force systems

Force Systems

Force Systems is a fundamental topic in Statics of Rigid Bodies. Understanding how forces interact, combine, and affect bodies is crucial for structural analysis and design.

Relevance to Philippine Standards: The concepts discussed here are foundational for understanding the provisions of the National Structural Code of the Philippines (NSCP) 2015, Volume 1, particularly in the analysis of forces and structural response. Section 203 of the NSCP details the combination of loads, which relies heavily on the principles of force resultants and components.

Scalars and Vectors

In mechanics, physical quantities are classified as either scalars or vectors.

• Scalars have magnitude only (e.g., mass, volume, time).
• Vectors have both magnitude and direction (e.g., force, velocity, displacement).

A force vector $\mathbf{F}$ is characterized by its magnitude, point of application, and line of action.

Rectangular Components (2D and 3D)

Two-Dimensional Force Systems

A force $\mathbf{F}$ acting in the $xy$-plane can be resolved into two rectangular components: $F_x = F \cos \theta$ $F_y = F \sin \theta$ where $\theta$ is the angle the force makes with the positive $x$-axis.

Conversely, if the components are known, the magnitude and direction of the force can be found: $F = \sqrt{F_x^2 + F_y^2}$ $\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$

Three-Dimensional Force Systems

In 3D space, a force is resolved into $x, y, z$ components: $\mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} + F_z \mathbf{k}$ The magnitude is given by: $F = \sqrt{F_x^2 + F_y^2 + F_z^2}$ The direction is defined by the direction cosines: $\cos \theta_x = \frac{F_x}{F}, \quad \cos \theta_y = \frac{F_y}{F}, \quad \cos \theta_z = \frac{F_z}{F}$

Resultants of Force Systems

The resultant of a system of forces is the single force that produces the same effect as the original forces combined.

For a system of concurrent forces (forces intersecting at a common point):

1. Sum the components along each axis: $R_x = \Sigma F_x$ $R_y = \Sigma F_y$
2. Calculate the resultant magnitude: $R = \sqrt{R_x^2 + R_y^2}$
3. Calculate the direction: $\theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

Example 1: Resultant of Concurrent Forces

Problem: Three forces act on a bracket: $F_1 = 100$ N at $30^\circ$, $F_2 = 150$ N at $120^\circ$, and $F_3 = 200$ N at $270^\circ$ (measured counter-clockwise from the positive x-axis). Determine the magnitude and direction of the resultant.

Solution:

1. Resolve forces into x and y components:

   • $F_{1x} = 100 \cos 30^\circ = 86.6$ N
   • $F_{1y} = 100 \sin 30^\circ = 50.0$ N
   • $F_{2x} = 150 \cos 120^\circ = -75.0$ N
   • $F_{2y} = 150 \sin 120^\circ = 129.9$ N
   • $F_{3x} = 200 \cos 270^\circ = 0$ N
   • $F_{3y} = 200 \sin 270^\circ = -200.0$ N

2. Sum the components: $R_x = 86.6 - 75.0 + 0 = 11.6 \text{ N}$ $R_y = 50.0 + 129.9 - 200.0 = -20.1 \text{ N}$

3. Calculate Magnitude: $R = \sqrt{(11.6)^2 + (-20.1)^2} = \sqrt{134.56 + 404.01} = \sqrt{538.57} \approx 23.2 \text{ N}$

4. Calculate Direction: $\theta = \tan^{-1}\left(\frac{-20.1}{11.6}\right) \approx -60.0^\circ$ Since $R_x > 0$ and $R_y < 0$, the resultant is in the 4th quadrant, $60.0^\circ$ below the positive x-axis.

Moment of a Force

The moment of a force about a point (or axis) is a measure of the tendency of the force to rotate a body about that point (or axis). $M_O = F d$ where $d$ is the perpendicular distance (moment arm) from the point of rotation $O$ to the line of action of the force.

Varignon's Theorem: The moment of a force about any point is equal to the sum of the moments of its components about the same point. $M_O = F_x y + F_y x$

Example 2: Calculation of Moment

Problem: A force of 500 N acts at the end of a 3m horizontal beam. The force makes an angle of $60^\circ$ with the horizontal. Calculate the moment of this force about the pivot point at the other end of the beam.

Solution: Using Varignon's Theorem is often easier than finding the perpendicular distance.

1. Resolve the force:

   • $F_x = 500 \cos 60^\circ = 250$ N (horizontal)
   • $F_y = 500 \sin 60^\circ = 433$ N (vertical)

2. Calculate moments of components:

   • The line of action of $F_x$ passes through the pivot (if the beam is thin), so its moment arm is 0. Moment = 0.
   • The moment arm for $F_y$ is the length of the beam, $d = 3$ m.
   • $M = F_y \times d = 433 \text{ N} \times 3 \text{ m} = 1299 \text{ N}\cdot\text{m}$

Result: The moment is $1299 \text{ N}\cdot\text{m}$ (Counter-clockwise if force is upwards).

Equivalent Force-Couple Systems

A force acting at a point $A$ can be moved to a different point $B$ if a couple is added. The couple moment is equal to the moment of the original force about point $B$. $M = F d$ where $d$ is the perpendicular distance between the lines of action of the force at $A$ and the new position at $B$.

This principle allows us to reduce a general system of forces (forces that are neither concurrent nor parallel) into a single resultant force acting at a specific point and a resultant couple moment.

Example 3: Reducing a Force System

Problem: A beam is subjected to a vertical force $F = 400$ N at $x=2$m and a couple moment $M = 200$ N$\cdot$m at $x=4$m. Replace this system with an equivalent single force.

Solution:

1. Resultant Force: Since there is only one force, $R = 400$ N (downward).
2. Resultant Moment about Origin (x=0): $M_R = \Sigma M_O = -(400 \text{ N})(2 \text{ m}) - 200 \text{ N}\cdot\text{m}$ (Assuming clockwise is negative) $M_R = -800 - 200 = -1000 \text{ N}\cdot\text{m}$
3. Location of Resultant Force: To place the single resultant force $R$ at a distance $d$ such that it produces the same moment: $-(R)(d) = M_R$ $-(400)(d) = -1000$ $d = \frac{1000}{400} = 2.5 \text{ m}$

Result: The equivalent system is a single 400 N downward force acting at $x = 2.5$ m.