If Fr is the force along r and Ft that along t at right angles to r,
| Fr = X cos θ + Y sin θ = | M | 2 cos θ, |
| r³ |
(12)
| Ft = −X sin θ + Y cos θ = | M | sin θ. |
| r³ |
(13)
For the resultant force at P,
| F = √ (Fr² + Ft²) = | M | √3 cos² θ + 1. |
| r³ |
(14)
The direction of F is given by the following construction: Trisect OP at C, so that OC = OP/3; draw CD at right angles to OP, to cut the axis produced in D; then DP will be the direction of the force at P. For a point in the axis OX, θ = 0; therefore cos θ = 1, and the point D coincides with C; the magnitude of the force is, from (14),
Fx = 2M / r³,