Magnitude and Point of Application Suppose there are vertical forces $P_1, P_2, \dots P_n$ that act at the points $A_1, A_2, \dots , A_n$ that are located in $xz$ plane. How can I find the magnitude and point of application of the resultant forces?
 A: I guess you're talking about rigid bodies.
Any system of forces can be reduced to an equivalent one composed of a force (applied to a point) and a torque which doesn't modify the equations for the rigid body. This reduction in not univoque, so the point of application is arbitrary, unless you want the torque to be null, wich means
$$\sum_{i=1}^n(\mathbf{r}_i - \mathbf{r}_Q) \times \mathbf{P}_i=0,$$
where $\mathbf{r}_Q$ is the position vector of the point of application $Q$ of the resultant force, $\mathbf{r}_i$ are the position vectors of the points of application $A_i$ of the $\mathbf{P}_i$ forces ($i=1,...,n$). So we got a vector equation for $\mathbf{r}_Q$.
In this case, since the forces are vertical and applied to the plane $xz$, $\mathbf{r}_P$ will have the following components:
$$ r_Q^x=\frac{\sum_{i=1}^n r^x_i P^z_i}{\sum_{i=1}^n P^z_i}, \quad r_Q^y=0, \quad r_Q^z=h \in \mathbb{R}. $$
The resultant force is the vector sum of the forces, so the magnitude will just be
$$R=\Biggl| \sum_{i=1}^n \mathbf{P}_i \Biggr|.$$
A: Suppose you pull on an object with your hand, if you lift up the left edge, it rotates to the right and vice versa. If you lift at both the left and the right edge, it is as though you lift it in the centre, i.e., it doesn't rotate.
But, if you pull twice as hard on the left side, will the object rotate? What does this say about where it appears you're pulling if you pretend you're using one hand?
The point of application is the weighted average of the positions.
$$s_x=\frac{\sum_{i=1}^n{||P_i||x_i}}{\sum_{i=1}^n{||P_i||}}$$
$$s_z=\frac{\sum_{i=1}^n{||P_i||z_i}}{\sum_{i=1}^n{||p_i||}}$$
Where $||Pi||$ is the length of the vector (magnitude of the force), i.e. its weight, and xi and zi are the vectors positions.
What do you think the total force is if you place all the forces at the centre, does this give the same total force if you distribute the forces?
