Confused about Tension

[dmission]

I'm a bit confused about how exactly tension works.

For example, there is one question in my book that says a man (50kg) is lifted by a helicopter upward at a rate of 5m/s (edit: accelerates upward at this rate). It then says that the total tension on the rope is 750N. I thought the tension on a rope was just equal to the force on one side, not both (is that only the case if the forces are equal or something?). Even if that's true, shouldn't it be 500N-250N, since he's being pulled downward with a force of 500, and upward with a force of 250? I'm wondering why they're added.

But, a second question in the same book seems to contradict that whole method. The next question says that when a 50kg mass is allowed to fall down, while connected to a pulley, it creates a 200N force upward. In then says that for that reason, the tension in the rope is 400. However, doesn't it also experience a force downward of 500N? So again shouldn't be either 200N + 500N or 500N-250N downward or something?

Would appreciate any clarification, as I've no idea what I'm doing anymore it seems Thanks!

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[MD Odyssey]

For example, there is one question in my book that says a man (50kg) is lifted by a helicopter upward at a rate of 5m/s. It then says that the total tension on the rope is 750N.

This is wrong, or at the least, incomplete. If the mass of the man is 50 kg, then there is a downward force due to gravity of 500 N. Since the man is moving upwards at a rate of 5 m/s, we can conclude that the net force on the object is 0, therefore the tension in the rope has to be 500 N.

I thought the tension on a rope was just equal to the force on one side, not both (is that only the case if the forces are equal or something?).

Tension depends upon whatever is connected to the rope. There isn't a general expression for tension, which is why we draw a free-body diagram for problems involving tension.

Even if that's true, shouldn't it be 500N-250N, since he's being pulled downward with a force of 500, and upward with a force of 250? I'm wondering why they're added.

I don't know either. You're problem is either ill-conceived or has a typo.

But, a second question in the same book seems to contradict that whole method. The next question says that when a 50kg mass is allowed to fall down, while connected to a pulley, it creates a 200N force upward. In then says that for that reason, the tension in the rope is 400. However, doesn't it also experience a force downward of 500N?

I'm confused by your problem. It sounds like you're missing part of the problem - a 50 kg mass falling has a force of 500 N downwards. Now, if you have a pulley or something above, then the rope has to be attached to something on the other end, but you're saying that there is a 400 N tension in the rope. If that's the case, then there is probably another mass of 10 kg on the other side, which would be what offsets the 500 N force, but this is a guess, since you're clearly omitting part of the problem.

I suggest that you draw a picture of the mass, label all the forces, and then sum them so that you can find what the next acceleration of the object is.

 

[Grom]

For example, there is one question in my book that says a man (50kg) is lifted by a helicopter upward at a rate of 5m/s. It then says that the total tension on the rope is 750N. I thought the tension on a rope was just equal to the force on one side, not both (is that only the case if the forces are equal or something?). Even if that's true, shouldn't it be 500N-250N, since he's being pulled downward with a force of 500, and upward with a force of 250? I'm wondering why they're added.

If he's travelling upward at a constant 5m/s, then the tension should be 500N as Odyssey said. If he's accelerating upwards at 5 m/s^2, then the tension should be 750N.

I didn't understand what the second question was.

 

[dmission]

This is wrong, or at the least, incomplete. If the mass of the man is 50 kg, then there is a downward force due to gravity of 500 N. Since the man is moving upwards at a rate of 5 m/s, we can conclude that the net force on the object is 0, therefore the tension in the rope has to be 500 N.

Sorry, I meant to write that he is accelerating upwards at a rate of 5m/s. Since he's 50kg, that's where the 250N is coming from. Since that force is going upward(?) though, shouldn't the total net force in the rope be 500N-250N=250N? (as opposed to adding them and getting 750N, which is the correct answers)

I'm confused by your problem. It sounds like you're missing part of the problem - a 50 kg mass falling has a force of 500 N downwards. Now, if you have a pulley or something above, then the rope has to be attached to something on the other end, but you're saying that there is a 400 N tension in the rope. If that's the case, then there is probably another mass of 10 kg on the other side, which would be what offsets the 500 N force, but this is a guess, since you're clearly omitting part of the problem.

I guess wasn't very clear again -- the answer (total tension in the rope) is 400N. All that is stated in the problem is that a 50kg mass falls (from a pulley) and creates a constant 200N force upward. It seems to be like, taking in the 200N into account, the net force on the rope should be 500N-200N in the downward direction, but the answer is 400N.

If he's travelling upward at a constant 5m/s, then the tension should be 500N as Odyssey said. If he's accelerating upwards at 5 m/s^2, then the tension should be 750N.

Sorry, that was what I meant to say. Do you think you could please clarify why a bit for me?


Thanks for the help.

 

[Calvinball]

Sorry, I meant to write that he is accelerating upwards at a rate of 5m/s. Since he's 50kg, that's where the 250N is coming from. Since that force is going upward(?) though, shouldn't the total net force in the rope be 500N-250N=250N? (as opposed to adding them and getting 750N, which is the correct answers)

Let's take a step back and imagine a much simpler scenario. You are holding a mass on a string, just holding it there. The tention on the string will be the mass of the object x acceleration of gravity, right? Now imagine that you suddenly yank the object upward, you would dramatically increase the tension on the string (or possibly break it!) by the amount of force necessary to accelerate the mass.

Another way of looking at this is that you have to add the amount of force necessary to hold the man and the force necessary to accelerate him (this is all upward force provided by the helicopter).

As to the question of why force is involved for acceleration but not constant velocity, refer to Newton's First Law of Motion.

 

[Integrated MCAT Course]

The confusion in the original question about the meaning of tension that should be addressed separately before talking about solving these kinds of accelerating frame problems or related static equilibrium problems.

Basically tension is a word which has a bit of a shifting definition in different referential frames in physics. In physics, semantic shifts are so rare, in other words, confusing words so few and far between that a person can't help but be self-doubting. You don't want to believe it. It's a bit like 'flow rate', for example, which might mean volume flux, flow speed, mass flow, or whatever, depending on the context. In simple dynamics problems, like this one, tension is used in an everyday context to mean 'the magnitude of the force the rope or cable supplies' or even 'the vector force on the rope or cable', sometimes, but that usage is verging in improper because when physicists talk to each other using the word tension, they are using it as a kind of quantity you aren't taught about in algebra based physics. Really proper usage of tension is as a tensor quantity. A tensor is an array or field of vectors. The tension is the tensor comprising the pair of oppositely oriented vectors at both ends of the cable, but the equal magnitude of either can also colloquially be called 'tension'. Mechanical deformation of solid bodies, or elasticity, is an MCAT topic where it helps to picture a tensor array of forces when you think of the array of forces that might produce a tensile stress, sheer stress, or volume stress.

Basically, in teaching algebra based physics, physicists allow themselves to use the word tension colloquially.

So how do you solve elevator problems, or similar accelerating frame or static equilibrium type problems? I think the key is to make a free body diagram, not of the tension on the rope or cable, but of the tension as one of the forces acting on the object.

Here's a question that may help demonstrate this type of problem solving. When you are stuck, make a free body diagram. It really does work. The solution will usually come clear in accelerating reference frame dynamics problems and static equilibrium problems if you take a few seconds to draw a free body diagram. A free body diagram shows all the forces acting on a body.