# Concurrent forces have no turning effect??

• posted

My textbook says that concurrent forces have no turning effect, but I don't quite understand this.

What are the conditions under which this applies?

If I hold a pencil horizontally and then drop it, the only considerable force acting on it is the force of gravity. If, while it's falling, I then hit it at one end, obliquely downwards, it spins. And yet the direction in which I applied the force and the direction of gravity are concurrent.

What am I missing??? (probably something very simple, but I will be grateful for any help).

Cheers,

Chris

• posted

Gravity acts as though it is applied at the center of mass. If you apply your force at the center of mass to be concurrent, it won't twist.

--Lynn

• posted

Seems like you don't understand what concurrent forces actually are. Look this

• posted

Jeff Finlayson wrote in news:11nuqs258qqvf5 @corp.supernews.com:

Well, I ain't arguing because it is defined that way, but that is one lousy piece of terminology, concurrent has a strong implication of simultaneous, not coincident in space, to your average English speaker.

Would it have hurt so much to use coincident, way back when?

Cheers

Greg Locock

• posted

That fits with the definition. See #3 below.

1. Happening at the same time as something else. See Synonyms at contemporary.
2. Operating or acting in conjunction with another.
3. Meeting or tending to meet at the same point; convergent.
4. Being in accordance; harmonious.
• posted

Jeff Finlayson wrote in news: snipped-for-privacy@corp.supernews.com:

Thanks Jeff.

That source says that concurrent forces "pass through a common point", and has a diagram showing three forces applied to the outside of an irregularly shaped object, with the three lines of force intersecting at a point inside the object.

In my example, the lines of force also intersect, namely at a point directly below the middle of the pencil.

Why aren't those two forces concurrent?

I thought concurrent forces were those acting along lines that intersect. This is in fact what it says in my textbook, but since the forces in my example fulfil this condition and cause a turning effect, I am certainly missing something!

Chris

• posted

They are. I believe that's another example, a simpler one.

Your example with the pencil does not meet the criteria, since the sum of the moments is not 0.

• posted

You are missing the other condition for equilibrium, that the vector sum of the forces is zero.

For your example, there is no real need to have the pencil in free fall. Just have it freely pivoted at the centre of mass. Then apply your oblique force at one end and look at the balance of forces. (You now also have a reaction force at the pivot.)

• posted

It's in the definition of concurrent. Concurrent forces act through a single point. Gravity can be considered to act through the center of ..er.. gravity, so if you whack it rhrough this point, it won't turn. Right?

Brian W

• posted

This is a slight digression from the OP's question; but "concurrent" (from Latin "running together") is a long-established term in geometry. Three or more lines are concurrent iff they all pass through a common point.

Ken Pledger.

• posted

Jeff Finlayson wrote in news: snipped-for-privacy@corp.supernews.com:

[...]

Aren't both moments about the point where the lines of action meet, and therefore their sum, equal to zero?

Chris

• posted

Gib Bogle wrote in news:dloftg\$cqq\$ snipped-for-privacy@lust.ihug.co.nz:

But surely in your three-force example, the line of action of the reaction force at the pivot would not run through the intersection of the lines of action of the other two forces (gravity and the oblique force, which would intersect directly below the pivot)...

Chris

• posted

Your non-concurrent pencil example is not a static problem. The moments sum to I*alpha. If you can't understand this then I can't help any more.

• posted

Jeff Finlayson wrote in news: snipped-for-privacy@corp.supernews.com:

OK, I think I can now help myself! :)

I believe the answer to my query is that

a) my textbook is wrong to state that concurrent forces always have no turning effect

b) concurrent forces will have a turning effect if the line of action of their resultant does not go through the centre of gravity (or centre of mass if there is zero gravity)

c) ...because in such cases, the resultant moment about the centre of mass will not be zero

The main point I was missing is that the moments must be measured about the centre of mass.

The resultant moment about a point where the lines of action of concurrent forces intersect (or about any other point on the line of action of the resultant force) will always be zero...

...but this is irrelevant to the issue of whether or not there's a turning effect.

(If I've got anything wrong above, I'd be grateful if someone could tell me!)

Cheers,

Chris

• posted

Quite! Jeff just won't accept that concurrent forces do not imply a turning moment. It's an idee fixe so he needs to stew awhile

Brian Whatcott

• posted

The turning effect is due to the inertial reaction of the mass, opposite the applied force, making a couple. If the body is in free space, like your falling pencil example, it acts from the object's center of mass.

But...if the object is not completely free this may not be true. If you strike a hanging pendulum, for example, it depends on the object's "center of percussion", which is usually in a completely different spot than its center of mass.

Think of a rod hanging down and hinged at the top. If you strike it on its right side, up near the top, the hinge will experience a force from the right. If you strike it at the bottom, the rod will try to spin clockwise and the hinge will experience a force from the left. If you strike it at its midpoint (center of mass) the whole rod will try to move to the left and the hinge will still experience a force from the right. There is a point on the rod where you can strike it and the hinge will not feel any force at all from the blow. This is the center of percussion.

Don Kansas City

• posted

Given three forces and points of body on which they act. For full equilibrium:

1) Displace force vectors, draw a triangle, force vector sum should be
2) Add up clockwise and anticlockwise moments with reference to any point, moment vector sum should be algebraically 0.

If 2) is satisfied and not 1) then the body translates and accelerates by Newton's First Law. Concurrency gives zero turning effect on a knot between three pulled strings, which may move.

If 1) is satisfied and not 2) then the body rotates at the same location. Example : A couple of two equal and opposite forces F separated by a distance d. Couple rotating body is F *d. An example is an armature of an electric motor.

• posted

...

These comments are misplaced and as stated are backwards with what I have posted elsewhere in this thread.

• posted

The lines of the (concurrent) forces only intersect when the pencil's body force (m*g) and the reaction are collinear. There is no rotatation when this is the case.

For concurrent forces the moments sum to 0 since there is no moments about the intersection point. As part of this, concurrent forces are the only loads acting on the body.

• posted

On Wed, 23 Nov 2005 11:12:13 -0600, Jeff Finlayson wrote: [Chris]

[Jeff]

My sincere apologies to Jeff. I was indeed preaching to the choir. The interspersed comments to Chris threw me. Chris WILL get it: Jeff evidently not only has got it, but is willing to SHARE it!

:-)

Brian Whatcott

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