Coplanar Forces and Equilibrium

A force has magnitude and direction, so forces combine like vectors, not like numbers. Almost every question in this topic reduces to one of two things: resolving forces into perpendicular components, or setting the total in each direction to zero.

Take $g = 9.81\ \text{m s}^{-2}$ throughout, and note that the weight of a mass $m$ is $mg$ newtons, acting vertically downwards.

Forces as vectors

A force $F$ at angle $\theta$ to the horizontal has components

$$F_x = F\cos\theta, \qquad F_y = F\sin\theta$$

and conversely a force with components $(F_x, F_y)$ has

$$|F| = \sqrt{F_x^2 + F_y^2}, \qquad \theta = \tan^{-1}\!\left(\frac{F_y}{F_x}\right)$$

Finding a resultant

Resolve every force, add the components, then recombine.

Forces of 5 N along the positive $x$-axis and 8 N at $60°$ to it.

$$R_x = 5 + 8\cos 60° = 9, \qquad R_y = 8\sin 60° = 6.93$$

$$|R| = \sqrt{9^2 + 6.93^2} = 11.4\ \text{N}, \qquad \theta = \tan^{-1}\frac{6.93}{9} = 37.6°$$

The conditions for equilibrium

A particle is in equilibrium when the resultant force is zero:

$$\sum F_x = 0 \qquad\text{and}\qquad \sum F_y = 0$$

A body of weight 100 N hangs from two light strings making $40°$ and $55°$ with the horizontal, on opposite sides.

Horizontally: $T_1\cos 40° = T_2\cos 55°$

Vertically: $T_1\sin 40° + T_2\sin 55° = 100$

Solving the pair gives $T_1 = 57.6\ \text{N}$ and $T_2 = 76.9\ \text{N}$.

Lami's theorem and the triangle of forces

When exactly three forces hold a particle in equilibrium,

$$\frac{P}{\sin\alpha} = \frac{Q}{\sin\beta} = \frac{R}{\sin\gamma}$$

where each angle is the one opposite its force — that is, between the other two.

Equivalently, three forces in equilibrium form a closed triangle when drawn head to tail, so the sine and cosine rules apply directly to it.

Friction

Friction acts along the surface, opposing motion or attempted motion, with

$$F \le \mu R$$

where $R$ is the normal reaction. At the point of slipping the friction is limiting, and $F = \mu R$.

The inclined plane

Resolve along and perpendicular to the plane, not horizontally and vertically. For a body of mass $m$ on a plane at angle $\alpha$:

$$R = mg\cos\alpha \qquad \text{component of weight down the slope} = mg\sin\alpha$$

A block of mass 8 kg rests on a plane at $25°$ with $\mu = 0.3$. Does it slide?

$$mg\sin 25° = 8(9.81)(0.4226) = 33.2\ \text{N}$$ $$\mu R = 0.3 \times 8(9.81)\cos 25° = 21.3\ \text{N}$$

The pull down the slope exceeds the greatest friction available, so the block slides, with

$$a = g(\sin 25° - \mu\cos 25°) = 1.48\ \text{m s}^{-2}$$

Moments

For a rigid body, forces summing to zero is not enough: it could still rotate. The moment of a force about a point is

$$\text{moment} = F \times d$$

where $d$ is the perpendicular distance from the point to the line of action. A body in equilibrium satisfies

$$\sum F = 0 \qquad\text{and}\qquad \sum \text{moments} = 0 \ \text{about any point}$$

A uniform beam $AB$ of length 6 m and weight 400 N rests on supports at $A$ and $B$. A person of weight 600 N stands 2 m from $A$.

Taking moments about $A$ removes $R_A$ from the equation:

$$6R_B = 400(3) + 600(2) = 2400 \quad\Longrightarrow\quad R_B = 400\ \text{N}$$

$$R_A = 1000 - 400 = 600\ \text{N}$$