Forces and Equilibrium: Vectors, Resultants, Equilibrium of a Particle

1. Vectors and Vector Addition

In mechanics, forces are represented by vectors. A vector has both magnitude and direction. We use unit vectors along the x, y, and z axes, denoted by i, j, and k respectively. A force vector F can be expressed in component form as F = $F_x \mathbf{i} + F_y \mathbf{j} + F_z \mathbf{k}$.

Vector Addition

To add two vectors, we simply add their corresponding components. For example, if A = $A_x \mathbf{i} + A_y \mathbf{j}$ and B = $B_x \mathbf{i} + B_y \mathbf{j}$, then the resultant vector R = A + B is:

R = ($A_x + B_x$)i + ($A_y + B_y$)j

This principle extends to addition in 3D space.

Resultant Vector

The resultant of a system of forces is the single force that produces the same translational motion as the system of forces. It is found by vector addition of all the individual force vectors.

2. Equilibrium of a Particle

A particle is in equilibrium if the net force acting on it is zero. This means the vector sum of all forces acting on the particle is equal to zero. Equilibrium can be static (at rest) or dynamic (moving with constant velocity).

For a particle in equilibrium, the following conditions must be met:

• The sum of all forces acting on the particle is zero: $\sum \mathbf{F} = \mathbf{0}$
• The sum of the moments about any point is zero: $\sum \boldsymbol{M} = \mathbf{0}$

Where F represents the force vector and M represents the moment vector.

3. Moments (Torque)

A moment (or torque) is a measure of the tendency of a force to cause rotation about a point. It depends on the magnitude of the force, the perpendicular distance from the line of action of the force to the pivot point, and the angle between the force and the perpendicular distance.

The magnitude of the moment is given by:

$$M = rF \sin \theta$$

where:

• M is the magnitude of the moment
• r is the perpendicular distance from the line of action of the force to the pivot point
• F is the magnitude of the force
• θ is the angle between the force and the perpendicular distance

The direction of the moment is perpendicular to the plane containing the force and the line of action of the moment arm, determined by the right-hand rule.

4. Equilibrium of a Particle: Applying Equilibrium Conditions

To analyze the equilibrium of a particle, we apply the two equilibrium conditions:

1. Sum of forces = 0: We resolve all forces into their x, y, and z components and set each component equal to zero.
2. Sum of moments = 0: We choose a convenient point about which to calculate the moments. We then calculate the moment of each force about that point and set the sum of these moments equal to zero.

By solving these two equations simultaneously, we can determine the unknown forces acting on the particle.

Force	Magnitude (F)	Direction	Moment (M)
Force 1	$F_1$	$\mathbf{F}_1$	$r_1 F_1 \sin \theta_1$
Force 2	$F_2$	$\mathbf{F}_2$	$r_2 F_2 \sin \theta_2$
...	...	...	...

5. Example Problem

A particle of mass m is in equilibrium. Two forces act on it: a horizontal force of 10 N and a vertical force of 20 N. The horizontal force acts at a distance of 2 m from a pivot point, and the vertical force acts at a distance of 3 m from the same pivot point. Determine the magnitude and direction of the horizontal force.

Solution:

1. Sum of forces = 0: Let the unknown horizontal force be $F_x$. The horizontal component of the known force is 10 N. So, $F_x + 10 \text{ N} = 0$, which gives $F_x = -10 \text{ N}$.
2. Sum of moments = 0: The moment due to the 10 N force is $2 \text{ m} \times 10 \text{ N} \times \sin(90^\circ) = 20 \text{ Nm}$. The moment due to the 20 N force is $3 \text{ m} \times 20 \text{ N} \times \sin(90^\circ) = 60 \text{ Nm}$. The net moment is $60 \text{ Nm} - 20 \text{ Nm} = 40 \text{ Nm}$. Since the particle is in equilibrium, the net moment must be zero. This implies that the moment due to the unknown force must be $-40 \text{ Nm}$. Therefore, $F_x \times 2 \text{ m} \times \sin(90^\circ) = -40 \text{ Nm}$, which gives $F_x = -10 \text{ N}$.

The magnitude of the horizontal force is 10 N, and it acts to the left.