- #1

- 91

- 0

## Homework Statement

Force F= (-6.93 N) + (7.85 N) acts on a particle with position vector r = (4.66 m) + (1.36 m). What are (a) the magnitude of the torque on the particle about the origin and (b) the angle between the directions of and ?

## Homework Equations

## The Attempt at a Solution

So I find (a) to be 46.0058 N*m which is correct and for (b) I use t/f*r

So 46.0058/sqrt(6.93^2+7.85^2)*sqrt(4.66^2+1.36^2) = .9050610046

sin-1(of that) = 64.8315006 = angle. But this is incorrect?