I have a question which asks to replace a force and a couple moment with ones of equivalent force and couple moment, I'll attach the original and I have also done one which I think is correct. The one done in pen it mine to clear it up lol - the changes aren't that different but seeing as the type of force has changed as well is it acceptable?

The force on the left has changed from a pulling to a pushing force and the force on the right has changed from a pulling to a pushing force.

Thanks!
http://i94.photobucket.com/albums/l101/fiddler2k6/original.jpg?t=1276177984

http://i94.photobucket.com/albums/l101/fiddler2k6/revised.jpg?t=1276177941

You only forgot to put the 3 m mark from O to the pushing force of 140 N and to put the moment as 40Nm.

I'm a little surprised that 's what your question is asking. There is not much 'work' to do to answer the question if that's the way to answer it.

If you could post the original question, it would be nice. And are there other parts in the question, or that it?

I was surprised how easy it was - if its what they want - as well. The question sheet is at work, I'll post it in the morning, though it isn't much different from what I posted originally. I think it was something like 'which has the same effect at point P'

-Cheers

But tell me, what are the points O ad P supposed to represent? Do they form the corners of a rectangular block/book and the x axis? Or is there no information about that too?

No information, we got given the question then the diagram I posted first. It might be clearer with the full question but there wasn't anything else.

Ok, I hope you'll get it right then :)

What they want is the values of a single force and a single moment placed at point P that would be equivalent to the forces and moments shown in the problem. You do this by "moving" the forces to P, and for each adding a complimentary moment at P, to mimic the moment that the force originally had about P. So - the 140N force applies a moment of 140 N x 15 m = 2100 N-m about point P, counter-clockwise. So you could move that force to point P and apply a counter-clockwise couple at point P of 2100 N-m and maintain the same overall efect on the system. The 40 N-m couple can simply be shifted to point P without any other effect. The 60 N force is a little more problematic - you need to calculate its couple about P, and to do that you're goint to have to calculate the cross product of the 60 N force vector and the vector from P to the point where that force is acting. Once you've done these translation, you vectorally add up the forces to find a single force equivalent, and also add the moments. The result is one force and one moment acting at P that is equivalent to the two forces and one moment presented in the problem. Hope this helps!

Ah, yes I understand now, the 'which has the same... '

I didn't notice that, thanks ebaines :)

Here's the original question.http://i94.photobucket.com/albums/l101/fiddler2k6/Question.jpg?t=1276263205

Thanks to you both, I'm having another look the ebaines' explanation to see what I get.

EDIT
Here's what I got:-
http://i94.photobucket.com/albums/l101/fiddler2k6/revised1.jpg?t=1276267113