the net force $$F_1$$ acting on $$M_1$$ is by Newton's second law

$$F_{1}=\frac {10M_{1}}{M_{1}+M_{2}}$$

Since the only net external force acting on $$M_1$$ is the tension in the string, this also equals the tension on the string, or

$$T=\frac {10M_{1}}{M_{1}+M_{2}}$$

The net force acting on $$M_2$$ is by Newton' second law

$$F_{2}=\frac {10M_{2}}{M_{1}+M_{2}}$$

Now we know that the string is experiencing the same acceleration $$a$$ at the two masses. Per Newton's second law on the string you have, where $$F_{net}$$ is the net force on the string,

$$F_{net}=M_{string}a$$.

If both $$F_{net}$$ and $$M_{string}$$ are zero, you can have any non-zero value of $$a$$ without violating Newton's second law.

Some observations:

If $$M_1$$ = 0, $$T$$ = 0, as expected.

If $$M_2$$ = 0, $$T$$ = 10N, as expected.

Hope this helps. • It was very helpful. Thanks. You said the two systems are equivalent, because they are accelerating at the same rate. But tension in the string in the two systems is not the same. You said they are equivalent because (I guess) you were only considering the external force(s)? Because tension is an internal force and gets cancelled by itself? – π times e Dec 4 '19 at 15:42
• @πtimese Yes, when you consider the string and the two masses as a system, the tension in the string is an internal force. But when you do a FBD on each mass individually, the tension is an external force on each. – Bob D Dec 4 '19 at 15:47
• You also explained that the acceleration of the string does not violate the Newton's 2nd law. I always wondered how Newton's 2nd law would explain the acceleration of the string, because its mass is zero and also because $F_{net}$ acting on it is zero, so it should not be accelerating. You explained it here, now I understand that it is not violating the 2nd law. You spared me the trouble of asking this question in an other post. Thanks a lot :) – π times e Dec 4 '19 at 15:48
• @πtimese Glad that helped. I also wondered about it myself awhile ago until I saw the explanation in another post. Great thing about this site is you keep expanding your knowledge by learning from others, then have the opportunity to "pass it forward". – Bob D Dec 4 '19 at 23:02
• Indeed. I am so glad I found this site, helps a lot – π times e Dec 5 '19 at 3:46

Consider a variation on the Atwood machine where instead of hanging straight down, the two masses $$m_1$$ and $$m_2$$ each sit on a separate frictionless plane at an angle of $$\theta_1$$ and $$\theta_2$$ to the vertical. The planes are joined along a ridge and the string connecting the two masses runs over the ridge, again without friction. The simple Atwood machine corresponds to the case $$\theta_1=\theta_2=0$$.

We set the values of $$m_1$$ and $$m_2$$ so that

$$m_1g \cos \theta_1 = 70 N\\m_2g \cos \theta_2 = 60 N$$

so there is a force of $$70$$ N on one end of the string and a force of $$60$$ N on the other end. The accelerations of $$m_1$$ and $$m_2$$ must be equal in magnitude and opposite in direction, so we have

$$\frac{70-T}{m_1}=\frac{T-60}{m_2}$$

where $$T$$ is the tension in the string. This gives us

$$T= \frac{60m_1+70m_2}{m_1+m_2}$$

If we let $$k=\frac{\cos \theta_2}{\cos \theta_1}$$ we can rewrite $$T$$ in terms of $$k$$:

$$T=\frac{4200(1+k)}{60+70k}$$

By varying $$\theta_1$$ and $$\theta_2$$ we can give $$k$$ any value we like between $$0$$ and $$\infty$$, and so $$T$$ can take any value between $$60$$ N and $$70$$ N.