The situation sketched above is often used for problems involving tension.

Let's say we know the masses $m$ and want to determine the tension $T$ in the rope (no friction, massless rope).

Since this is a vector problem, I expect that there exist two distinct methods which lead to the solution:

  1. Use algebra (this yields $T = \frac{1}{2}mg$, see the older answer by Chester Miller for example).
  2. Draw force arrows and combine them to get the tension force arrow.

I don't get the second method to work.

So my question is: is there a way to determine the tension force arrow by drawing free body diagrams (i.e. without calculating anything)? If not, why not? I have the notion stuck in my head that vector problems should be solvable by drawing arrows.

I'm also interested in the situation where the plane is inclined. The algebraical solution is very similar (only the net force is smaller) but again, I don't see how to determine the tension from the free body diagrams alone.

  • $\begingroup$ 'Using force arrows' isn't not using a calculation, it is performing the calculation graphically (and generally to low precision, but that is an implementation detail). $\endgroup$ – dmckee Aug 23 '17 at 5:02

I think I finally figured out what you are asking. You would like to do an overall force balance on the system using vectors, and have it give you the same result as your scalar analysis. Your problem is that it seems there are only vertical forces on m2 and only horizontal forces on m1, so how can they all properly cancel.

For mass 1, I think we can agree that the normal force of the table top cancels with the weight of mass 1. So we will exclude this from our vector force balance. Our system will be the two masses and the rope. The key to doing this correctly is to recognize that the pulley exerts a net force on our system. From a force balance on the pulley, we find that the pulley axle exerts a force of $(T\mathbf{i_x}+T\mathbf{i_y})$ on the pulley, and this force is transmitted via the pulley to the rope. So the overall force balance on our system is: $$m_1a\mathbf{i_x}+m_2a(-\mathbf{i_y})=(T\mathbf{i_x}+T\mathbf{i_y})-m_2g\mathbf{i_y}$$ If we resolve this force balance into components, we obtain the two scalar force balances in our previous answers.

  • $\begingroup$ Nice! I think this is the answer. Thanks a lot for sticking around and taking the time to understand my concern even though our terminologies were different. I just drew the vector arrows which correspond to your equation and was able to determine $T$ and $a$ graphically. $\endgroup$ – Marc Aug 23 '17 at 10:48
  • $\begingroup$ I need to think a bit more about the force balance on the pulley, though. We have a horizontal tension force vector acting tangentially on the pulley at 12 o'clock and a vertical tension force vector acting tangentially on the pulley at 3 o'clock. I don't get why we can replace these two force vectors by their sum acting on the pulley at 1:30 o'clock. $\endgroup$ – Marc Aug 23 '17 at 10:49
  • $\begingroup$ +1 This answer is correct @Marc. What is being done here is that the mass-pulley-rope system is considered as a control volume on which external forces act. The inertial forces $-m\mathbf{a}$ are to be exerted on respective masses to maintain the entire system in equilibrium. $\endgroup$ – Deep Aug 23 '17 at 12:46
  • $\begingroup$ @Marc if you do a force balance on the section of massless rope in contact with the pulley, the force that the pulley exerts on this section of rope must add up to minus the two tangential forces on the two ends of the rope section. $\endgroup$ – Chet Miller Aug 23 '17 at 13:01


Lower mass: $ma=mg-T$

Upper mass: $ma=T$

These represent two linear algebraic equations in the two unknowns a and T. Adding the two equations together gives: $2ma=mg$.
So, $a=g/2$. Substituting a into the second equation gives: $T=mg/2$

  • $\begingroup$ Thanks for your answer, Chester Miller. Your solution is very similar to the algebraical one of the OP. My question is whether I can determine the magnitude of tension without calculating anything but by only drawing forces in the free body diagram. $\endgroup$ – Marc Aug 23 '17 at 1:38
  • $\begingroup$ As I said, you need separate free body diagrams for the two masses, not just one overall free body diagram. You can't do it with just the one diagram, since the tension T is internal to that. $\endgroup$ – Chet Miller Aug 23 '17 at 1:57
  • $\begingroup$ Yes, we need a separate diagram for each of the masses. But is there a way to get the length of the tension force arrow without calculating anything? Since this is a vector problem, we should have two distinct ways of solving it: by using algebra and by drawing arrows. I can't see how to make the second method work and this confuses me. $\endgroup$ – Marc Aug 23 '17 at 2:19
  • $\begingroup$ Isn't using lengths of force arrows the same thing as writing down the force balance equations? However, in a problem involving acceleration, you don't know how long to make its force arrow until you solve for the acceleration. If there were no acceleration, you could use force arrows. $\endgroup$ – Chet Miller Aug 23 '17 at 2:26
  • $\begingroup$ That doesn't seem to be true in general. The problem of a body on the inclined plane involves acceleration as well and there, the net force arrow -the downhill force- can be determined nicely by geometry. $\endgroup$ – Marc Aug 23 '17 at 2:36

It does not look like you can get to the answer simply by drawing vector diagrams. The reason is that each body has two unknown forces acting on it: tension $\mathbf{T}$ and the so-called inertial force $-m\mathbf{a}$. Vector diagram allows you to find only one unknown force given all other forces; for example, if acceleration were given then you may construct vector diagram for either body to find tension in the rope. If, as in this case, there is more than one unknown force vector, then drawing a vector diagram will only give you their resultant, which in your case is $\mathbf{F}\equiv\mathbf{T}+(-m\mathbf{a})$. But as you can see, knowledge of this resultant is useless; for the body being pulled horizontally it is a zero vector (because tension and inertial force are the only two forces acting on it), and for the body falling vertically the resultant equals $m\mathbf{g}$.

On second thought, your question is also interesting because it asks whether we can solve simultaneous vector equations (which, algebraically, is an easy thing to do) for more than one unknown vector by drawing multiple vector diagrams (two free body diagrams in your case) and combining them in some manner. I am not sure if there is any general method to do so. The reason is clear from the free body diagrams below. You cannot cancel, for example, tension force acting on one body with that acting on the other body because they are two different vectors (that's why I have used different symbols for them, for example $\mathbf{T}_1$ and $\mathbf{T}_2$, although their magnitude may be the same).

enter image description here

However, interestingly, you may rotate one of the free body diagrams by $90^\circ$, clockwise and counter-clockwise in turn, and then add force vectors acting on both bodies. This will give you each unknown force in turn, once you account for the fact that due to absence of friction $|\mathbf{T}_1|=|\mathbf{T}_2|$ and due to rope being inextensible $|\mathbf{a}_1|=|\mathbf{a}_2|$. The operation of $90^\circ$ rotation probably has no physical meaning and should be seen only as a mathematical trick; I am not even sure if it is general.

  • $\begingroup$ Thanks for engaging with my question, Deep! I also find this problem interesting because I think that there's something to be learned about vector equations in general. Regarding your answer: the last paragraph sounds like that you think a graphical solution is possible, if we rotate the free body diagrams (and maybe invoke Newton's third law?). However, I don't see yet how this leads to a $T$-arrow which is half as long as the $mg$-arrow. Can you elaborate? $\endgroup$ – Marc Aug 23 '17 at 6:07
  • 1
    $\begingroup$ @Marc If you rotate the upper free body diagram (i.e. for horizontally moving body) by $90^\circ$ counter-clockwise and add those rotated force vectors to those in the lower free body diagram, then since $|\mathbf{a}_1|=|\mathbf{a}_2|$, $-m\mathbf{a}_1$ will be equal and opposite of $-m\mathbf{a}_2$ and therefore the two will cancel. However tension force vectors $\mathbf{T}_1$ and $\mathbf{T}_2$ will be pointing in the same direction and add up. Since $|\mathbf{T}_1|=|\mathbf{T}_2|$ we may write $\mathbf{T}_1=\mathbf{T}_2=\mathbf{T}$ say, which then gives $2\mathbf{T}=m\mathbf{g}$. $\endgroup$ – Deep Aug 23 '17 at 12:33

I am here telling you how to find the direction of tension using free body diagrams. For calculating the magnitude, you have to perform some mathematical calculations.

So, to know where the tension is directed, the most important thing is to know what it is and why it is. The mass hanged pulls the string. This be easily validated that when a slack string, becomes taut when a mass is hanged. So no by Newton's third law, the string must also pull the mass and that is what tension is.

Now with this information, you can easily determine the direction of tension using free body diagrams.

As in your example, the mass on surface is pulling the string towards left. This can be checked by considering the fact that if there would have been no mass on surface, the string would easily pass by due to mass that is hanging. Therefore, as the mass is pulling string towards left, the string pull mass towards right as the direction indeed shows. You can apply the same reasoning for determining the direction of tension for hanged mass.

I hope I made myself clear. If you still face any problems, feel free to comment.


  • $\begingroup$ Thanks for your answer. So you are saying that I can't determine the magnitude of tension using only free body diagrams? Why is this? $\endgroup$ – Marc Aug 23 '17 at 1:17

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