question about forces

Because you're assuming the rope is weightless and there is no friction between the rope and the ring. Then there are no forces on the rope in the tangential direction.

Robert Israel snipped-for-privacy@math.ubc.ca Department of Mathematics

University of British Columbia Vancouver, BC, Canada

1) With no ring movement, the tension on the two ends is related to their angles from the horizontal. 2) With side force applied to a ring on a string, the force may be applied to accelerating the ring, or may transfer to the string by friction.

On this basis, I suppose that it is not necessary that the string tensions either side of the ring are equal.

Brian Whatcott Altus OK

Dear Chris Nellist:

If the ring is "smooth", how much force parallel to the ring surface can the rope exert on the rope and vice versa? I believe they added "smooth" that tidbit for that purpose.

Force from the ring can only act perpendicular to a line through the mounting points.

David A. Smith

"N:dlzc D:aol T:com \(dlzc\)" wrote in news:e_QHf.32588$jR.23577@fed1read01:

None.

I.e. its weight.

Sorry, but I still don't understand why the two tensions have to be equal in magnitude!

Let's say the lefthand part of the string is at 20 degrees to the vertical, and the righthand part is at 40 degrees to the vertical, and the horizontal force exerted on the ring is directed horizontally to the right.

I realise people are saying that when the horizontal force is exerted, the ring will move, without friction, until the tensions are equal, but I don't understand why.

Chris

"Chris Nellist" wrote in message news:Xns97694DBFCE8B6pnellist@195.92.193.157...

Your two mounting points are at the same height and the force on the ring is horizontal so you don't need to worry about the vertical plane at all. You can draw yourself a 2D diagram :

X X \ / \ / o | call the force you exert on the ring R1 - draw it's direction ; now say that the string can't exert a net force normal to R1 or the ring will just slip till it reaches a stable position where the two components normal to R1 cancel. Now calculate the angle between each part of the string and R1.

Best of luck - Mike

Dear Chris Nellist:

...

Well, plus your sideways force that you were worried about...

Consider an infinitessimal slice of rope just to the left of the ring, then consider another adjacent slice to the right.

This will look like poop, but here goes:

+--+ | | TT | | +--+

"T_left" = "T_right", or the rope-slice would accelerate, right? "T_right" equals "T_left" for the next slice, right? As you move to the right, you will always end up with the same value for T.

If you are always oriented along the axis of the rope, ring-surface-normal components (perpendicular to the slice faces) don't factor in, only the differential angle each infinitessimal slice is distorted into provides the counter to the ring-surface-normal forces. And that leaves a single value for T, that must do it all.

Tension must be the same on both side of the ring, unless there is friction at the ring surface. This is what is contrary to your common sense. It would be much easier for you to believe what I am saying if the ring had been a pulley. Your real world experience doesn't believe the "smooth ring", not so much the "tension doesn't change thoughout the rope", I think.

I hope that helps.

David A. Smith

All of the devices in such problems (forces, friction, etc.) are used mathematically to predict behavior. You will often unconciously switch between real world (push-pull, harder to push) and model world (forces and friction), and sometimes you will see what looks like a difference in the two environments and wonder why.

In the particular problem you presented-

The rope is one element. Period. Not two pieces, just one. So if you slowly pull on one end of a rope, that pull is felt equally on all points of the length of the rope

Now take that same rope and place a ring onto the rope - is there any change in the rope as one if a part of the rope is touched? No, none. The rope is still one element.

Put a horizontal force on the ring. Where exactly is that force reacted? By the rope. Can the ring move? Yes, it can slide anywhere along the rope until the forces balance due to geometry, not baloance due to the ring being stopped by a knot. Now, is the rope still one element or did someone tie a knot in it so when the force is put on the knot, it becomes two separate pieces of rope separated by a knot, the knot being a place where the ring is prevented from moving? No, the ring is free to move in the example, so the rope is still one element.

Can you have two different tensions in one element? You can't by definition. Can you have two different tensions in one rope? Only if you add a knot/loop to add to the tension on one side of the knot, thus making the rope two elements.

So since the rope in the example is one element, the tension anywhere along the rope is the same as anywhere else along the rope --

- to have otherwise, you must restrain the ring so it can put other than a tranverse load into the rope. E.g., by tying the ring in place. (it will add tension to only one side because the held ring's force is unbalanced relative to the balanced geometry of a rope with a free-sliding ring. With a held ring, there is a component along the rope -rather than all force transverse to the rope- that is unbalanced - and that must be satisfied by sum of forces = 0 along the rope. No axial component put into the rope, tension equal. Axial component pu tinto the rope, tension unequal. )

Sounds reasonable.....

>

With the ring in the middle everything is symmetrical about the plane of the ring.