# Tension in strings which accelerate 3 masses

This is a seemingly simple undergraduate problem that I cannot understand with my background (I have studied pure mathematics. now laugh freely).

Suppose three masses $$M_1>M_2>M_3$$ lie in rest on a frictionless plane. THe masses are tied with a single rope in a row, like $$M_3--M_2--M_1--$$. The rope is pulled at its free end by a force $$F$$. What is the acceleration of every mass? What are the tension of the pieces of rope between $$M_1$$ and $$M_2$$, and $$M_2$$ and $$M_3$$?

So I know that $$F=(M_1+M_2+M_3)a$$ where $$a=a_1+a_2+a_3$$ is the acceleration of the system. For the mass $$M_1$$, the forces are... $$F$$ and $$T1$$, right? So for $$M_1$$, the acceleration $$a_1$$ satisfies $$M_1 a_1=F-T_1$$.

Is this right?

How do I formulate the equation for $$M_2,M_3$$?

And what good is the equation $$a=a_1+a_2+a_3$$? Or perhaps I can use this as $$F/(\sum M_i)$$?

• RE: "What is the acceleration of every mass?" -- The problem is a bit ambiguous, but I assume that the solution is supposed to be for a time when all three masses have the same velocity at any given time, thus are all accelerating at the same rate. Also assume that the rope is massless and that gravity can be ignored. – MaxW Nov 1 '19 at 20:37
• Just to be absolutely specific. You say "The rope is pulled at its free end by a force F", sorry to be pedantic, but should we assume that the other end is also free, or tethered to the plane? My question is absurd I know, but a possible case. – A Rogue Ant. Nov 1 '19 at 20:44
• There is no need to have $a_1$, $a_2$, $a_3$. All masses are tied together with a tensioned string so they have the same acceleration $a$. – user1936752 Nov 1 '19 at 21:05
• The question is unclear. First you say there is a "SINGLE rope in a row". Then later you ask what are the tensions in each of the PIECES of rope. To me that means there are pieces of rope that interconnect the masses. Which mays more sense than saying there is a "SINGLE rope in a row", whatever that means. – Bob D Nov 1 '19 at 21:16
• @MelaniesWoes Careful. They will all have the same acceleration but if the masses are different each mass will have a different net force, and that will make the tensions in each piece of rope different. You need to proceed methodically. Do a free body diagram of each mass. – Bob D Nov 1 '19 at 21:45

First of all, in this kind of problems is essential to specify what our mechanical system is, so we can distinguish between external and internal forces. This way we can apply Newton's laws without confusion. Let me consider the three masses plus the rope as our system, so an external agent is applying a force $$F$$ upon it, producing two tensions (internal forces): $$T_{32}$$ between $$M_3$$ and $$M_2$$ and $$T_{21}$$ between $$M_2$$ and $$M_1$$, according to your drawing. The total mass of the system is indeed the sum of the three of them, assuming the rope is massless. Applying Newton's Second Law about some horizontal axis give us: $$$$F = (M_1+M_2+M_3)a_{s} ,$$$$ where $$a_{s}$$ is the system's acceleration. Here is something important to clarify about your reasoning: the system's acceleration is not the sum of the individual accelerations. In fact, after $$F$$ is transmitted through the rope, the objects acquire the same acceleration despite the different masses. An intuitive way to think about this is to suppose the accelerations are $$\textbf{not}$$ equal. In this scenario the rope would bend, so tension would not act anymore and the objects would crash, and this of course doesn't happen. Getting back to the problem, the accelerations are: $$$$a_s=a_1=a_2=a_3=\frac{F}{M_1+M_2+M_3}.$$$$ Now we can apply Newton's Second Law to every mass individually, but instead of doing Free Body diagrams and stuff, let's do it in an intuitive way; developing intuition in Physics is much more valuable. If we remove $$M_3$$ from the system that you drew, we note that the rope would have another free end, but this one would fall to the ground, and applying the force $$F$$ at the other end would not disturb this end at all in that sense, physically meaning that tension is zero in that chunk of rope. This way, $$T_{32}$$ is only dependent upon $$M_3$$, and we get the equation: \begin{align} T_{32}&=M_3 a_s \\ &=\frac{M_3}{M_1+M_2+M_3}F. \end{align} Applying the same reasoning for $$T_{21}$$, we note that removing $$M_3$$ and $$M_2$$ from the system will make this chunk of rope hit the ground, and $$F$$ won't tense it. This way, $$T_{21}$$ only depends upon the masses $$M_3$$ and $$M_2$$, so we obtain: \begin{align} T_{21}&=(M_3+M_2)a_s \\ &=\frac{M_3+M_2}{M_1+M_2+M_3}F. \end{align} Note: Of course you can also analyze the forces acting upon every individual mass and solve for the tensions. This way, the equation $$M_1 a_1 = F-T_1$$ that you point out is correct.