understand that, when there is no resultant force and no resultant torque, a system is in equilibrium

Published by Patrick Mutisya · 14 days ago

Equilibrium of Forces – Cambridge A-Level Physics 9702

Equilibrium of Forces

Objective: Understand that a system is in equilibrium when both the resultant force and the resultant torque are zero.

Key Concepts

  • Resultant (net) force – the vector sum of all forces acting on a body.

  • Resultant (net) torque – the algebraic sum of all torques about a chosen axis.

  • Static equilibrium – the state in which a body remains at rest or moves with constant velocity.

Conditions for Equilibrium

For a rigid body to be in equilibrium, the following two conditions must be satisfied simultaneously:

  1. The vector sum of all external forces is zero:

    \$\sum \vec{F}= \vec{0}\$

  2. The algebraic sum of all external torques about any axis is zero:

    \$\sum \tau = 0\$

Both conditions must hold; satisfying only one is insufficient.

Resultant Force

The resultant force is obtained by adding all individual forces vectorially:

\$\vec{F}{\text{res}} = \sum{i=1}^{n} \vec{F}_i\$

If \$\vec{F}_{\text{res}} = \vec{0}\$, the translational motion of the centre of mass is unchanged (Newton’s first law).

Resultant Torque

Torque about a point O is defined as \$\tau = \vec{r}\times\vec{F}\$, where \$\vec{r}\$ is the position vector from O to the line of action of the force. The resultant torque is:

\$\tau{\text{res}} = \sum{i=1}^{n} (\vec{r}i \times \vec{F}i)\$

When \$\tau_{\text{res}} = 0\$, the rotational motion about the chosen axis is unchanged.

Illustrative Example

Consider a uniform beam of length \$L\$ supported at its ends and carrying a weight \$W\$ at its centre.

For equilibrium:

  • Vertical forces: \$RA + RB - W = 0\$

  • Torques about point A: \$R_B L - W\frac{L}{2}=0\$

Solving gives \$RA = RB = \dfrac{W}{2}\$, satisfying both conditions.

Suggested diagram: Free‑body diagram of a beam in static equilibrium showing forces \$RA\$, \$RB\$, and weight \$W\$.

Problem‑Solving Strategy

StepActionWhat to Check
1Draw a clear free‑body diagram (FBD).All forces and points of application are shown.
2Choose a convenient axis for torque calculation.Eliminate as many unknown forces as possible.
3Write the equilibrium equations:

\$\sum \vec{F}=0\$

\$\sum \tau =0\$

Separate into horizontal, vertical, and rotational components.
4Solve the simultaneous equations for the unknown forces.Check that each solution satisfies both force and torque equations.
5Verify units and physical plausibility (e.g., no negative reaction forces unless indicated).Consistent with the system’s constraints.

Common Mistakes

  • Using the wrong sign for torques (clockwise vs. anticlockwise).

  • Neglecting forces that act at an angle; resolve them into components.

  • Choosing an axis that does not simplify the torque equation, leading to unnecessary unknowns.

  • Assuming equilibrium automatically means the body is rigid; deformation can introduce additional internal forces.

Practice Questions

  1. A uniform ladder of length \$4\,\$m and weight \$200\,\$N rests against a smooth wall, making an angle of \$30^\circ\$ with the ground. The foot of the ladder is \$1.5\,\$m from the wall. Determine the forces exerted by the ground on the ladder (horizontal and vertical components) assuming the ladder is in static equilibrium.

  2. A rectangular plate of mass \$5\,\$kg is supported by three vertical strings as shown (strings at the corners A, B, and C). The plate is horizontal and the distances between the strings are known. Find the tension in each string when the plate is in equilibrium.

  3. A torque wrench applies a force of \$30\,\$N at the end of a \$0.4\,\$m handle. What is the torque about the centre of the bolt? If the required tightening torque is \$12\,\$Nm, is the applied torque sufficient?