# Static of a Material Point

For a body to be in rectilinear motion with constant velocity or at rest, the sum of the forces acting on it must be zero.

The construction of a building must be planned in such a way that the set of forces (weight, normal…, among others) that act on it must have a null value as a resultant force; if not, the building could collapse.

A material point subjected to the action of various forces will be in equilibrium if the sum of these forces is zero.

we have a material point P under the action of four forces (F1, F2 and F3 and F4).

Decomposing the vectors will make it easier to obtain the resulting vector.

Therefore, we have in the x direction the vectors: F1x, F2x and F3x.

Such that: F1x = F1 – F2x = F2.cos45° – F3x = F3.cos30°

And in the y direction, we have the vectors: F2y, F3y and F4.

Such that: F2y = F2.sin45° – F3y = F3.sin30° – F4 = F4

As the material point is in equilibrium, we have that Fr = 0.

Then: Frx = F1x + F2x – F3x = 0

F1 + F2.cos45 ° – F3.cos30° = 0

F1 + F2.(√2)/2 – F3.(√3)/2 = 0 in the x direction – equation 1

Fry = F2y + F3y – F4 = 0

F1 + F2.sen45 – F3.cos30° = 0

F1 + F2..(√2)/2 – F3.(1/2) = 0 in the y direction – equation 2

We then have the sum of the forces in the x direction and in the y direction, by which we arrive at equations 1 and 2.

Remembering that in this circumstance the applied forces were reduced to the two-dimensional plane (Ox – Oy), however they may be in a three-dimensional plane (Ox – Or – Oz).