# Forces with direction and magnitude

1. Jul 13, 2006

### dopey9

A 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

i got F[1] as:
x=3*(root5)*cos(theta)
y= 3*(root5)*sin(theta)

i got F[2] as:
x=(root5)*cos(theta)
y= (root5)*sin(theta)

and then F[1] + F[2] as:
F[x] = 4*(root5)*cos(theta)
F[y] = 4*(root5)*sin(theta)

i was just wondering if i have done it right ???

from there i need to show the couple of the total force about the point with position vector i is zero..................does any understand this part or know how to do it ??????

2. Jul 13, 2006

### HallsofIvy

But you don't know theta, do you? So those formulas a pretty much useless. Yes, you could calculate theta from the vectors they give you but that would be like going east around the world to get to a destination 1 mile west of you!

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

For the second question I recommend that you first look up the definition of "couple of a force about a point".

3. Jul 14, 2006