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Forces with direction and magnitude

  1. Jul 14, 2006 #1

    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


    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
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  3. Jul 15, 2006 #2
    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.
  4. Jul 15, 2006 #3


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    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.

  5. Jul 15, 2006 #4

    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
  6. Jul 15, 2006 #5


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    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!

  7. Jul 15, 2006 #6
    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
  8. Jul 17, 2006 #7
    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
  9. Jul 17, 2006 #8


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    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)
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