Exhaustive Analysis of Static Equilibrium and Force Vectors

The fundamental condition for the static equilibrium of a particle or a point-mass is that the vector sum of all external forces acting upon it must be exactly zero. This is represented by the resultant force formula: $F_R = 0$ .
In a two-dimensional Cartesian plane, this vector equation can be broken down into two independent scalar equations representing the sum of force components along the horizontal ( $x$ ) and vertical ( $y$ ) axes:
- $\sum F_x = 0$
- $\sum F_y = 0$
If the resultant force $F_R$ is not zero, the object will undergo acceleration according to Newton's Second Law ( $F = m \times a$ ). Therefore, in the context of the transcript, equilibrium implies the absence of net acceleration and a balanced state of all interacting vectors.

The system under analysis involves three distinct forces, designated as $F_1$ , $F_2$ , and $F_3$ .
Specific Relationship Between $F_1$ and $F_2$ :
- The transcript specifies a mathematical ratio between the magnitudes of the first two forces, where the second force is $20\%$ larger than the first: $F_2 = 1.2 \times F_1$ .
- The spatial orientation between these two forces is defined as being perpendicular, or at a right angle: $\text{Angle}(F_1, F_2) = 90^{\circ}$ .
By placing $F_1$ along the positive x-axis for the purpose of calculation, the forces can be expressed in vector component form:
- $\mathbf{F_1} = F_1 \, \mathbf{i} + 0 \, \mathbf{j}$
- $\mathbf{F_2} = 0 \, \mathbf{i} + (1.2 \times F_1) \, \mathbf{j}$

To find the required magnitude of the third force ( $F_3$ ) that will maintain equilibrium, one must first determine the resultant of the first two forces ( $F_1$ and $F_2$ ).
Using the Pythagorean theorem for perpendicular vectors, the magnitude of the resultant $R_{12}$ is:
- $R_{12} = \sqrt{F_1^2 + F_2^2}$
- Substituting the known value of $F_2$ : $R_{12} = \sqrt{F_1^2 + (1.2 \times F_1)^2}$
- $R_{12} = \sqrt{F_1^2 + 1.44 \times F_1^2}$
- $R_{12} = \sqrt{2.44 \times F_1^2}$
- $R_{12} \approx 1.562 \times F_1$
For the entire system to be in equilibrium ( $F_R = 0$ ), the third force $F_3$ must be equal in magnitude to this resultant but opposite in direction:
- $|F_3| = R_{12} = 1.562 \times F_1$

The angle $\theta$ refers to the orientation of the forces within the coordinate system required to cancel out the net force.
The angle of the combined resultant of $F_1$ and $F_2$ relative to $F_1$ (the x-axis) is calculated using the arctangent function:
- $\tan(\phi) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{F_2}{F_1}$
- $\tan(\phi) = \frac{1.2 \times F_1}{F_1} = 1.2$
- $\phi = \arctan(1.2) \approx 50.19^{\circ}$
Direction of Force $F_3$ : Since $F_3$ must directly oppose the resultant of $F_1$ and $F_2$ , its angle $\theta$ relative to force $F_1$ must be situated in the third quadrant (180 degrees away from the resultant angle):
- $\theta = 180^{\circ} + 50.19^{\circ} = 230.19^{\circ}$

The transcript highlights that for the condition $F_R = 0$ to be satisfied given $F_2 = 1.2 \times F_1$ at $90^{\circ}$ , the system must adhere to the following rigorous constants:
1. The third force $F_3$ must possess a magnitude exactly equivalent to $\sqrt{2.44}$ times the magnitude of $F_1$ .
2. The vector direction of $F_3$ must precisely bisect the plane such that it negates both the horizontal component provided by $F_1$ and the vertical component provided by $F_2$ .
3. Any deviation in the angle $\theta$ or the magnitude ratios would result in a non-zero net force, breaking the state of static equilibrium.