Balancing forces on a particle.

[Final_HB]

A particle is under the influence of 5 forces. Three of these forces are reversed, and the particle remains in equilibrium. Prove that the particle will remain in equilibrium even if these three forces were removed altogether.

 

My thinking for this is:
With three of the forces gone, the 2 forces (F₁ and F₂ ) left must balance the system if its to remain at equilibrium.
If change of forces occurs, the system will react by moving to a new state of equilibrium.
No reaction means that the 3 forces in the question cancel out each other, and no shift occurs.
If the three forces balance to 0, there is no need to have them as acting on the particle.

Right?

If im right, Is there any maths way to say this, or a better way to explain what im trying to say. Thank you in advance.

[swansont]

Reversed is a - sign, since forces are vectors. Cancel is a subtraction. Does that help you write down an equation (or two)?

[Final_HB]

I think so.

 

something like:

F₁+F₂+F₃+F₄+F₅=0

 

And when its reversed:

F₁₊F₂-F₃-F₄-F₅=0

let the two equal each other and cancel?

[studiot]

You can't say "let the two equal each other" because the problem is not specific about the five forces to begin with.

 

That is it could be any 5 forces in equilibrium.

 

However the second statement introduces an additional condition about 3 of them.

 

What must be the case if you can reverse these 3 forces with no effect on the equilibrium?

[Final_HB]

Fair enough...

The sum of the three forces is 0.

[studiot]

So can you now complete the problem?

[Final_HB]

Cause the sum is 0, they can be just taken out of the formula and so

 

 

F₁+F₂=0

 

is the only forces needed for equilibruim

[studiot]

There is more to be obtained from this problem.

 

Firstly I would say that not only is the resultant of the three forces zero but that means they are in equilibrium.

 

However it does not mean that any three forces of five in equilibrium can be reversed and equilibrium maintained.

 

 

What happens if you reverse F1, F2 and F3? Can you remove them maintaining equilibrium?

 

What happens if you reverse F3, F4 and F5? Can you remove them maintaining equilibrium?

 

You have correctly noted that if the remaining two are in equilibrium their resultant must be zero.

 

This means that they must be colinear (equal and opposite along the same line of action).

 

Can you see what would happen if they were equal and opposite, but did not act along the same line?

[swansont]

You can't say "let the two equal each other" because the problem is not specific about the five forces to begin with.

 

That is it could be any 5 forces in equilibrium.

 

However the second statement introduces an additional condition about 3 of them.

 

What must be the case if you can reverse these 3 forces with no effect on the equilibrium?

 

 

But it's true that F_n and -F_n will cancel, and that the numbering of the forces is arbitrary. Simply call the three that are reversed force 3, 4 and 5. Then it's perfectly reasonable to say F₁+F₂+F₃+F₄+F₅=0 and F₁₊F₂-F₃-F₄-F₅=0 , as Final_HB has done

 

[studiot]

Hi Swansont, you are not understanding my example.

 

5 forces are in equilibrium. F1, F2,F3,F4,F5

 

If you reverse F1, F2 and F3 they system is no longer in equilibrium.

 

F1, F2 and F3 do not form an equilibrium system by themselves.

 

So I am emphasising that you cannot just choose any 3 of the 5 at will.

 

You have to choose and equilibrium set.

 

So you are partitioning the system into two equilibrium sub-systems.

[swansont]

Hi Swansont, you are not understanding my example.

 

5 forces are in equilibrium. F1, F2,F3,F4,F5

 

If you reverse F1, F2 and F3 they system is no longer in equilibrium.

 

F1, F2 and F3 do not form an equilibrium system by themselves.

 

So I am emphasising that you cannot just choose any 3 of the 5 at will.

 

You have to choose and equilibrium set.

 

So you are partitioning the system into two equilibrium sub-systems.

Yes, you have to choose the set that the problem says you can reverse. But given that constraint, the solution works.