Two parallel forces `F_(1)` and `F_(2)` acting in the same direction are applied to a rigid body.
Porve that
(a) the magnitude of the resultant force is equal to the sum of magnitudes of the forces being added,
(b) the resultant force is parallel to the forces being added and acts in the same direction,
(c ) the resultant force passes through the centre of the parallel forces, i.e. through the point which divides the distance between the points of application of both forces into sections inversely proportional to the magnitude of the forces.

Explanation

Apply to the body two forces `T_(1)` and `T_(2)` equal in magnitude and opposite in direction (Fig.).

Adding up the forces `F_(1) and T_(1) and F_(2) and T_(2)`, respectively, we obtain two forces `R_(1) and R_(2)` intersecting in point B. Translate the forces `R_(1) and R_(2)` to this point and resolve them again into the former components. The forces `T_(1) and T_(2)` are in equilibrium, but the forces `F._(1)=F_(1), and F._(2)=F_(2)` act in the same direction and their resultant force `R=F._(1)+F._(2)=F_(1)+F_(2)`.
The position of point O, the centre of parallel forces, may be found from the similarity of triangles. From the condition `DeltaA_(1)OB ~DeltaF._(1)R._(1)B and DeltaA_(2)OB ~ DeltaF._(2)R._(2)B`
we obtain
`(l_(1))/(T) =(OB)/(F_(1)) and (l_(2))/(T)=(OB)/(F_(2))`
from which `l_(1)F_(1) = l_(2)F_(2) `, which was required to prove.