Prove Polygon of Forces Not Necessary for Equilibrium

[batballbat]

If any number of forces acting on a point is in equilibrium then the forces does not necessarily form a polygon of forces? How can we prove it?

 

[bp_psy]

hint: A counter example is proof enough.

 

[batballbat]

so if the forces acting on the point do not form a polygon they do form an incomplete rectilinear figure with one side incomplete the side being the resultant of the forces. But as the forces are in equilibrium the resultant must be zero. But the resultant is not zero. THen they must form a polygon. can somebody help me with this?

 

[DaveC426913]

so if the forces acting on the point do not form a polygon they do form an incomplete rectilinear figure with one side incomplete the side being the resultant of the forces. But as the forces are in equilibrium the resultant must be zero.

Okay.

But the resultant is not zero.

Why do you say this?

THen they must form a polygon. can somebody help me with this?

If the net force is zero then all the force vectors will sum to zero, i.e. the final arrow will end at the point i.e. it will form a closed figure.

 

[batballbat]

so why does my book say that the converse of polygon of forces is not true?

 

[Studiot]

so why does my book say that the converse of polygon of forces is not true?

This looks like homework.

No doubt your text or teacher has told you the conditions for equilibrium and this is a test to see if you have noted them.

Post again with your thoughts on these conditions and we can help you answer.

Note a force polygon is not the only way a bunch of forces can be in equilibrium.

 

[sophiecentaur]

If the polygon is not complete, how can there be equilibrium? The polygon can always be reduced to a triangle or, in this case, two non cancelling vectors. This even applies to non-coplanar vectors.

 

[batballbat]

According to the book
Polygon of Forces: If any number of forces, acting on a particle, be represented, in magnitude and direction, by the sides of a polygon, taken in, order, the forces shall be in equilibrium.

Note: the converse of polygon of forces is not true

 

[Studiot]

You didn't answer my question about homework, but never mind this time.

Polygon of Forces: If any number of forces, acting on a particle, be represented, in magnitude and direction, by the sides of a polygon, taken in, order, the forces shall be in equilibrium.

Note: the converse of polygon of forces is not true

The quote from the book is all true.

It says If the forces do form a polygon then theyare in equilibrium.

It also says that the converse is not true ie

If the forces do not form a polygon then they are not in equilibrium.

So that implies that there is at least one more way in which the forces can be in equilibrium.

What does you book say about parallel forces?

 

[batballbat]

You didn't answer my question about homework, but never mind this time.



The quote from the book is all true.

It says If the forces do form a polygon then theyare in equilibrium.

It also says that the converse is not true ie

If the forces do not form a polygon then they are not in equilibrium.

So that implies that there is at least one more way in which the forces can be in equilibrium.

What does you book say about parallel forces?

But here we are dealing with forces acting on a single point.

 

[Studiot]

But here we are dealing with forces acting on a single point.

What difference does that make?

 

[batballbat]

so can we conclude that
if any number of forces of acting on a point be in equilibrium then either they form a polygon or they are parallel?

 

[Studiot]

if any number of forces of acting on a point be in equilibrium then either they form a polygon or they are parallel?

Yes, exactly so.

Consider a charged weight hanging from a string (in equilibrium)

We have gravity acting downwards and the string tension acting upwards.
Now introduce another charged object so that the weight is attracted downwards and also push upwards from underneath so equilibrium is maintained.

There are 4 concurrent parallel forces acting on the weight, which is in equilibrium.

go well

 

[batballbat]

i searched in google and found a book which contains the following reasoning:
THe converse of the polygon of forces in not true. A number of polygons can be found whose sides represent the forces, all the polygons are similar; but any polygon with its sides parallel to the forces will not represent them.

can somebody explain this to me?

 

[sophiecentaur]

i searched in google and found a book which contains the following reasoning:
THe converse of the polygon of forces in not true. A number of polygons can be found whose sides represent the forces, all the polygons are similar; but any polygon with its sides parallel to the forces will not represent them.

can somebody explain this to me?

I don't know what all this can mean; the mechanics of the OP are obvious - aren't they? But all of this may just be a bit of a geometrical 'joke'. If someone gives you a set of vectors (i.e. arrow lines of a given length and each one orientated in a given direction) then you need to connect them in a particular way / order if they are to form a closed polygon. Many of the possible arrangements will not form one closed polygon, when you get to the end - they may form a number of polygons. But, if there is a solution to this joining up puzzle, then the forces must be in equilibrium. Depending on the symmetry of the situation, there may be more than one solution.
I must say, there is no distinction between a polygon and a set of parallel vectors - they constitute a polygon with no volume - that's all.