# Free body diagram of a rope

Consider that a rope having mass $2Kg$ is attached to two blocks as shown in the following diagram:

The whole system is accelerating upward.

When drawing the free body diagram of the rope, what are all the different forces that I must consider? Clearly, it experiences a downward pull of (3+2)g due to gravity. Since it's accelerating in the upward direction, there must be a force other than $Fo$ in the upward direction that is causing it to accelerate. What is this force? Additionally, if the rope was massless, would this situation be possible?

(This is not a homework question for I don't understand the basic idea behind drawing a free body diagram for a rope and require the answers to address this issue and not the problem in itself.)

• Are you sure $F_0$ is not the accelerating force? If it is not, than, in this problem, you would be allowed to add in another force to account for the acceleration. Aug 24, 2016 at 12:18
• No, no, I'm not sure. What about the $5Kg$ block? Doesn't it exert any force on the rope?
– user106570
Aug 24, 2016 at 12:20
• The downward force exerted by the 3 kg mass on the rope is > 3g because the 3kg mass is accelerating. The force is $3(g+a)$. Similarly, the upward force exerted by the 5 kg mass on the rope is $F_0-5(g+a)$. Aug 25, 2016 at 3:44

to understand the problem and the basis behind free body diagram check the following three FBDs of sytem, Mass A and B and finally the rope.

The net force on the system is Fnet = (F0-Mg) Accorging to Newtons Second law for a system of mass M; Fnet =Ma Thus acceleration of system will be a = (F-Mg)/M. A body will accelerate in a given direction if there is any net force in that direction.

Note here that the top of the rope is pulling the 5Kg mass downward with a force Ttop (i.e. tension at the top of the rope) At the same time the rope is pulling the 3Kg mass with a force Tbottom (i.e. tension at the bottom) Two points to note here: 1. In this problem , as the joint system of all three masses is acceleration with some acceleration ‘a’ each FBD should give a result that each mass is also moving with same acceleration This would be better explained by numerical values if force F0 was known 2. Also note the directions of tension forces by rope in each case as the rope can only pull and not push (unless it’s a rigid rod).

Newtons Third law: There exist equal and opposite reaction force for every action force. And the reaction force acts on the adjacent body in contact. It is can be easily seen that Ttop is greater than not only Tbottom but the sum of (Tbottom and 2g) Additionally the combination of all the FBDs of subsystems of a bigger system must result into FBD of the system itself. Check the same for yourself. Note that I have assumed that rope is perfectly rigid otherwise problem requires completely different treatment.

Now moving up if rope was massless Ttop = Tbottom. Tension throughout the rope would be same. That value then can be obtained by applying the Second Law on any of the two FBDs given above.

• Thanks! Additionally, if the rope was massless and Ttop=Tbottom, would the rope still be able to accelerate?
– user106570
Aug 26, 2016 at 0:04

If the rope is not being deformed (that is, if it's straight, and unstretched), then it operates the same as a box. Also, the mass of the top box contributes to the gravitational force, so your total downward force is $98 N$ (10 kilograms total multiplied by 9.8 $\frac{m}{s^2}$). Then there's just your %F_0$force pulling upward. Summing up all the forces (gravitational and applied) and dividing by the total mass gives your upward acceleration. • Yes, yes, doing all that gives me the magnitude of the force Fo, since acceleration is already given in the question but my doubt is not that. What are ALL the forces acting on the centre of this rope? – user106570 Aug 24, 2016 at 12:23 • Acting on the rope is$F_0$and the weight of the rope and the block underneath it. Aug 24, 2016 at 12:23 • What about the$5Kg$block? Doesn't it pull on the rope? – user106570 Aug 24, 2016 at 12:25 • Yeah, with the applied force,$F_0$. Aug 24, 2016 at 12:30 • Oh God, I am hugely confused about ropes and cables! I don't quite know which direction I must assign to the tension or even if there is tension in the first place! Can you please help? – user106570 Aug 24, 2016 at 12:32 The upward force exerted by the 5 kg block on the rope is$F_0-5(g+a)$where a is the acceleration of all three bodies. The downward force exerted by the 3 kg block on the rope is$3(g+a)$. So, the force balance on the 2 kg rope is:$$2a=F_0-5(g+a)-3(g+a)-2g$$Solving this equation for$F_0$gives:$$F_0=10(g+a)$$ • Won't the force exerted by the rope on the upper block be$Fo$+$m(a-g)$? – user106570 Aug 26, 2016 at 0:08 • No. The force balance on the upper block is$F_0-T-5g=5a$, where T is the tension in the rope at the top of the rope. Solving for T gives$T=F_0-5(g+a)\$. The force that the rope exerts on the upper block is -T (or T pointing downward). Aug 26, 2016 at 0:23