I need some help to win a bet.
Here is the deal. Take a string with a breaking strength of 30 pounds, Drape it across a table so that each end hangs off a side. Now hang a weight of 20 pounds on each end. Assume no friction, does the string break?
I would say yes. Friction way affect it, but assuming no friction, you have 40 lbs of force pulling on a string that can only handle 30. I can’t see why it wouldn’t break.
No. The load on the string is 20 pounds.
Ahhh, lovely physics! More counter-intuitive examples to provoke and annoy.
Whack a nail into your frictionless table, and tie two strings to it, hanging the weights off each one. 20 pounds on each, right? So they don’t break, and there’s 20 pounds of tension in each string.
There is no real difference between this and the first situation. Tie those strings together instead of to a nail, they each have 20 pounds of tension so nothing moves. The tension can’t magically increase.
Another way to look at it is - you have 40 pounds weight pulling downwards, and TWO strings each pulling upwards with 20 pounds of tension. The fact that the strings join in the centre doesn’t give you 40 pounds of tension in the string.
Here we go with more WAG’s. Assume you tie two spring scales to the string, one on each end. You and a friend each pull horizontally with twenty lb. of force. Will the string break?
OTOH, your string is lying on a horizontal table with weights hanging over the end. Conveniently, there is no friction, so the force is exerted equally throughout the string (and the weights are exactly equal, so the whole thing doesn’t slide to the heavy side).
Here’s the kicker: All the force is exerted downward. The force is distributed evenly through the string, and the majority of it is absorbed by the table. No single point on the string sees forty lb, and it does not break.
You are correct. My bad. It’s early here, and I wasn’t thinking right. Oh well.
I went the same direction you did until I really thought about it. It is counter intuitive. Interesting question.
How do we get the string to stay on top of a frictionless table???
Another way sto think about it: the amount of fource at each point must balance out. So where the string is tied to the string, there’s 20 lbs. pulling down, so there must be twenty lbs. of tension pulling up. If there were 40 lbs., the weight would move up.
PS Which side did you take? Or are you asking our help to decide which side to take?
I took the correct side. I didn’t want to sway answers so I made the question neutral. We have been debating this now for 2 days and I still can’t get him to believe me. You have to have a 20 pound force in the opposite direction just to keep the weight off the ground. Total tension on the string…. 20 pounds!
Funny, I just attended a lecture on String Theory by the guy who wrote “the Elegant Universe” and he didn’t cover this… WHAT A WASTE OF AN HOUR!!
If the string had just the single weight on one end, and the other end was nailed into the table, obviously it would only have 20 lbs. of tension. Yet the nail is providing a 20 lb. force in the opposite direction of the weight. In the problem, you’ve just replaced the nail with an actual 20 lbs weight. It’s providing the exact same force that the nail was.
I know you understand this principle already. I’m just trying to give you some weapons to convince your coworkers. I once spent a whole afternoon trying to explain the Monty Hall problem to some of my coworkers and still failed. Oh, well. Just remember, we’re doing Cecil’s work.
Well being a sailor and working with pulleys (correctly called “blocks” on a boat) I can tell you the tension on the string is 20 Lbs. You pull one end with 20 lbs of force and who is pulling the other side is immaterial.
Examples (Assuming no friction):
A pulley hanging from the ceiling with a line through it. You pull down on one side to lift 20 lbs on the other. Tension on the line: 20lbs. Tension on the ceiling: 40 lbs
Another example:
A pulley on the ceiling with two sheaves and an object on the floor weighing 60 lbs with a pulley with one sheave. The line is made fast to the object, goes up, around one sheave, down, around the lower sheave, up, around the other sheave and down to you. Note there are 3 segments going to the weight and one to you. You pull with 20 lbs and that is the tension all along the line (ignoring friction). the 3 segments pull the weight up with a force of 60 lbs and you and the weight are pulling the ceiling down with 80 lbs.
That is the whole principle: the line with 20 lbs can lift several times that.
Check out http://www.howstuffworks.com/pulley.htm
Another question that baffled me for a bit in college:
- Take a length of very thin thread that can only take 10lbs of tension and tie it to a doorknob.
- Tie a 10lb dumbbell to the thread and let it hang from the doorknob.
- Take another length of that same thread and tie it to the dumbbell and let it hang down.
- Now here’s the fun part: What will happen if you slowly pull straight down on that bottom string? What will happen if you yank down fast and hard on the bottom string?
Answer:
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If you pull very slowly the top string will break and the 10lb weight will fall to the floor under the force of gravity (be careful it doesn’t hit your hand).
If you yank down fast and hard the bottom string will snap because if it didn’t then the 10lb dumbbell would be pulled along with it. The weight would have to be accelerated to a high rate of speed in a very short period of time requiring more than 10lbs of force. Thus, the bottom string snaps and the top string holds.
Hmm. Not so sure about that. The top string is already at its breaking tension, and however hard you jerk on that bottom string, the tension in it will rise in a finite time. So you could break BOTH strings. On the other hand, you could argue that the top string is “protected” from the sudden jerk in the bottom string by the inertia of the dumbbell. Works better if the dumbbell weighs 9.5 pounds.
I’m going to start a thread on physics riddles. Re-post this there if you like!
The string might break at one of the points where it makes a 90 degree bend over the edge of the table.
Strength as a material property is given in terms of stress (force per unit area) not force. The breaking load of 20 pounds given for the string is for the simple case of a uniform distribution of stress over the string’s cross-section. Bending over the edge of the table affects the uniformity of the stress within the string, and at the bend there will be regions with higher stress than in a uniformly loaded string-- even though the total load carried by the string is still 20 pounds. These locations of high stress are where the break starts.
Whether the string breaks will depend on a number of factors, such as the radius of the bend at the table edge, the strain to failure of the material used to make the string, and the diameter of the string. But I would guess that if you tried this experiment with a string loaded to two-thirds of its breaking load, there is a good chance that it would break near the corner of a table unless the table edge is rounded.
There are several reasons that stress is non-uniform at the bend. The corner of the table applies a lateral load to the string which causes a multi-axial state of stress where the string contacts the table. Even though the table is frictionless, it still provides a normal force which acts on the side of the string where the string bends over the edges. The magnitude of this load is 20*sqrt(2) or about 28 lbs, and it is concentrated over a very small area. There are also bending stresses in the string which are related to radius of the bend. Another source of stress non-uniformity is the stress concentration factor as result of the change in geometry at the bend.
Consider this more extreme example: Take two 20 lb weights tied together with this string of breaking strength 30 lb. Then try lifting them with a frictionless razor blade by laying the string over the edge of the blade. There is a good chance the string will break where the razor touches it, but the chances of the string surviving go up as the razor is made more blunt or replaced by a rounded surface.