Thread: Tension basics

It seems I am really having a difficult time understanding tension.

I understand that if two people are pulling on opposite ends of a rope each exerting a force of 25 N, the tension of the rope is 25 N, but when the proportions are changed, I admit I start to get confused. What kind of tension would exist in a rope where one person is pulling in his direction with a force of 25 N and the other guy is pulling with a force of 20 N in the opposite direction?

Using my own common sense, I would think that the tension of the rope would be equal to the lesser of the two forces, but aparently that is not correct.

Assuming that the rope is uniform (i.e. every centimeter of it has the same mass), the tension would be a linear function of distance. Greatest at the point where the greater force is applied, gradually decreasing towards the opposite end.
And the rope would be accelerating at a rate equal to , where and are the forces and [/tex]m[LaTeX ERROR: Compile failed]

Originally Posted by Demen Tolden

What kind of tension would exist in a rope where one person is pulling in his direction with a force of 25 N and the other guy is pulling with a force of 20 N in the opposite direction?

It is an impossible situation. You cannot pull with a force of 25 N unless it is resisted by a force of 25 N in the opposite direction.

Just try to break a string that is not anchored at the other end, or does not at least have a weight that can be accelerated by the force you apply. It cannot be done.

Originally Posted by Harold14370

(...) does not at least have a weight that can be accelerated by the force you apply.

Any nonzero weight can, Harold.

With a small weight, the acceleration will be so great that in a split second from the start you will find it impossible to pull at it with your hand. But a rocket might do the trick. Or some other equally dynamic "tractor".

Originally Posted by Harold14370

Originally Posted by Demen Tolden

What kind of tension would exist in a rope where one person is pulling in his direction with a force of 25 N and the other guy is pulling with a force of 20 N in the opposite direction?

It is an impossible situation. You cannot pull with a force of 25 N unless it is resisted by a force of 25 N in the opposite direction.

Just try to break a string that is not anchored at the other end, or does not at least have a weight that can be accelerated by the force you apply. It cannot be done.

I think Demen is thinking of this situation. Imagine two people holding the rope at the ends, one person at each end. They are not pulling, just holding the rope. Now imagine one of them being accelerated to the left by a force of 25 N and the other being accelerated to the right by a force of 20 N. If they are still holding the rope, there will be a net leftward force on the system of 5 N. Now what will the tension at each end of the rope be?

Originally Posted by Demen Tolden

It seems I am really having a difficult time understanding tension.

I understand that if two people are pulling on opposite ends of a rope each exerting a force of 25 N, the tension of the rope is 25 N, but when the proportions are changed, I admit I start to get confused. What kind of tension would exist in a rope where one person is pulling in his direction with a force of 25 N and the other guy is pulling with a force of 20 N in the opposite direction?

Using my own common sense, I would think that the tension of the rope would be equal to the lesser of the two forces, but aparently that is not correct.

Think of a suspended pulley with a rope over it. You hang both your 25 pound weights on each end of the rope. The rope is under 50 pounds of tension.

So if you change one weight to 20 pounds while the other is still 25 pounds. The ratio is 20/25 of 50 pounds. Or 40 pounds. But when you first hang your weights before they move, you are correct there is a 45 pound tension. That is what I would calculate for.


Sincerely,


William McCormick

Originally Posted by William McCormick

Think of a suspended pulley with a rope over it. You hang both your 25 pound weights on each end of the rope. The rope is under 50 pounds of tension.

No, it isn't. The tension would be 25.

Originally Posted by William McCormick

Think of a suspended pulley with a rope over it. You hang both your 25 pound weights on each end of the rope. The rope is under 50 pounds of tension.

I have specifically been told this is not true. A rope with 25 N of force exerted at both ends has a tension of 25 N.

Originally Posted by Harrold

It is an impossible situation. You cannot pull with a force of 25 N unless it is resisted by a force of 25 N in the opposite direction.

I think maybe this kind of set up of a problem may make things more difficult to understand since force is not very clearly able to be interpreted as mass and acceleration. Maybe its easier to think of a pully system with a massless rope that doesn't stretch and a frictionless pulley of course. This way we have very specific masses and accelerations.



Lets say our blocks have the following masses and gravity is:




So we know that the force exerted from the first weight is:


and the force exerted from the second weight is:


Obviously, the first block is accelerating upward, and the second block is accelerating downward. How do we find the tension in the rope then?

This is what I am reasoning out:
First of all, if m2 is exerting a stronger force than the resisting force, m1, then if we subtract the force of m1 from m2, we should be able to find the force at which the system is exerting.... on itself? (If we think of the system as having a mass, which it does by having the two blocks moving at a uniform acceleration, and this acceleration, then it seems that there is a force exerted on the system as a whole.) Then if we divide this force by the collective masses of these objects, we find the acceleration of the system.




Hmm... what to do to find tension...

Could we then say that the total force of the first weight equals the m1g subtracted from the tension?:

Originally Posted by Demen Tolden

So we know that the force exerted from the first weight is:


and the force exerted from the second weight is:


Obviously, the first block is accelerating upward, and the second block is accelerating downward. How do we find the tension in the rope then?

Important note: the force exerted by a hanging body on the rope is not necessarily equal to the weight of the body.

The rope tension is the same for both blocks, and it is also the same as the force exerted by each of them on the rope. In spite of the fact that the blocks don't weigh the same.

The net force acting on each block is the difference between its weight (gravitational pull) and rope tension. This net force accelerates the block. The acceleration is downward for the heavier block because its weight is greater then the rope tension. The reverse is the case for the lighter block.

Can you do the maths now?

Ok, I'll try starting over by first redefining terms and then rewriting equations.



(in the j direction)
tension
sum of forces on block 1 (in j direction)
sum of forces on block 2 (in j direction)
sum of forces on the system
acceleration

sum of forces on block 1:


Newton's second law:


by substitution:



sum of forces on block 2:


Newton's second law:


by substitution:



by substitution:








I think I just proved that I am making a mistake.

Originally Posted by Demen Tolden

sum of forces on block 1:

It should be because the first block is accelerating in the direction of the tension exerted by the rope on the block.

Originally Posted by Demen Tolden

sum of forces on block 2:

And this should be . The tension and the weight of the block are pulling the block in opposite directions, with the block moving in the direction of its weight.

I changed g to be negative so it would better correspond to a coordinate system and also that a net force would be a sum of all forces rather than having to decipher what force should be positive and which one negative.

For example:




They come out to be the same in both cases, I just changed it because I thought it would be easier.

----------------------------------------------------------------
Actually, the way that the F2 equation is set up, its negativity indicates a downward direction while the positive value of the F1 equation indicates an upward direction.

Guys, before you post anymore. Just take a look at the pulley with a rope through it, a 25 pound weight at each end of the rope.

The pulley is going to be supporting 50 pounds. That means that the rope is supporting the 50 fifty pounds.

I do understand the other ways you can look at this.

Especially if you are a slave owner and very cheap, and want to save on rope. Ha-ha. I deal with them all the time. Ha-ha. They will swear to anything to save a $1.98

But the bottom line is that with two weights, 25 pounds each, the rope will have a fifty pound tension. And you can measure the weight at the pulley, it will be 50 pounds, plus the rope and pulley. So certainly since the rope is the only thing touching the weights. The rope is supporting the 50 pounds of weights.


Sincerely,


William McCormick

Think of it this way william. If you have a massless rope held firmly in your hand that is attached to nothing, and you pull on the rope, there will never be any tension in the rope since there is nothing pulling back against you.

Now if you tie the other end of the rope to something that will resist the force of your pull, say a car, then you can create tension in the rope. The tension is created by two forces. Those forces are the force of your pull, and the static friction on the car which is a force directed away from you and the rope. If the car is not moving in one direction or the other, it must have a balanced force applied to it, meaning the force of the pull is exactly equal to the force of the friction. So to create a somewhat simplistic diagram of this situation:

<---- 25 N Static Friction] Stationary Car [25 N Rope tension ------->

If the tension were to overcome the static friction, the object would accelerate in the direction of the rope. If we interpret this using the Atwood's system (the pulley system above), the rope would actually pull BOTH blocks strait upward, which would be an interesting thing to see, but not very plausable in the reality I live in.

---------------------------------------------------------------------------------
I also realized that there is a particular problem I was consistantly making with these tension problems that is kind of what Jane was stating, but I discovered that if you set up your axis to have your tensions the same, meaning you have the same positive-up, negative-down set of axis applied to both weights, your accelerations will often have opposite signs, and vice versa. With most of these problems, it is not possible to have both your tensions and accelerations be equal.

Originally Posted by Demen Tolden

Think of it this way william. If you have a massless rope held firmly in your hand that is attached to nothing, and you pull on the rope, there will never be any tension in the rope since there is nothing pulling back against you.

Now if you tie the other end of the rope to something that will resist the force of your pull, say a car, then you can create tension in the rope. The tension is created by two forces. Those forces are the force of your pull, and the static friction on the car which is a force directed away from you and the rope. If the car is not moving in one direction or the other, it must have a balanced force applied to it, meaning the force of the pull is exactly equal to the force of the friction. So to create a somewhat simplistic diagram of this situation:

<---- 25 N Static Friction] Stationary Car [25 N Rope tension ------->

If the tension were to overcome the static friction, the object would accelerate in the direction of the rope. If we interpret this using the Atwood's system (the pulley system above), the rope would actually pull BOTH blocks strait upward, which would be an interesting thing to see, but not very plausable in the reality I live in.

---------------------------------------------------------------------------------
I also realized that there is a particular problem I was consistantly making with these tension problems that is kind of what Jane was stating, but I discovered that if you set up your axis to have your tensions the same, meaning you have the same positive-up, negative-down set of axis applied to both weights, your accelerations will often have opposite signs, and vice versa. With most of these problems, it is not possible to have both your tensions and accelerations be equal.

A weight is sitting on the ground, I have a rope attached to it. I go to pick it up. I apply twenty five pounds of force to the rope. The weight applies 25 pounds to the rope in the opposite direction. There is 50 pounds of tension on that rope.

Years ago, this was talked about. But tests were always done on chain and most fabric rope, with load values. So this was calculated in. To a load rating. However that is not the tension that is on the rope or chain.

However if you are doing the calculating then you have to realize that there are two opposing forces at work, to hold a weight off the ground. There is the weight, and there is counter force to hold it there.

Sure if you measure the counter force it will be 25 pounds and if you measure the weight it will be 25 pounds. But in my book that is 50 pounds of tension on the rope.

As I mentioned earlier with the two 25 pound weights over a pulley. If you check the weight of the suspended pulley it will be 50 pounds. Meaning that in order to suspend 25 pounds you need 50 pounds of tension in your rope.

What people do not realize is that when you put something on a scale, you do not just put pressure on the scale. You also put 25 pounds of extra force on the floor under the scale.

So you create 25 pounds of measurable force on top of the scale, that moves the scale. And you create a 25 pound force on the floor. The floor is required to create the opposing force to make the scale move.

The pulley, rope and two 25 pound weights just make it more obvious. That you are dealing with 50 pounds of force. Two and Two is Four.

These guys in the movie, although I do not trust their cable. And I will not stand under the units while they are hoisted.

Do take the boom up very high so the unit just glides right into place with no effort. Very nice company to work with.

http://www.Rockwelder.com/WMV/Crane/Crane.html

Sincerely,


William McCormick

From Demen's:
(m1*g+T)/m1=(m2*g+T)/ m2


'g' was extracted incorrectly. The acceleration for the T is not 'g' but 'a' instead.

Continuing on:

m1*m2*g+T*m2=m1*m2*g+T*m1

2*m1*m2*g=T(m1-m2)

Thus,

T=2*m1*m2*g
(m1-m2)

Notice also that you'll end up with the correct units,i.e., Newtons.

http://www.Rockwelder.com/Flash/Mrbi...umpingiron.htm


Check that out. It pretty clearly shows what is happening when you lift 25 pounds. Your body or the chain, rope or cable has to take 50 pounds of tension or hold 50 pounds of pressure.

Or else Mr. Bill is a perpetual motion machine. And they are against the law now in America.


Sincerely,


William McCormick

Originally Posted by Demen Tolden

Originally Posted by William McCormick

Think of a suspended pulley with a rope over it. You hang both your 25 pound weights on each end of the rope. The rope is under 50 pounds of tension.

I have specifically been told this is not true. A rope with 25 N of force exerted at both ends has a tension of 25 N.

Find the fellow that told you that and straighten him out. Many people that go to college lose the real world. Perhaps forever. So if he gets angry or violent just walk away.


Sincerely,


William McCormick

My physics teacher actually told me that. The class in which he teaches is part of the Minnesota Transfer Curriculum, which is a previously agreed upon selected group of classes which is transferable to most Minnesota colleges including the University of Minnesota where I will finish up my chemical engineering degree. If you are right, and we are being taught nonsense, then we are in for a host of falling bridges and buildings and all kinds of apoctalyptical craziness once my class graduates with their respective degrees in three years time!

My physics book even says something similar! Here, read this: "When a cord (or a rope, cable, or other such object) is attached to a body and pulled taut, the cord pulls on the body with a force T directed away from the body and along the cord. The force is often called a tension force because the cord is said to be in a state of tension (or to be under tension), which means that it is being pulled taut. The tension in the cord is the magnitude T of the force on the body. For example, if the force on the body from the cord has magnitude T = 50 N, the tension in the cord is 50 N."

And with the publisher, Wiley, such a popular and well respected name (http://gov.wiley.com/WileyCDA/Section/id-306226.html) imagine all the people in the world that are reading this!

Watch out! Here comes the apocalypse!
(Please don't take offense, I just felt like trying to be funny. )

Originally Posted by Demen Tolden

My physics teacher actually told me that. The class in which he teaches is part of the Minnesota Transfer Curriculum, which is a previously agreed upon selected group of classes which is transferable to most Minnesota colleges including the University of Minnesota where I will finish up my chemical engineering degree. If you are right, and we are being taught nonsense, then we are in for a host of falling bridges and buildings and all kinds of apoctalyptical craziness once my class graduates with their respective degrees in three years time!

My physics book even says something similar! Here, read this: "When a cord (or a rope, cable, or other such object) is attached to a body and pulled taut, the cord pulls on the body with a force T directed away from the body and along the cord. The force is often called a tension force because the cord is said to be in a state of tension (or to be under tension), which means that it is being pulled taut. The tension in the cord is the magnitude T of the force on the body. For example, if the force on the body from the cord has magnitude T = 50 N, the tension in the cord is 50 N."

And with the publisher, Wiley, such a popular and well respected name (http://gov.wiley.com/WileyCDA/Section/id-306226.html) imagine all the people in the world that are reading this!

Watch out! Here comes the apocalypse!
(Please don't take offense, I just felt like trying to be funny. )

They will just change the definition of tension, or claim that was an engineering tension definition. Based on the load ratings from the past.
I would be the greatest counterintelligence agent the world ever knew. Ha-ha I have seen them all.

But my pop and I used to talk about this stuff when I was a rather small kid. While we would be eating supper. My pop told me they know, they don't care. And to be honest this was not the hottest subject we had upon us at the time. There were much more lethal issues.

Most of these types of misunderstandings happened when college engineers, started a project using lobbyists to get the grant or contracts. Often the Garage mechanic shops would offer for free, their findings after seeing the project or having the project sent to them by the government for inspection. Most of the time the good Garage mechanic shops were ignored. Even though they offered total proof, of a failed project.

Big money is what gets contracts from Senators.

It was already time for a total recall.

Grumman called the downing of the Concord before it was built. They said that debris would get sucked into the engines. And Grumman, for free, offered the GE electric engine design they had used already, to the builders of the Concord. Who were more afraid of losing funding then changing plans. And perhaps because of the outrageous fuel economy that it would have gotten.

Do not expect the media to come to your, or any Americans aid. They are afraid of total proof more then a possible conspiracy that cannot be proven. This tension thing is nothing, compared to things that have already occurred.

Rope and cable are not engineering materials. But today most do not even know that. I rode on a chain elevator and I felt very safe.




Sincerely,


William McCormick

duhhhh, I thought the math is enough.

Originally Posted by CoolEJ

duhhhh, I thought the math is enough.

If you were being helpful, anyone could come and make sense of that. I read that and I don't see any help.

You have to list out each variable and explain what it is doing. How it applies to some model or example. Short of that, and I am guessing at what you mean.

I work across twenty or thirty different fields. You think "m" means anything?

If people don't know,what we are trying to explain, and they certainly do not appear to, you need to create an understanding of what you are talking about.

None of us here are that smart, or we wouldn't be here. We would be on other planets chatting about the excitement, on the new planets.

I posted really simple math earlier, and not to many got that. I even included a written explanation.


Sincerely,


William McCormick

It is clearly written that that equation came from Demen's and I just continued and corrected it.

You do not need an IQ 999 for that. Common sense will do.
You just need to check out his previous post.

Originally Posted by CoolEJ

It is clearly written that that equation came from Demen's and I just continued and corrected it.

You do not need an IQ 999 for that. Common sense will do.
You just need to check out his previous post.

The original question was what was the tension on a rope with weights of 20 and 25 respectively. I am using pounds because I think in pounds.

The ratio is 20/25 of a possible, stall weight ratio of 25/25, and a dead weight of 50 pounds.

So if at stall you add the 25 pound weight or pull to the 20 pound weight, for a split second you will get 45 pounds of tension. As soon as the rope moves you will get 40 pounds of tension. For the rest of its fall.


Theoretically if you wanted to work out the ratio of some ideal condition or acceleration over some specific period.

You would take the difference of the equality 25/25 and 20/25. What you have left is 5/25 of potential velocity, or weight or pull differential.

If you have a maximum free fall speed of 32 feet per second you just divide 5 by 25 that equals 0.2 so you multiply 0.2 times 32 feet per second. And the velocity is 6.4 feet per second based on a free fall speed of 32 feet per second.

However in actual testing I have noted that a weight differential can apparently reach a near maximum free-fall velocity, no matter the differential. Often only friction and getting the large masses up to speed matter.

Because after all you are dropping 5 pounds in that scenario, the only thing that might cause it to take longer to get up to a maximum free-fall velocity is friction and the mass of the equalized weights, reaching that speed.

It is common sense.


Sincerely,


William McCormick

If you figure that same scenario out, a 20 and 25 pound weight over a pulley, using horse power, to lift the twenty pound weight, with 25 foot pounds a second.

It comes out that the heavier weight will move downwards 75 feet a minute. Somehow I believe it will go a bit faster.


Sincerely,


William McCormick

Originally Posted by CoolEJ

T=2*m1*m2*g
(m1-m2)

Well, this can't be right. If m1=m2 then this would give infinity for the tension, and we know it should be mg.

Here's how I would work this out. First find the acceleration of the system, and that will be the net force divided by the sum of the masses.

a=F/m = g(m2-m1)/(m1+m2)
For m1=1.5 kg and m2=4 kg, this would be

a=g(2.5/5.5)

Now consider only the 4 kilogram mass. It is accelerating downward at a rate less than the acceleration of gravity. The net downward force is its weight minus the tension in the rope.

F=m2g-T = m2a =m2 g(m2-m1)/(m1+m2)

Then T = m2g-m2g(m2-m1)/(m1+m2)
T= m2g [1-(m2-m1)/(m1+m2)]
T=m2g (1-2.5/5.5) = 2.1818g = 21.38 newtons.

William, shut up.

Originally Posted by Harold14370

Originally Posted by CoolEJ

T=2*m1*m2*g
(m1-m2)

Well, this can't be right. If m1=m2 then this would give infinity for the tension, and we know it should be mg.

Here's how I would work this out. First find the acceleration of the system, and that will be the net force divided by the sum of the masses.

a=F/m = g(m2-m1)/(m1+m2)
For m1=1.5 kg and m2=4 kg, this would be

a=g(2.5/5.5)

Now consider only the 4 kilogram mass. It is accelerating downward at a rate less than the acceleration of gravity. The net downward force is its weight minus the tension in the rope.

F=m2g-T = m2a =m2 g(m2-m1)/(m1+m2)

Then T = m2g-m2g(m2-m1)/(m1+m2)
T= m2g [1-(m2-m1)/(m1+m2)]
T=m2g (1-2.5/5.5) = 2.1818g = 21.38 newtons.

William, shut up.

Harold is this a valid definition for a Newton?






Sincerely,


William McCormick

How did you get this Harold?
a=F/m = g(m2-m1)/(m1+m2)

Is F the net force on the system?
What is m under F?

And how did this come up: g(m2-m1)/(m1+m2) ?

sorry, i may be missing something.

Originally Posted by CoolEJ

How did you get this Harold?
a=F/m = g(m2-m1)/(m1+m2)

Is F the net force on the system?
What is m under F?

And how did this come up: g(m2-m1)/(m1+m2) ?

sorry, i may be missing something.

I guess I skipped a step or two. I thought of the system as being equivalent to two weights connected by a rope in a straight line. There is a force pulling the system to the left and another pulling to the right. The force pulling left is the weight of the smaller object, and that is m1*g. The force pulling right is the weight of the larger object m2*g. The net force is the difference between these two forces, g(m2-m1). The total mass of the system is (m1+m2).

Therefore the acceleration, a, is the net force, g(m2-m1), divided by the total mass, (m1+m2).

Originally Posted by William McCormick

Harold is this a valid definition for a Newton?

William, yes that is a valid definition. If you are given the mass in kilograms and want to find the weight, i.e., the force it exerts because of gravity, you multiply it by the acceleration due to gravity, which is 9.8 meters per second per second. This gives you the force in kilogram-meters per second per second, which is newtons.

Going back a little bit, yes a rope with 25N on each side will exert a tensional force of 25N, unless you add a pulley. The tension in each parallel part of the rope will become 12.5N. If you now hang the whole thing off another rope, 50N of tensional force will be measured in it. This may be whats causing the confusion.

Originally Posted by GHOST

Going back a little bit, yes a rope with 25N on each side will exert a tensional force of 25N, unless you add a pulley. The tension in each parallel part of the rope will become 12.5N. If you now hang the whole thing off another rope, 50N of tensional force will be measured in it. This may be whats causing the confusion.

http://www.rockwelder.com/WMV/2025ff/2025ff.html

I did an actual test. I cut two pieces of 5/8" square bar Aluminum 60-61. One 6 inches long the other 4.8 inches long. They work out to a 20/25 ratio of weight differential.

I start with the weight just touching a large aluminum block that is mounted to the wood. I hold a wire carrying a 60 hertz signal. The signal is grounded by the block, while I hold it.

As soon as I let go, the frequency counter starts recording cycles from the power company. When the weight gets to the bottom. There is a plate laying over two wires that makes a connection. When the weight hits the plate, it disconnects the loop or circuit and shuts off the counter.

I only finished this up late last night. I just wanted to make a recording to show you guys how I was doing the experiment. In case you see anything I could improve. The pulley is machined from Delrin ™ it is a very hard plastic. It makes great bearings. Hobbyists use it all the time in their model cars and planes.

I had tried smaller pulleys already made. Even some small tape recording drive pulleys and they were not at all in the range of the Delrin pulley. They created a totally different effect. In fact the speeds I was getting were closer to Horse Power predictions.

That one test that I recorded was actually, a wobbly release. When you release it smoothly you get about 117 cycles. Or 3/60ths less then two seconds to fall 6 feet.

If you remove all the friction I do not know what would happen. I am going to look into an experiment, that I can remove all the friction.

But there is a strange acceleration ratio. It starts off very slowly and then really starts to move. I believe with longer drops you could get a better idea of what is going on.

I would not just trust any formula unless there is an accompanying experiment. Done on video. Because there are so many variables that can play into it.

The weights start off slow, but then start to really accelerate, once you get the weight moving.

Over a longer run, or if I time the second half of the fall, I could probably get a better idea of the average acceleration.

You would have to know how long the drop is to get an actual idea of how fast the weights will be moving. It is due to the initial startup of the weights. Once moving the ration of force lessons, and you get an actual increase in acceleration once you get the weights moving.


This is well known in crane work. Good crane men do not make jerky movements. Because you can easily increase the actual stress on a crane by four fold, even though the weight is well under the cranes maximum load.



Sincerely,


William McCormick

Originally Posted by GHOST

Going back a little bit, yes a rope with 25N on each side will exert a tensional force of 25N, unless you add a pulley. The tension in each parallel part of the rope will become 12.5N. If you now hang the whole thing off another rope, 50N of tensional force will be measured in it. This may be whats causing the confusion.

Check out this animation.

How much force is pulling at Mr. Bill?

http://www.Rockwelder.com/Flash/Mrbi...umpingiron.htm

Mr. Bill is the weak link in the rope supporting two 25 pounds weights. How much pull is Mr. Bill fighting?




There is 50 pounds of tension on Mr. Bill


Sincerely,


William McCormick

I for one would like to see this happen to Mr Bill:



What a good riddance that would be. If only.

Originally Posted by JaneBennet

I for one would like to see this happen to Mr Bill:



What a good riddance that would be. If only.

Jane you are despicable.

Actually he would fall over towards the weights pulling at him, in both directions.

Without the other respective 25 pounds to keep him up.

Very good scientific observation Jane.




Sincerely,


William McCormick

I just recently came across a hospital "A" frame setup. Complete with chain hoist.

The funny thing was that the same chain they now trust for 750 pounds on the hospital "A" frame. Used to be 500 pound load rated chain. So all this misunderstanding is hitting the real world quick.

I know I can supersede that chains rating even further. We do it all the time. But we know the danger. We don't go near it, and expect it to fall. We also do not jostle it.

But if someone really heavy, like 600 pounds heavy, coughs or convulses, he might break the chain. Because he could easily create 1000 pounds while moving around.

Sincerely,


William McCormick

Originally Posted by JaneBennet

I for one would like to see this happen to Mr Bill:



What a good riddance that would be. If only.

Or do you mean that you wish there were two of me? Ha-ha.


Sincerely,


William McCormick

http://www.rockwelder.com/WMV/2025ff/2025ff.html

After doing this test, it is the same as the earlier post. I did some more testing and came up with, 1.95 seconds total during a six foot drop. Based on smooth drops. Or 1 57/60ths

I came up with 1.316 seconds for the top part of the drop from six foot to three foot. Or 1 19/60ths.

I came up with 0.63 seconds for the bottom part, from three feet to zero feet. Or 38/60ths.

That means that the weight is covering the same distance in half the time, during the second part of the drop. By actual test.


That would mean that in six feet the velocity is 3.0769230769230769230769230769231 feet per second I Edited this.

That would mean that in the first three feet the velocity is 2.2796352583586626139817629179331 feet per second. I Edited this.

That would mean that in the last three feet the velocity is 4.7619047619047619047619047619048 feet per second. I Edited this.


These would be the average speeds for each section of the fall. Or lift however you look at it.

I had a lot of trouble with that cutting and pasting. Sorry. And then I double checked them and got into all kinds of trouble. I have them now though. They do not have to work out to six feet traveled.



Sincerely,


William McCormick

I definitely got some better understanding of velocity, from doing this experiment.

I also got how little velocity can mean to the actuality. If you do not get the whole picture. Or see it from many angles.

This was a great exercise for me. Did anyone note anything wrong or anything I overlooked?


Sincerely,


William McCormick

Originally Posted by William McCormick

I definitely got some better understanding of velocity, from doing this experiment.

I also got how little velocity can mean to the actuality. If you do not get the whole picture. Or see it from many angles.

This was a great exercise for me. Did anyone note anything wrong or anything I overlooked?


Sincerely,


William McCormick

With a weight of 20 on one side of the pulley and 25 on the other side, you are accelerating a mass of 45 (the sum of the weights) with a force of 5 (the difference). This means the acceleration should have been 5/45, or 0.111 times the acceleration due to gravity. The acceleration due to gravity is 32.2 feet per second per second.

Using the formula s=1/2at^2, or a=2s/t^2
we plug in your data points.
The data from your experiments were
s=3 ft, t=1.316 seconds
s=6 ft, t=1.95 seconds

The first test gives a=3.464 which is 3.464/32.2=0.107g
The second test gives a=3.156/32.2=0.098g

So that's pretty close to the 0.111g we expected. The difference is probably due to the mass of the pulley and friction.

Originally Posted by Harold14370

Originally Posted by William McCormick

I definitely got some better understanding of velocity, from doing this experiment.

I also got how little velocity can mean to the actuality. If you do not get the whole picture. Or see it from many angles.

This was a great exercise for me. Did anyone note anything wrong or anything I overlooked?


Sincerely,


William McCormick

With a weight of 20 on one side of the pulley and 25 on the other side, you are accelerating a mass of 45 (the sum of the weights) with a force of 5 (the difference). This means the acceleration should have been 5/45, or 0.111 times the acceleration due to gravity. The acceleration due to gravity is 32.2 feet per second per second.

Using the formula s=1/2at^2, or a=2s/t^2
we plug in your data points.
The data from your experiments were
s=3 ft, t=1.316 seconds
s=6 ft, t=1.95 seconds

The first test gives a=3.464 which is 3.464/32.2=0.107g
The second test gives a=3.156/32.2=0.098g

So that's pretty close to the 0.111g we expected. The difference is probably due to the mass of the pulley and friction.

What about from three feet to six feet.

Looking at the feet per second. It wobbled my mind there for bit, as I was double checking them. Then I realized that the velocity is only an average speed over any of those distances.


Sincerely,


William McCormick

Originally Posted by William McCormick

What about from three feet to six feet.

In order to calculate the acceleration from the time from 3 to 6 feet, you would need to know the velocity at the 3 foot mark, which you did not measure directly. We already knew the time for the bottom part from the first two tests. It's the difference between the time for a 0 to 3 foot drop and the time for a 0 to 6 foot drop, 1.95-1.316=0.634. So measuring the time to drop from 3 to 6 feet did not give us any new information.

a=(Vf-Vi)/t

that's one of the basic formula in Physics

Physics in everyday terms, is the science of matter[1] and its motion. It is the science that seeks to understand very basic concepts such as force, energy, mass, and charge. More completely, it is the general analysis of nature, conducted in order to understand how the world around us behaves.
______________________________
Carhartt Insulated Gloves investment real estate

Originally Posted by GHOST

Going back a little bit, yes a rope with 25N on each side will exert a tensional force of 25N, unless you add a pulley. The tension in each parallel part of the rope will become 12.5N. If you now hang the whole thing off another rope, 50N of tensional force will be measured in it. This may be whats causing the confusion.

Here is what you are not seeing. The tension on the rope holding the pulley holding/supporting, two twenty five pound weights, is under 100 pounds of tension. So indeed the rope holding both 25 pound weights is under half that, 50 pounds of tension.

The pulley is a load of 50 pounds. The two twenty five pound weights are a load of 50 pounds. However the tension is as per your formula. Except that you double the value of the load to get the tension for the rope supporting the load.

Sincerely,


William McCormick

William, think about the example where the scale measured 25 newtons of tension when tied in between the two 25 newton blocks hanging off the side of a table by a pulley.

Ok, now it makes sense, if you think about picking just one of them up with your arm. The 25 newton block exerts a force in the -j(downward) direction and your arm counteracts that with 25 newtons of force in the +j(upward) direction.

So think about the first system again with the two blocks. The tension is only 25 newtons because the second block is acting like your arm does in the second system.

This is why you can't relate tension directly to weight like your trying to do. It works in some cases but not in others. You have to look at the problem form the general description of tension.

Originally Posted by GenerationE

William, think about the example where the scale measured 25 newtons of tension when tied in between the two 25 newton blocks hanging off the side of a table by a pulley.

Ok, now it makes sense, if you think about picking just one of them up with your arm. The 25 newton block exerts a force in the -j(downward) direction and your arm counteracts that with 25 newtons of force in the +j(upward) direction.

So think about the first system again with the two blocks. The tension is only 25 newtons because the second block is acting like your arm does in the second system.

This is why you can't relate tension directly to weight like your trying to do. It works in some cases but not in others. You have to look at the problem form the general description of tension.

The tension on the device, (rope, chain, cable, rod) is always twice the load.

A hanging scale only measures half the tension. Because you do not want to know what the ceiling is lifting. Or what the tension is. You just want to know how much of a load the scale has in it.

If you had two hanging scales and you placed a 25 pound weight in each scale. You would say that there is 50 pounds combined weight hanging from the ceiling. A downward force of 50 pounds. So when I configure the weights and scales as they are below. Nothing should change. Now one scale measures one weight and the other scale measures the other weight.

I am sure you would agree that if we hung the both scales with a 25 pound weight in each from the ceiling, I would need a ceiling that could lift more then 50 pounds. I think we could also agree that the ceiling would have a load of fifty pounds upon it. All I have done is take the load from the ceiling and canceled it out by connecting the two scales. I am not cheating in anyway. Both scales record just as they did before.

Now the scales are doing what they are designed to do. Measure the weight below them. And they are.

The thing to remember, the ceiling with two scales with a 25 pound weight in each, is going to be fighting 50 pounds of downward pressure. From the two hooks of the scale.



Don't get me wrong I can see just about every way you can get this confused. However if someone does not get it right soon. The next generation of engineered stuff is going to be from out of this world.



Sincerely,


William McCormick

Here's a simple experiment, William. Put three scales on there. What would they read?

William could put scales in there until the cows come home, and he's never going to measure anything higher than 25 lb. This idea that the tension is twice the weight is strictly a figment of William's imagination.

as always..

Originally Posted by MagiMaster

Here's a simple experiment, William. Put three scales on there. What would they read?

I am totally aware of that. I mentioned it myself.

However we use two separate scales to measure two 25 pounds weights, accurately. We all agree totally that there is a fifty pound hanging weight, upon the ceiling the scales are suspended from. Twenty five pounds from each hook of the scale attached to the ceiling. No one is arguing that.

Now I suspend both scales from each other. And there is only a 25 pound force in total? Not a fifty pound force? Wow.

You guys would get your hands cut off in certain Arab villages, if you tried that there.

Heck if you engineered anything, thinking like that you might get your hands cut off. Ha-ha.






Sincerely,


William McCormick

I meant three in a row. Why are they now attached to the ceiling?

What would you say if I built a perfectly symmetrical spring scale? How can that be backwards?

Originally Posted by William McCormick

Guys, before you post anymore. Just take a look at the pulley with a rope through it, a 25 pound weight at each end of the rope.

The pulley is going to be supporting 50 pounds. That means that the rope is supporting the 50 fifty pounds.

I do understand the other ways you can look at this.

Especially if you are a slave owner and very cheap, and want to save on rope. Ha-ha. I deal with them all the time. Ha-ha. They will swear to anything to save a $1.98

But the bottom line is that with two weights, 25 pounds each, the rope will have a fifty pound tension. And you can measure the weight at the pulley, it will be 50 pounds, plus the rope and pulley. So certainly since the rope is the only thing touching the weights. The rope is supporting the 50 pounds of weights.



Sincerely,


William McCormick

Wrong, as usual.

There is 25 lb of tension in the rope resisting each of the 25 lb weights on the two ends of the rope. There is 50 lb being held up by the support of the pulley, supporting both of the 25 lb weights simultaeously. This neglects the weight of the rope itself.

Originally Posted by MagiMaster

I meant three in a row. Why are they now attached to the ceiling?

What would you say if I built a perfectly symmetrical spring scale? How can that be backwards?

http://www.Rockwelder.com/Flash/Pressure/Pressure.htm

A spring scale either floor or ceiling mounted, calculates and shows the weight of the object hanging from the scale, or laid upon the scale. It does not measure the counter force. Because you do not want to pay, for the tension. You just want to pay for the weight upon the scale.

We know there is twenty five pounds of force upon each ceiling mounted scales hook, when they have a 25 pound weight upon each of them. When we combine the two separate forces, of the ceiling scale, we know that together the two hooks create fifty pounds of weight, force.

In the case of the two weights on the pulleys. We also know that there is 25 pounds of force, created by each weight, as accurately measured by the ceiling scales.
When we hook the two hooks together, and they support themselves as did the ceiling, that had a 50 pound force upon the ceiling. I think it should be obvious that there is 50 pounds upon the hooks and in fact all the ropes have a 50 pound tension upon them. All the ropes supporting just one 25 pound weight.

But if you need more disasters, then so be it. I have warned many.

I am surprised no one knows this.

Sincerely,


William McCormick

What I am saying is that the ceiling is stopping two twenty five pound weights, and the scales the weights are suspended from, from hitting the floor. With a force of 50 pounds.

So when I put both hooks together, the same force is necessary to keep the two twenty five pound weights and the scales they are suspended from, from hitting the floor.

Can we agree upon that?


Sincerely,


William McCormick

Originally Posted by DrRocket

Originally Posted by William McCormick

Guys, before you post anymore. Just take a look at the pulley with a rope through it, a 25 pound weight at each end of the rope.

The pulley is going to be supporting 50 pounds. That means that the rope is supporting the 50 fifty pounds.

I do understand the other ways you can look at this.

Especially if you are a slave owner and very cheap, and want to save on rope. Ha-ha. I deal with them all the time. Ha-ha. They will swear to anything to save a $1.98

But the bottom line is that with two weights, 25 pounds each, the rope will have a fifty pound tension. And you can measure the weight at the pulley, it will be 50 pounds, plus the rope and pulley. So certainly since the rope is the only thing touching the weights. The rope is supporting the 50 pounds of weights.



Sincerely,


William McCormick

Wrong, as usual.

There is 25 lb of tension in the rope resisting each of the 25 lb weights on the two ends of the rope. There is 50 lb being held up by the support of the pulley, supporting both of the 25 lb weights simultaeously. This neglects the weight of the rope itself.

So then which scale do you feel is wrong?


Sincerely,


William McCormick

Try this William. Take a piece of fishing line, cord, string or whatever. Something of, say, 50 lbs. test.

First, tie 5 feet of it to the ceiling, or some other fixed point and hang 45 lbs. on it. It shouldn't break. Hang 90 lbs. on it and it should. Be careful not to drop the weight since the shock load will exceed the base load.

Now take another 5 feet of the same line and hang it from a pulley. Carefully put 45 lbs on each side. Will it break or not?

Originally Posted by MagiMaster

Try this William. Take a piece of fishing line, cord, string or whatever. Something of, say, 50 lbs. test.

First, tie 5 feet of it to the ceiling, or some other fixed point and hang 45 lbs. on it. It shouldn't break. Hang 90 lbs. on it and it should. Be careful not to drop the weight since the shock load will exceed the base load.

Now take another 5 feet of the same line and hang it from a pulley. Carefully put 45 lbs on each side. Will it break or not?

Weight, load rating, and test, are not tension, on the device carrying the load. You always double the tension for whatever weight you are supporting.

Sincerely,


William McCormick

Originally Posted by MagiMaster

Try this William. Take a piece of fishing line, cord, string or whatever. Something of, say, 50 lbs. test.

First, tie 5 feet of it to the ceiling, or some other fixed point and hang 45 lbs. on it. It shouldn't break. Hang 90 lbs. on it and it should. Be careful not to drop the weight since the shock load will exceed the base load.

Now take another 5 feet of the same line and hang it from a pulley. Carefully put 45 lbs on each side. Will it break or not?

You guys have to start to understand what this misunderstanding did to shock loads. Or hammer loads.

When you have something like chain that is load rated, it is designed to hold, or be used within a certain range of velocity, with its rated load. It can only hold that rated load if a small amount of slack is allowed to cause the load to fall, a small distance.

What takes place in engineering since the late sixties. Is that if that device or similar device or structure is to be used in a different area or on a different project. They use the load values, and say to themselves, "well, this load is going to be subject to heavy jostling. So we will get a chain with twice the load rating and we should be fine.

The only problem is that they are using the weight itself, to calculate above and beyond the dead weight. They are not using the actual tension created, by the shock load.

They need to perform a new actual test. And when they do, it often shows what I am saying.

When they take the highest recorded scale value of a weight dropped on a scale a certain distance. They are only measuring half the tension created by the weight. Because that is what a scale does.


Sincerely,


William McCormick

In our Physics class, we have laboratory exercises to confirm what he have learned in lecture.

So far, labs said lecture was correct.

Originally Posted by CoolEJ

In our Physics class, we have laboratory exercises to confirm what he have learned in lecture.

So far, labs said lecture was correct.

I know some of the lab exercises of today, they are meek. Even in my day some of the hand on classes in my rather insane school were getting meek.

If you go into the real world, you will throw out most of what you were taught in college. If you want to make quality stuff.

Look at a scale, before it can register a load, the load has to apply weight to the scale, if it is a ceiling scale, the ceiling has to apply a counter force, before the scale can register. This takes time. The movement in the scale will negate any test results. Because it takes time to move the scale.

The only way that I know of, to actually test a load, and the system holding it, under a shock or hammer effect. Is to create an actual test when you are done with whatever you are building, and destroy some part or all parts of it, destructively testing it.

Because different loads, either flexible or stiff, different supports either flexible or stiff, can cause standard, chain, rope, cable, and rods, to break prematurely. But that is another aspect of it.

I have suspended things from truly secure pieces of large steel. And I have instantly snapped, lifting equipment that I know would hold that load with a less solid support.

I have used chain, to pull down large trees, with an ordinary 150 Ford pickup truck. Root and all pull up. Because the chain does not flex, the tree does. People watch and often cannot believe it. I can actually with chain pull all the roots out as well as pull it down.

One thing that many sheet metal guys know, is the wave effect in metal. If you are lifting a long piece of extruded metal, "I" beam, "U" channel, bar, and someone strikes the long extruded piece of metal. And creates a wave in the metal, the force could be many, many times greater then the weight. I am not just talking two fold either. Exponential force. This effect can break your neck if you carry sheet metal over your head. Metal ducts, signs can all kill.

Today what most of the cranes do is, allow for, a cushioning in their cable systems, or boom. This way that tremendous force never totally hits any one part of the system. The problem with allowing spring is that, it can hurl objects, and or operators. It can create a dead blow effect on the boom or crane housing.

It can pull a crane over.

Why don't you share some of these convincing experiments you saw? Maybe I cannot hack them apart?


Sincerely,


William McCormick

William, I was telling you to specifically avoid shock force, not measure it. I agree that an ordinary spring scale will not easily measure shock force, but since that's not what I was suggesting you measure, that's a moot point, isn't it?

Originally Posted by MagiMaster

William, I was telling you to specifically avoid shock force, not measure it. I agree that an ordinary spring scale will not easily measure shock force, but since that's not what I was suggesting you measure, that's a moot point, isn't it?

I know you did. Never said you did not.

However they have never understood shock load in any college that I know of today. I hope it changes. If it does change, they will also see tension as I do, twice the weight being supported.

I wanted everyone reading to look at this from all the angles. To see the flaws in current views.

Many have experience with shock loads collapsing things, or cables breaking. It is a very memorable moment usually. Often having to do with shock loads. That today are not necessarily taken into consideration properly.

As I mentioned about your post, load rating, test, and weight, are not tension. Even by definition they are not tension.

Sincerely,


William McCormick

Load rating is how much non-shock weight the thing is designed to hold up, right?

Originally Posted by MagiMaster

Load rating is how much non-shock weight the thing is designed to hold up, right?

No, not originally. Load rated meant that the device, chain, rod, was tested at the rated load. That did not just mean held there, like a new born baby. It meant dropped from a certain height, or lifted at a certain speed.

Years ago, many used to think they overrated chain to much. But not when you consider its use in day to day life. In its actual applications.

A 500 pound load rated chain will lift more dead weight then 500 pounds. But it is no longer load rated for that amount.

Some chain will survive a tremendous drop at its rated load. But then it must be thrown out, because it elongates.

This elongation and the time it takes, could cause damage to a crane or other device. Because of the increased duration of very high force.

Often elongated chain will not move through a chain hoist.


Sincerely,


William McCormick

Well then, my experiment is still vaild, but you'll have to find the lines non-shock breaking point by testing.

Originally Posted by MagiMaster

Well then, my experiment is still vaild, but you'll have to find the lines non-shock breaking point by testing.

But what was your experiment meant to show?

Sincerely,


William McCormick

Would you agree that too much tension is what causes the line to break? In that case, what you say would imply that 90 lbs on one end of a line fixed to the ceiling would be that same as 45 lbs on each side of a line around a pulley. Either both should break the line, or neither should.

Originally Posted by MagiMaster

Would you agree that too much tension is what causes the line to break? In that case, what you say would imply that 90 lbs on one end of a line fixed to the ceiling would be that same as 45 lbs on each side of a line around a pulley. Either both should break the line, or neither should.

No, 90 pounds of weight from the ceiling creates 180 pounds of tension in the chain, rope, cable, rod. That is the same as two 90 pound weights over a pulley.

Your tension is always going to be double your weight suspended from a ceiling by a single line.

Your tension is going to be equal to, two counterweights, of equal weight over a pulley, totalled up.

That is why you can trampoline on, roofs made with the collegiate truss design.


Sincerely,


William McCormick