Understanding Forces in an Anchor System

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In summary, the conversation revolves around determining the distribution of load between two parallel ropes holding up a 70 kg load, with one person trying to prove that the load will be split equally between the ropes. The use of symmetry is discussed, with the conclusion that if the system is perfectly symmetrical, the load will be split equally between the ropes regardless of their length. However, without further information, it is not possible to determine if the magnitudes of T1 and T2 are equal. The question of whether a perfectly symmetrical system can be defined in this situation is also raised, with the possibility that calculus may be needed to solve the problem accurately.
  • #1
rambo5330
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Homework Statement


This will probabaly seem like a very simple question and maybe I'm wayy over thinking it.
This goes back to forces on an anchor system.. let's say I have two pieces of rope holding up a 70 kg load.. each rope is attached to a separate anchor.. the two ropes are perfectly parallel so there is no angle between the two..
how do i know that's the load will be split between the two.. what's the proof.

I have T1 and T2 acting up in the same direction... (not sure on magnitude).. and mg acting down
so
T1 + T2 = mg = 686 N

but how do i know T1 = T2 which it clearly does...


Homework Equations

 

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  • #2
No, not that obvious.
If the problem says nothing more, then the most we can deduce is T1+T2 = mg. Only under some certain other conditions (which means the problem must provide more info), such as the system is symmetrical, can we state T1=T2.
 
  • #3
it just seems odd to me that if i were to take that exact same question and instead of having each rope parallel to each other i make an angle of 140 degrees between the two.. i can tell everything. Since it adds a horizontal component and the system is at equilibrium TIx must equal T2x and since the load has only Y components and no horizontal.
T1x = T2x therefore T1 = T2

but i take that same situation and remove the angle between the two ropes and now we can no longer deduce that T1 = T2 ? something seems off
 
  • #4
rambo5330 said:
Since it adds a horizontal component and the system is at equilibrium TIx must equal T2x

How do you deduce this?
 
  • #5
As per the diagrams attached on this reply.
Situation a.) shows what i am trying to figure out.. and situation b.) shows a more complex situation that its clear to see that the two horizontal components must equal each other since the object is not accelerating the sum of forces on the x access must = 0(bad example here because it is perfectly symetrical) but regardless of the angles there .. if the load is hanging straight down.. it has no horizontal component so the forces created by the tension in the x-axis must cancel each other out.. thus making them equal.

its not quite as saying T1x = T2x off the start .. in a non symetrical system youd end up with T1 = T2cos(25) / cos(45) then you can sub into the Y access and solve.

this is more complex but yet way easyier for me to find T1 and T2 but for a set up like a.) what's the math behind proving T1 = T2 ... it must.. even if the rope for T2 is 10ft and T1 is 2ft they should still be splitting that load 50/50 but i can't prove it...
 

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  • #6
If you flip the original system right to left, can you tell the difference between rope 1 and rope 2? Assuming you did not see it flip?

If your answer is no, that means you can deduce by symmetry that the two split the load evenly, since there is no reason why any other possibility should be plausible.
 
  • #7
well say for example rope1 was is 10ft and rope 2 is 4 feet.. so they are hung from diffrent heights but in the end they attach to the load at exactly the same point parrallel to each other.. the lengths are different so you can definitely see a difference between rope 1 and rope 2 but .. the force must be the same.. i can see it in my head.. doesn't mean its right but I am 99% sure it is.. i just don't know how to create a proof on paper that can deduce...
T1 = T2
 
  • #8
There's a difference as far as the rest of the system goes. The way the load is distributed among the ropes will be very different, but if you focus your attention on the box and its contacts with the two ropes, can you tell a difference?
 
  • #9
if you focus in right at the box no you cannot both ropes would leave the connection at the box from roughly the same point perfectly parallel..
so i can see the force would have the exact same direction. but how do i know the magnitudes would be equal
 
  • #10
Oh so x denotes horizontal direction. I thought it's vertical.
So how do you deduce this: T1x = T2x therefore T1 = T2? (post #3)
Anyway we can never know whether the magnitudes are equal if no further information is provided.
 
  • #11
yes third post was a bad example i was looking at a symmetrical system at that point..

so what dictates a perfectly symmetrical system in a situation where the two ropes a perfectly parallel with each other... if one rope is longer than the other.. yet they are parallel the force should still be split equally but it doesn't look symmetrical to me..
 
  • #12
There has to be a way that physics can solve this sort of a question short of sticking a load cell on the ropes to determine that the load would in fact be split equally... I'm definitely missing something here. maybe its calculus I have no idea. but if i take those two ropes that form an angle of 140 degrees and now I make the angle of 0.00000001 degreess then you could easily determine that the resultant vectors created by T1 and T2 would be essentially half the weight of the load each(assuming a symmetrical set up like you said earlier)... now if i make the angle 0 degrees why is this so much more complex... this should be simple to explain. or maybe i just answered my own question... if the system is symetrical rope length doesn't not matter because the freebody diagram is only concerened with the connection point at the load and the direction of the forces, and it is the magnitude of these forces that will dictate the length of these vectors (but we do not know the magnitude to start so the lengths are arbitrary to begin with)?
I would like some more input please
 
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  • #13
do the length of ropes affect the magnitude?
i don't think so
because i think they have experienced the same force..
what have made T1 and T2 unequal?
can please list out the factors??
 
  • #14
thats also a good way of looking at it. what makes T1 and T2 unequal... well I am not sure in this case.

the two ropes are parallel with each other... and attached to the load at the same spot. one rope is longer than the other. the forces are only in the Y axis.

I cannot think of any other factors?
 
  • #15
You started out with the 2 parallel ropes attached at different points, but since then you have been focusing on 2 ropes attached to the same point. Assuming inextensible cords, the tensions T1 and T2 are the same in both cases, if the loading is symettrical in case 1 and 2. You can show this by summing torques about any point for case 1, and by joint equilibrium of forces for case 2.

In case one, for 2 parallel ropes attached at different points and not symmetrical, then you must use moments, T1(d1) =T2(d2), where d1 and d2 are the respective perpendicular rope distances from the center of mass of the block, which must be given. For example , if T1 was attached directly above the center of the block, and a parallel rope T2 was attached at the right corner of the block, then T1 would be equal to the blocks weight, and T2 would be zero. For the second symmetrical case, T1cos theta = T2 costheta, T1 = T2 (you run into dividing by 0 when theta is 90 degrees, but in the limit, T1 still equals T2).

For inextensible rigid ropes, the length of the ropes make no difference.
 
  • #16
very good answer thanks so much for looking at my problem... There was some clarification issues on my behalf but you covered each situation perfectly. The initial question I posted "attached to separate anchors" was misleading because i meant separate anchors for connection to a wall or whatever surface one above the other and connected to the load at the same point..in the diagram i separated T1 and T2 for clarification that there were two different forces but I think it lead to confusion.

I will play around with your answer thanks !

EDIT: after playing around with this I am more concerned with case 2 and the method of joint equilibrium. can I get pointed in the right direction on joint equilibrium? I did some googling and found it to be a common concept for determining forces on trusses .. is this the same method?
 
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  • #17
rambo5330 said:
after playing around with this I am more concerned with case 2 and the method of joint equilibrium. can I get pointed in the right direction on joint equilibrium? I did some googling and found it to be a common concept for determining forces on trusses .. is this the same method?
Yes.
 
  • #18
is this not a joint sum of forces (note T1 and T2 are separated for clarity, you can think of them as attaching to the same point)? sorry to keep bugging you on this one. I really want to make sense of it and have not heard the term joint equilibrium before but from what I gathered on what I was reading this is all it was a summation of forces at each joint... I'm haveing trouble yielding T1 = T2 .
 

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  • #19
rambo5330 said:
is this not a joint sum of forces (note T1 and T2 are separated for clarity, you can think of them as attaching to the same point)? sorry to keep bugging you on this one. I really want to make sense of it and have not heard the term joint equilibrium before but from what I gathered on what I was reading this is all it was a summation of forces at each joint... I'm haveing trouble yielding T1 = T2 .
This is a special case of symmetry where theta, the angle between each force and the horizontal, is 90 degrees, so you lose the identity of the triangle as in your '140 degree' example above. In general, for equal angles (the symmetrc case) from sum of forces in x direction = 0, then T1cos theta = T2 cos theta, ot T1=T2. So pretend that theta is 89.999999 degrees instead of 90 degrees, for the parallel ropes case, so you don't have to divide by 0 , and then T1 = T2.
 
  • #20
okay nice! that's logic I've been messing around with as well and it seemed everything was going good and then I thought...well if T1 and T2 are 90 degrees and have 0 horizontal component then we assume 89.999999 degrees etc... can the same not be said for the load pulling downward giving it a slight horiztonal component as well?
so F_x= T2cos(theta) - T2cos(theta) + T3 Cos(theta)
or because the force acting on the load is gravity we know it will always be perpendicular in this example?
 
  • #21
rambo5330 said:
or because the force acting on the load is gravity we know it will always be perpendicular in this example?
Yes, gravity acts vertically downward with no x component.
 
  • #22
Thanks very much for your help. My head feels so much better.
 

1. What is a simple force?

A simple force is a type of force that acts on an object without causing any rotation. It is a force that acts in a single direction, such as pushing or pulling an object without causing it to spin or turn.

2. How is a simple force different from other types of forces?

A simple force is different from other types of forces because it only acts in one direction and does not cause any rotation or movement around an axis. Other types of forces, such as torque, can cause an object to rotate or turn.

3. What are some examples of simple forces?

Some examples of simple forces include a person pushing a door open, a book resting on a table, or a car being pulled by a tow truck. Any force that acts in a single direction without causing rotation is considered a simple force.

4. How do simple forces affect an object's motion?

Simple forces affect an object's motion by causing it to accelerate in the direction of the force. If the force is greater than the object's mass, it will cause the object to move in the direction of the force. If the force is less than the object's mass, it may cause the object to slow down or stop.

5. How can we calculate the magnitude of a simple force?

The magnitude of a simple force can be calculated using the formula F=ma, where F is the force, m is the mass of the object, and a is the acceleration caused by the force. This formula is known as Newton's Second Law of Motion.

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