Every force is a vector with magnitude (in newtons) and direction. In 2D:
$$\mathbf{F} = F\cos\theta\,\mathbf{i} + F\sin\theta\,\mathbf{j}$$
Resultant force: $\mathbf{R} = \sum\mathbf{F}_i$ (vector sum).
Equilibrium: $\mathbf{R} = \mathbf{0}$, i.e., $\sum F_x = 0$ and $\sum F_y = 0$.
Example 1: Three forces on a particle:
$\mathbf{F}_1 = 4\mathbf{i}$, $\mathbf{F}_2 = 3\mathbf{j}$, $\mathbf{F}_3 = -(4\mathbf{i}+3\mathbf{j})$.
Resultant $= \mathbf{0}$. The particle is in equilibrium. \checkmark
Example 2: A 5 kg block on a smooth incline at $30^\circ$. Find the normal force and the force needed to hold it stationary.
Weight: $\mathbf{W} = -5g\mathbf{j} = -49\mathbf{j}$ N (taking up as positive $y$).
Along incline (down positive): $W\sin30^\circ = 24.5$ N downward.
Normal: $N = W\cos30^\circ = 42.4$ N.
Holding force $= 24.5$ N up the incline.
When three concurrent forces are in equilibrium:
$$\frac{F_1}{\sin\alpha_1} = \frac{F_2}{\sin\alpha_2} = \frac{F_3}{\sin\alpha_3}$$
where $\alpha_i$ is the angle opposite to force $F_i$ (angle between the other two).
$$\mathbf{F} = m\mathbf{a} \Rightarrow \sum F_x = ma_x, \quad \sum F_y = ma_y$$
Example 3: A 2 kg particle is subject to $\mathbf{F}_1 = 6\mathbf{i}+2\mathbf{j}$ N and $\mathbf{F}_2 = -2\mathbf{i}+4\mathbf{j}$ N.
$\mathbf{F}_{\text{net}} = 4\mathbf{i}+6\mathbf{j}$ N.
$\mathbf{a} = \mathbf{F}/m = 2\mathbf{i}+3\mathbf{j}$ m/s$^2$. $|\mathbf{a}| = \sqrt{4+9} = \sqrt{13}$ m/s$^2$.
With constant force (constant acceleration), apply vector suvat:
$$\mathbf{v} = \mathbf{u} + \mathbf{a}t, \qquad \mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$$
Example 4: A boat ($m = 500$ kg) starts at rest. Force $\mathbf{F} = 1000\mathbf{i} + 200\mathbf{j}$ N.
$\mathbf{a} = (2\mathbf{i}+0.4\mathbf{j})$ m/s$^2$.
After 5 s: $\mathbf{v} = 5(2\mathbf{i}+0.4\mathbf{j}) = 10\mathbf{i}+2\mathbf{j}$ m/s. Speed $= \sqrt{104} \approx 10.2$ m/s.
Example 5: A mass $m = 3$ kg hangs from two strings making angles $30^\circ$ and $45^\circ$ with the vertical. Find tensions $T_1$, $T_2$.
Vertical: $T_1\cos30^\circ + T_2\cos45^\circ = 3g$
Horizontal: $T_1\sin30^\circ = T_2\sin45^\circ$
From horizontal: $T_1 = T_2\sqrt{2}$. Substitute into vertical equation to find $T_2$.
KEY TAKEAWAY: Forces are vectors; equilibrium means the vector sum is zero. Always resolve into components and set up a system of equations.
EXAM TIP: Label all forces (weight, normal, tension, friction) on a diagram before writing equations. Resolve in the most convenient directions (e.g., along and perpendicular to an incline).
COMMON MISTAKE: Taking moments when resolving forces is sufficient. For a particle (as opposed to a rigid body), moment equations are not needed.
