1. ## 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???????

2. Couple is given by :$\displaystyle r X F$, where r is a vector from the position of particle to the point of application
and F is the force vector.
Couple can be zero (if any of following conditions arise):
1)magnitude of r is zero
2)magnitude of F is zero
3)F and r are parallel or antiparallael

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