Question Video: Understanding Tension Forces between Two Identical Objects on a Horizontal Surface Physics • 9th Grade

Two identical objects are connected to each other by a rope as in the diagram. A second rope is connected to one of the objects. The masses of the ropes are negligible. A short time after constant force ๐น is applied to the end of the second rope, both objects uniformly accelerate in the direction of ๐น across a smooth surface. Tension ๐‘‡โ‚ is produced in the rope that the force is applied to, and tension ๐‘‡โ‚‚ is produced in the rope that connects the objects. Which of the following statements correctly represents the relationship between ๐‘‡โ‚ and ๐‘‡โ‚‚? [A] ๐‘‡โ‚ = ๐‘‡โ‚‚ [B] ๐‘‡โ‚ = ๐‘‡โ‚‚/2 [C] ๐‘‡โ‚ + ๐‘‡โ‚‚ = 0 [D] ๐‘‡โ‚ = 2๐‘‡โ‚‚

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Video Transcript

Two identical objects are connected to each other by a rope as shown in the diagram. A second rope is connected to one of the objects. The masses of the ropes are negligible. A short time after a constant force ๐น is applied to the end of the second rope, both objects uniformly accelerate in the direction of ๐น across a smooth surface. Tension ๐‘‡ one is produced in the rope that the force is applied to, and tension ๐‘‡ two is produced in the rope that connects the objects. Which of the following statements correctly represents the relationship between ๐‘‡ one and ๐‘‡ two?

Before we consider these statements, letโ€™s take a look at our diagram. We see here two masses, which weโ€™re told are identical, connected by this rope. Then thereโ€™s another rope here, which has a constant force ๐น applied to the end of it, pointing to the right. Under this influence, weโ€™re told that both objects accelerate uniformly in that direction. And note that thereโ€™s no friction force opposing this acceleration because weโ€™re told that the movement is across a smooth surface. So we have two ropes, one under tension ๐‘‡ one and the other under tension ๐‘‡ two. And we want to pick out what relationship correctly describes them.

Here are our options. (A) ๐‘‡ one equals ๐‘‡ two, (B) ๐‘‡ one equals ๐‘‡ two divided by two, (C) ๐‘‡ one plus ๐‘‡ two equals zero, (D) ๐‘‡ one equals two times ๐‘‡ two.

Now, hereโ€™s one way we can think about this scenario. We essentially have a system where that system consists of these two identical masses in the two ropes. We have an external force ๐น being applied to the system and causing it to accelerate. This can remind us of Newtonโ€™s second law of motion, which tells us that the net force acting on an object of mass ๐‘š is equal to that mass multiplied by the objectโ€™s acceleration. Now, in our case, if we think only of forces acting in a horizontal direction, we can say that thereโ€™s one external force acting on our system. Thatโ€™s the force ๐น. That force is transmitted through the first rope and then pulls on the first mass, then transmitted through the second and pulls on the second mass.

Effectively then, this force is pulling our whole system, both masses and both ropes. Therefore, ๐น is equal to our systemโ€™s mass times its acceleration. Now, weโ€™re told that the two ropes in our scenario are massless, but weโ€™re not told the masses of these two objects. We do know, though, that theyโ€™re identical. So, just to give them a name, letโ€™s say that they each have a mass ๐‘š. This means that the total mass of our system is two times ๐‘š. Again, the mass of our ropes is considered zero. So, if we call the acceleration of our two objects ๐‘Ž, then we can say that ๐น is equal to two times ๐‘š times ๐‘Ž.

But then, looking at our diagram, we see that this force ๐น is being applied to the end of our first rope. And therefore, the tension in this rope is equal to that applied force. This means we can write that two times ๐‘š times ๐‘Ž is also equal to ๐‘‡ one. Because we donโ€™t know ๐‘š or ๐‘Ž, we canโ€™t go about calculating a numerical value for ๐‘‡ one. But all we want to do is compare it to the other tension force ๐‘‡ two to arrive at an expression for that variable. Instead of considering our two masses and the two ropes, letโ€™s just consider the second mass and the rope under tension ๐‘‡ two. Focusing in here, we can say that ๐‘‡ two is the only horizontal force acting on this second mass. And therefore, by Newtonโ€™s second law, itโ€™s equal to the mass of this object, which is ๐‘š, times its acceleration ๐‘Ž.

And note that this objectโ€™s acceleration is equal to the acceleration of the system overall. This is because both of our masses move together and accelerate equally. We see then that the tension force ๐‘‡ two is equal to this unknown quantity ๐‘š times ๐‘Ž and that the tension force ๐‘‡ one is equal to two times that same quantity. So, if we replace ๐‘š times ๐‘Ž here with ๐‘‡ two, which is equal to that product, then we find that two times ๐‘‡ two is equal to ๐‘‡ one. And we see that that corresponds to answer choice (D). The correct relationship between these two tension forces is that ๐‘‡ one is equal to two times ๐‘‡ two.

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