Resultant force

Bound vector

A force applied to a body has a point of application. The effect of the force is different for different points of application. For this reason a force is called a bound vector, which means that it is bound to its point of application.

Forces applied at the same point can be added together to obtain the same effect on the body. However, forces with different points of application cannot be added together and maintain the same effect on the body.

It is a simple matter to change the point of application of a force by introducing equal and opposite forces at two different points of application that produce a pure torque on the body. In this way, all of the forces acting on a body can be moved to the same point of application with associated torques.

A system of forces on a rigid body is combined by moving the forces to the same point of application and computing the associated torques. The sum of these forces and torques yields the resultant force-torque.

Associated torque

If a point R is selected as the point of application of the resultant force F of a system of n forces F_i then the associated torque T is determined from the formulas

${\displaystyle \mathbf{F}=\sum _{i=1}^n{\mathbf{F}}_i,}$

and

${\displaystyle \mathbf{T}=\sum _{i=1}^n({\mathbf{R}}_i-\mathbf{R})\times {\mathbf{F}}_i.}$

It is useful to note that the point of application R of the resultant force may be anywhere along the line of action of F without changing the value of the associated torque. To see this add the vector kF to the point of application R in the calculation of the associated torque,

${\displaystyle \mathbf{T}=\sum _{i=1}^n({\mathbf{R}}_i-(\mathbf{R}+k\mathbf{F}))\times {\mathbf{F}}_i.}$

The right side of this equation can be separated into the original formula for T plus the additional term including kF,

${\displaystyle \mathbf{T}=\sum _{i=1}^n({\mathbf{R}}_i-\mathbf{R})\times {\mathbf{F}}_i-\sum _{i=1}^{n}k\mathbf{F}\times {\mathbf{F}}_i=\sum _{i=1}^n({\mathbf{R}}_i-\mathbf{R})\times {\mathbf{F}}_i,}$

because the second term is zero. To see this notice that F is the sum of the vectors F_i which yields

${\displaystyle \sum _{i=1}^{n}k\mathbf{F}\times {\mathbf{F}}_i=k\mathbf{F}\times (\sum _{i=1}^n{\mathbf{F}}_i)=0,}$

thus the value of the associated torque is unchanged.

Torque-free resultant

It is useful to consider whether there is a point of application R such that the associated torque is zero. This point is defined by the property

${\displaystyle \mathbf{R}\times \mathbf{F}=\sum _{i=1}^n{\mathbf{R}}_i\times {\mathbf{F}}_i,}$

where F is resultant force and F_i form the system of forces.

Notice that this equation for R has a solution only if the sum of the individual torques on the right side yield a vector that is perpendicular to F. Thus, the condition that a system of forces has a torque-free resultant can be written as

${\displaystyle \mathbf{F}\cdot (\sum _{i=1}^n{\mathbf{R}}_i\times {\mathbf{F}}_i)=0.}$

If this condition is satisfied then there is a point of application for the resultant which results in a pure force. If this condition is not satisfied, then the system of forces includes a pure torque for every point of application.

Wrench

The forces and torques acting on a rigid body can be assembled into the pair of vectors called a wrench. If a system of forces and torques has a net resultant force F and a net resultant torque T, then the entire system can be replaced by a force F and an arbitrarily located couple that yields a torque of T. In general, if F and T are orthogonal, it is possible to derive a radial vector R such that ${\displaystyle \mathbf{R}\times \mathbf{F}=\mathbf{T}}$, meaning that the single force F, acting at displacement R, can replace the system. If the system is zero-force (torque only), it is termed a screw and is mathematically formulated as screw theory.

The resultant force and torque on a rigid body obtained from a system of forces F_i i=1,...,n, is simply the sum of the individual wrenches W_i, that is

${\displaystyle \mathsf{W}=\sum _{i=1}^n{\mathsf{W}}_i=\sum _{i=1}^n({\mathbf{F}}_i,{\mathbf{R}}_i\times {\mathbf{F}}_i).}$

Notice that the case of two equal but opposite forces F and -F acting at points A and B respectively, yields the resultant W=(F-F, A×F - B× F) = (0, (A-B)×F). This shows that wrenches of the form W=(0, T) can be interpreted as pure torques.