Equilibrium of Non-concurrent Force Systems

Translational Equilibrium: The sum of all external forces in every direction must be zero

$$\sum F_x = 0,\quad \sum F_y = 0,\quad \sum F_z = 0 \quad \text{(for 3D systems)}$$

Rotational Equilibrium: The sum of all moments about any point must also be zero

$$\sum M = 0$$

In engineering mechanics, a non-concurrent force system refers to a group of forces whose lines of action do not intersect at a single point. These forces may act on different parts of a rigid body and can include forces and moments (torques). To achieve equilibrium in such systems, a body must satisfy both conditions above.

General Resultant of Non-concurrent Force Systems:

$$R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2 + (\sum F_z)^2}$$

For Forces Lying on a 2D Plane Only

$$\tan(\theta) = \frac{\sum F_y}{\sum F_x}$$

Directional Cosines:

$\cos(\theta_x) = \frac{F_x}{F}$

$\cos(\theta_y) = \frac{F_y}{F}$

$\cos(\theta_z) = \frac{F_z}{F}$

The location of the resultant force is determined by Varignon's Theorem.

$$ R(d_x) = F_1(x_1) + F_2(x_2) + \cdots + F_n(x_n) $$ $$ R(d_y) = F_1(y_1) + F_2(y_2) + \cdots + F_n(y_n) $$

Problem 1:

a. Calculate the location of the resultant from point D.
b. Prove that the resultant force intersects point E (as indicated below)
c. Determine the distance d_x and d_y from point D to the resultant force and indicate the location relative to D.
d. Determine the distance from point B to the resultant force.
e. Resolve the system into a resultant-couple system.

See images:

Below is a scaled drawing of the figure (modeled using AutoCad). Solution

We can obtain the moment caused by the resultant force about point D so that we can transfer the resultant so that it intersects point D. The moment (couple) will be applied at point D for clarity. Solution