Static Equilibrium Problems

A static equilibrium problem asks you to find an unknown force, tension, or distance for an object that is at rest and staying at rest. Nothing is moving in a straight line and nothing is spinning, so every push, pull, and twist on the object must cancel out exactly. Beams held up by cables, ladders leaning on walls, and signs hanging from brackets are all classic static equilibrium examples.

Solving one of these problems always means combining two separate ideas: translational equilibrium, which says the forces balance, and rotational equilibrium, which says the torques balance. A static equilibrium problem is really just those two conditions applied at the same time to the same object.

For an object that is not accelerating and not rotating, both of the following must be true:

\( \sum F_x = 0 \) and \( \sum F_y = 0 \) (translational equilibrium)

\( \sum \tau = 0 \) (rotational equilibrium, torques taken about any chosen pivot)

Together these give you up to three independent equations, which is exactly enough to solve for up to three unknowns, such as a cable tension and the two components of a hinge reaction.

1. Sketch the object and draw every force acting on it, including weight, normal forces, tensions, and reaction forces at supports.
2. Pick a pivot point for torques. Choosing the point where an unknown force acts removes that force from the torque equation entirely, since its lever arm is zero.
3. Break any angled force into horizontal and vertical components before you start adding.
4. Write \( \sum F_x = 0 \) and \( \sum F_y = 0 \) using the components.
5. Write \( \sum \tau = 0 \) about your chosen pivot, using \( \tau = r F \sin\theta \) for each force.
6. Solve the equations together, usually starting with the torque equation since it often has only one unknown.

A uniform 4 m beam is attached to a wall by a hinge at one end and held horizontal by a cable from the other end, making a 60° angle with the beam. The beam itself weighs 100 N, acting at its centre, and a 50 N box hangs from the far end. Find the cable tension and the hinge reaction.

Taking torques about the hinge removes the unknown hinge force from the equation, since its lever arm at the hinge is zero:

\( \sum \tau = 0 : \; T\sin(60^\circ)(4) - (100)(2) - (50)(4) = 0 \)

\( T\sin(60^\circ)(4) = 400 \; \Rightarrow \; T\sin(60^\circ) = 100 \; \Rightarrow \; T = \dfrac{100}{\sin(60^\circ)} \approx 115.5 \) newtons.

Now apply force balance. Vertically, \( H_y + T\sin(60^\circ) = 150 \), so \( H_y = 150 - 100 = 50 \) newtons. Horizontally, \( H_x = T\cos(60^\circ) \approx 57.7 \) newtons.

Combining the components gives the size of the hinge reaction, \( H = \sqrt{H_x^2 + H_y^2} \approx \sqrt{57.7^2 + 50^2} \approx 76.4 \) newtons, directed at about 41 degrees above the beam.

Most mistakes in static equilibrium problems come from forgetting to resolve an angled force into components before summing, or from choosing a pivot point that leaves too many unknowns in the torque equation. It also helps to double check that \( \sum F_x = 0 \), \( \sum F_y = 0 \), and \( \sum \tau = 0 \) are all satisfied once you have numbers, since a mistake in any one equation usually shows up as an imbalance in another.

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