# Couple on rod in electric field

1. Feb 17, 2015

### PhysicStud01

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I can understand why the resultant force is not zero as the field is not uniform. but for a couple, shouldn't the forces be equal. so how can B be correct?

also, at some point, the rod will be horizontal. won't the resultant force be non-zero and act horizontally while the torque is zero. the rod would stay horizontal?

can someone help if I'm misunderstanding something

2. Feb 17, 2015

### haruspex

The forces are certainly not equal. Where lines of force are closer together, the field is stronger.
However, I don't see an easy way to prove there is a couple about the mass centre of the rod.

3. Feb 17, 2015

### PhysicStud01

the only situation where I think the forces are equal is when it is vertical, but then the resultant force is zero even if there's a couple, right?

4. Feb 17, 2015

### haruspex

There would certainly be no torque if the centre of the rod is perpendicular to the field (at that point). But there would still be a net force.

5. Feb 17, 2015

### PhysicStud01

is it not the reverse? when the rod is vertical? the forces are equal and oppostite, so the net force is zero. but torque would be present right? the forces act at the end of the rod?

6. Feb 17, 2015

### haruspex

Whoops - didn't notice the opposite charges. Sorry about that.

7. Feb 17, 2015

### haruspex

So let me correct my answer... with the rod 'vertical' (I'd still rather describe it as perpendicular to the field at the rod's midpoint) there is a net force, but not in the direction you may be thinking of.

8. Feb 17, 2015

### PhysicStud01

yewah, when i looked again, the directions are not exactly opposite.
so, anything about the couple. the answers actually say couple, not moment or torque. is there a difference?

i thought that the answer given may be wrong, but the same answer is given as the correct one elsewhere.

9. Feb 17, 2015

### haruspex

You mean, back at the original question, right?
The question is not worded correctly.
The net of any system of forces acting on an object can be represented as the combination of a force through the mass centre and a couple.
If the force is nonzero then it is always possible to represent this pair by a single force that does not necessarily pass through the mass centre. Therefore the question ought to specify forces that act through the mass centre.
It is clear there is a net force because the two individual forces are not collinear. They are also different in magnitude, which is also enough in itself to ensure there is a net force.
There must be a net torque about the mass centre because the two forces act in the same sense about it. It follows that if the forces are represented as a force through the mass centre and a couple then the couple is nonzero.

10. Feb 17, 2015

### PhysicStud01

could you be a bit clearer please? because i read it and could not clearly understand some of them.
specially:
The net of any system of forces acting on an object can be represented as the combination of a force through the mass centre and a couple.

and
There must be a net torque about the mass centre because the two forces act in the same sense about it. It follows that if the forces are represented as a force through the mass centre and a couple then the couple is nonzero.

does a torque always imply a couple? a couple consists of 2 forces of equal magnitudes, right?

11. Feb 17, 2015

### PhysicStud01

can someone please explain how there's a couple here

12. Feb 17, 2015

### haruspex

First, let me correct one statement:
I was thinking only of 2D systems, as in this case. In general:
If the force is nonzero and the net torque is perpendicular to the net force then it is always possible to represent this pair by a single force that does not necessarily pass through the mass centre.​
E.g. if we consider two equal forces at right angles but not passing through a common point, the resultant linear force and the resultant torque are parallel.

You asked me to elaborate on this:
The net of any system of forces acting on an object can be represented as the combination of a force through the mass centre and a couple.
Suppose the forces Fi act on an object, with the line of action of force Fi being displaced by vector Ri from the object's mass centre.
The net force $F = \Sigma F_i$, and the net torque about the mass centre is $T = \Sigma F_i \times R_i$.
If we choose the line of action of F to be through the mass centre then it has no torque about the mass centre, so the system of forces is completely represented by the combination of F and T.
The pure torque component, T, can be represented as a couple: If C is any vector at right angles to T, and D is at right angles to both (e.g. D = C x T) then we can represent T as the combination of two forces kC, -kC with lines of action displaced by vectors +D , -D from the mass centre. $T = 2kC\times D$.

If instead we choose the line of action of F to be displaced by R from the mass centre then it produces torque $F \times R$. The remaining torque to represent the original system is $T-F \times R = \Sigma F_i \times R_i - \Sigma F_i \times R = \Sigma F_i\times(R_i-R)$.
If T is orthogonal to F and F is nonzero then $(T \times F)\times F$ is a vector parallel to T. Setting $R = kT \times F$ for some suitable k we get $T-F \times R = 0$, so we can represent the entire system of forces by a single force, maybe offset from the mass centre.
See http://en.wikipedia.org/wiki/Resultant_force, though I don't think the explanation there is great.

13. Feb 17, 2015

### PhysicStud01

thanks. but is there any simpler reasoning to this?

14. Feb 17, 2015

### haruspex

I can't offer a simpler explanation (maybe others can), but perhaps you're a bit confused by the teminology.
The torque due to a force depends on your reference point. If it's a point in the line of action of the force then the torque is zero.
Similarly, the net torque of a set of forces generally depends on the reference point, but there is an interesting special case. If the vector sum of the forces is zero then the torque is independent of reference point. This may be referred to as a pure torque, or sometimes as a "screw". The simplest example is an equal and opposite pair of forces with different lines of action (so antiparallel). This is referred to as a couple.
I was taught the memorable fact that "every couple reduces to a screw".
For an unequal pair of parallel forces, you can always represent the resultant by a single force that has the same net torque about every reference point. (See if you can calculate how to do that.). For intersecting forces, obviously the net force through their point of intersection is equivalent.
But in three dimensions there is a third possibility, forces that act along "skew lines in space", neither parallel nor intersecting. If equal in magnitude, these are equivalent to a linear force (halfway between the lines of action) plus a screw parallel to it. Think, the forces exerted on a screwdriver, pushing and turning at the same time.

15. Feb 17, 2015

### PhysicStud01

ok. let's forget about the 3d case.

how does that apply in this specific question. how do I then represent the couple of forces on the rod? are they the components of forces at +Q and -Q? I believe that what you mention above are actually mathematical theories of vectors? but how to interpret it in this physical example (in the question)?

thanks

16. Feb 18, 2015

### haruspex

In this specific question, there are two forces, unequal in magnitude. Consequently the net linear force (vector sum), F, is nonzero.
They are coplanar but not parallel. Consequently their lines of action intersect. The net force must be completely equivalent to F acting through that point of intersection. There is no need to add a couple.
However, if we require to express the resultant of the forces as F through the mass centre plus a screw (couple, if you prefer) as necessary then consider the midpoint of the rod. Each of the contributing forces has a clockwise moment about that point, so it will be necessary to add a clockwise screw.

17. Feb 18, 2015

### PhysicStud01

so, I need to consider the resultant force at the centre of the rod, not the individual forces at the ends? it's not a gravitational field - in which case the force can be considered to act at the centre. but it's an electric force, right, which acts on charges?

Also, for a couple, should the magnitudes of the forces be equal and the forces antiparallel, which are not the case here?

18. Feb 18, 2015

### haruspex

The question does not say that you should consider the resultant linear force as acting through the mass centre, but unless you take that view you cannot arrive at answer B. (I consider the question poorly worded, but I can see where it's coming from. What they're really asking is whether the rod will both move as a whole and rotate.)
For the answer to be a couple (and no linear force) the two forces would need to be antiparallel and equal in magnitude.

19. Feb 18, 2015

### PhysicStud01

but then, is answer B truly correct, or is A more likely

20. Feb 18, 2015

### haruspex

The correct answer is that it can be represented as a single linear force with no couple, but it can also be represented by a linear force (vectorially the same, but different line of action) plus a couple.