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Now to me it would seem logical that as a result of these TWO forces, the overall force experienced by the rope is 2xG. But is that actually correct? Thanks for your help:)

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- #1

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Now to me it would seem logical that as a result of these TWO forces, the overall force experienced by the rope is 2xG. But is that actually correct? Thanks for your help:)

- #2

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Actually, theNow to me it would seem logical that as a result of these TWO forces, the overall force experienced by the rope is 2xG.

Of course, what you are really asking is: Shouldn't the rope

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- #4

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I don't understand. What force is neglected in what calculation? Your FBD shows two forces: N (up) and G (down). They better add to zero.Still don't get it:( I drew a FBD of the rope. If it¨s correct, then howcome one force is neglected in the actual calculation?

If you want to consider rope tension, draw a FBD of the hanging mass. Two forces act on it: The rope tension T (up) and its weight G (down). The mass is in equilibrium, so the net force on it must be zero, thus the rope tension must equal the weight (not

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- #6

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The pull from the ceiling/ring contributes to the rope tension

If you cut the ring so that the ceiling no longer exerts a force on the top end of the rope, then there will be no tension in the rope. It just falls.

Here's a related example that often gives folks trouble. Compare these two scenarios:

(1) Attach a rope to a wall and pull the free end with force F. What's the tension in the rope?

(2) Two people hold opposite ends of a rope and both pull with force F. What's the tension in the rope?

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Did you draw yourself a FBD of the hanging mass, as I suggested in post #4?

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Did you draw yourself a FBD of the hanging mass, as I suggested in post #4?

Yep, and I understand the bottom part of the rope. The problem is the top...

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- #11

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Well, there's only one tension throughout the rope--the same everywhere.Yep, and I understand the bottom part of the rope. The problem is the top...

To analyze the forces at the top, consider the "rope + hanging mass" as a single system. What forces act on it?

By the way,

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But hey, maybe some rainy day I'll finally get it...

They've promissed rain for next week. i think I'll just wait for that;)

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Up to you, but the "analysis" needed to understand forces at the top of the rope should take you about 1 second of work.They've promissed rain for next week. i think I'll just wait for that;)

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Suppose you tie a goat to a stake in the ground and he pulls as hard as he can. The tension in that rope is one "goat" of force.

Now replace the stake in the ground with another goat identical to the first. We're inclined to think that the tension must be two "goats" of force. But that's not right, because the rope doesn't know what's the origin of the force holding it there. It's still just one goat's worth of tension. It could be the stake in the ground, or the second goat, or a big magnet. It doesn't matter to the rope.

Crowell has a nice discussion of this if you google for "light and matter."

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That example is equivalent to the one I tried to give in post #6. (But I'm happy to plug Crowell's books--I recommend them. )Here's an idea that comes from the Light And Matter text by Crowell...

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Your question pertains to the subject of "Statics"; the study of structures in static (non-moving) equilibrium. By the nature of your problem, no component of the assembly is moving. Good - so let's study it using the principles of statics.

One can take a slice (called a section) anywhere on the structure (say, the rope), and consider the forces on each side of the section. Because the structure is static, the forces will ALWAYS be equal and opposite. (If they weren't equal and opposite then there would be a net differential. Such a differential would propel that section into movement, as in F=ma.

What you had done in your opening statement was to sum the forces around both sides of the section; therein you got zero. Good, as it should be. When the sum of the forces at a section equal zero, then you have proven there is no net force remaining - thus the section is in static equilibrium. So, there must be no movement.

The above could be merely an exercise in accounting. If a lender borrows $100 from a bank. Then the lender has +$100 while the bank has -$100; together this financial system has $0 money overall. It would be wrong to take the absolute values of both moneys and say the system has $200 in it.

Note: if there is movement, then one has moved into the realm of "dynamics" - the study of structures in movement. In that case, there would be a net force once you have summed around a section.

One could ask what is the point of determining the forces at a section for statics, if you know that the sum of the forces equal zero. Well, what is of interest is the magnitude of the force on one side of the section. That level of force is often what is of interest, such as when designing the necessary strength of the rope in supporting the intended weight below.

One can take a slice (called a section) anywhere on the structure (say, the rope), and consider the forces on each side of the section. Because the structure is static, the forces will ALWAYS be equal and opposite. (If they weren't equal and opposite then there would be a net differential. Such a differential would propel that section into movement, as in F=ma.

What you had done in your opening statement was to sum the forces around both sides of the section; therein you got zero. Good, as it should be. When the sum of the forces at a section equal zero, then you have proven there is no net force remaining - thus the section is in static equilibrium. So, there must be no movement.

The above could be merely an exercise in accounting. If a lender borrows $100 from a bank. Then the lender has +$100 while the bank has -$100; together this financial system has $0 money overall. It would be wrong to take the absolute values of both moneys and say the system has $200 in it.

Note: if there is movement, then one has moved into the realm of "dynamics" - the study of structures in movement. In that case, there would be a net force once you have summed around a section.

One could ask what is the point of determining the forces at a section for statics, if you know that the sum of the forces equal zero. Well, what is of interest is the magnitude of the force on one side of the section. That level of force is often what is of interest, such as when designing the necessary strength of the rope in supporting the intended weight below.

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Don't know if this will help, but here goes. Have you ever been fishing & weighed a fish with one of those spring-scales? You hold the top ring of the scale, and hang the fish from the bottom ring. The spring stretches & a pointer tells you how much the fish weighs.

Now (for the ideal massless scale) what's the force on your hand pulling up? its the weight of the fish, right? Whats the force on the bottom ring of the scale? Its "one fish, pulling down" right? What your original question amounts to, is "why does the scale read '1 fish' instead of 2."

Now (for the ideal massless scale) what's the force on your hand pulling up? its the weight of the fish, right? Whats the force on the bottom ring of the scale? Its "one fish, pulling down" right? What your original question amounts to, is "why does the scale read '1 fish' instead of 2."

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