Forces with direction and magnitude

1. Jul 14, 2006

dopey9

Forces/couple!

particle of mass m kilograms is acted on by two forces F[1] and F[2] with magnitudes 3*sqr-root 5 newtons and sqr-root 5 newtons and directions parallel to the vectors i+2j and i-2j respectively.
The particle is initially at a position given by the vector 2i+j

i was told to calculate the cartesian components of F[1] and F[2] and hence calculate the total force F[1] + F[2], acting on the particle in component form

MY SOLUTION

The vector i+ 2j has length sqrt{1^2+ 2^2}= sqrt{5}[
A vector in that direction, with length is just 3 times that: 3i+ 6j. That's the first force vector.
Similarly, the vector i- 2j also has length root5 so that is the second force vector. The total force, then, is F1+ F2= (3i+ 6j)+ (i- 2j)= 4i+4j

from there i need to show the couple of the total force about the point with position vector i is zero.....so the total force is 4i+4j ............................

I would appreaciate a explanation to how to tackle this Q, so I can do it myself................but im assuming that it is a single force and a couple cannot be put in equilibrium by a single force so it must be zero? is that correct???????

Last edited: Jul 14, 2006
2. Jul 15, 2006

Mindscrape

I'm not really sure where you are getting your calculations from. They might be right, but you would have to maybe explain them a little more because I don't really understand where you are getting things such as the length is just 3 times the slope of the force. The way I would approach the problem is to take your two forces and resolve them into component form, so {i_1; j_1; i_2; j_2}, and then add them together.

3. Jul 15, 2006

nrqed

This all sounds correct to me
This is where I am losing you. You said earlier that the particle is at the position 2i +j. So why are we talking about position "i" here?
And by
couple" what do you mean? The torque? I am not sure what 'couple" means.

4. Jul 15, 2006

dopey9

..........

but this is what i have taken from a past paper...it says "the couple of the total force about the point with position vector i is zero" ....its confusing me too...thankz for the help tho

5. Jul 15, 2006

nrqed

Ok. By "past paper" do you mean it's a question taken from a book?!
Is there a drawing? Could you post the entire question as it is phrased there? Is this for an engineering class or a physics class? Have you covered torque?

I am asking all this simply because I wish I could be more helpful!

Patrick

6. Jul 15, 2006

dopey9

Dont worry about it ...iv left it out..thankz for help...but if u do want the full question ...the first one i posted...and the last one i jus posted ...and thats all i was given..there was no diagram thankz anyway

7. Jul 17, 2006

civil_dude

A couple is two equal and opposite forces times the perpindicular distance between them. Thus the body on which the couple acts will rotate, but not translate.

Last edited: Jul 18, 2006
8. Jul 17, 2006

Office_Shredder

Staff Emeritus
Ok, I'm ignoring everything in the solution, although some of it indicates you're answering another question

To break a vector into cartesian components, first you need to know the angle of the vector with respect to the x-axis. You would do this by taking arctan(y/x) of the direction it's in (y=j component, x=i component). Once you know the angle (call it a), you can find the x and y components by doing fcos(a) and fsin(a). Then add them together (keep in mind one force has positive and one negative y direction)