understand that, when there is no resultant force and no resultant torque, a system is in equilibrium
Equilibrium of Forces – Cambridge AS/A‑Level Physics (9702)
Objective
To understand and apply the two simultaneous conditions for static equilibrium of a rigid body:
- Translational equilibrium: the vector sum of all external forces is zero, \$\displaystyle\sum\vec F = \vec 0\$.
- Rotational equilibrium: the algebraic sum of all external torques about any axis is zero, \$\displaystyle\sum\tau = 0\$.
Action‑able Review of the “Equilibrium of Forces” Notes vs. Cambridge AS & A Level Physics (9702) Syllabus
| Syllabus Requirement (Topic 4) | How the Notes Measure Up | Gap / Issue | Suggested Improvement |
|---|---|---|---|
| State both equilibrium conditions (net force = 0, net torque = 0) and explain why they must hold simultaneously. | Both conditions are listed, but the link to Newton’s 1st law and the “any axis” requirement is brief. | Insufficient emphasis on “any axis” and on the physical meaning of each condition. | Expand the introductory paragraph, add a short box linking each condition to Newton’s laws, and stress “any axis”. |
| Define resultant (net) force and torque, give vector expressions, and relate them to centre of mass/centre of gravity. | Definitions and formulas are present, but the connection to the centre of mass is only mentioned for force. | Torque‑related discussion of CG is missing; also no explicit statement that the weight can be treated as a single force through the CG. | Combine force and torque sections, include a paragraph on how the CG simplifies torque calculations. |
| State and use the principle of moments (clockwise = anticlockwise) for equilibrium problems. | Principle of moments is covered, but the “any axis” caveat is omitted. | Students may think the principle only applies to a particular axis. | Add a note: “The principle holds for any axis you choose.” |
| Explain couples and give the torque formula \$ \tau = Fd \$. | Couple definition and formula are included. | Missing example showing that a couple produces a net torque even when \$\sum\vec F = 0\$. | Insert a short worked example of a door with two equal opposite forces. |
| Use vector‑triangle method for three (or more) coplanar forces. | Method is described with a diagram. | No step‑by‑step guide on constructing the triangle. | Provide a concise algorithm (draw, tip‑to‑tail, close the triangle). |
| Provide a systematic problem‑solving strategy and highlight common pitfalls. | Both are present. | Strategy table could be reordered for better flow; common mistakes lack examples. | Re‑order the table, add brief illustrative “what‑if” notes for each mistake. |
1. Fundamental Conditions for Equilibrium
Translational equilibrium – \$\displaystyle\sum\vec F = \vec 0\$
The centre of mass experiences no linear acceleration (Newton’s 1st law).
Rotational equilibrium – \$\displaystyle\sum\tau = 0\$ about any axis
The angular momentum of the body does not change (Newton’s 2nd law for rotation).
Both equations must be satisfied at the same time; satisfying only one leads to motion (translation or rotation).
2. Resultant (Net) Force
- Definition: vector sum of all external forces acting on the body
\$\vec F{\text{res}} = \sum{i=1}^{n}\vec F_i\$
- If \$\vec F_{\text{res}} = \vec 0\$, the centre of mass moves with constant velocity (including rest).
- Weight \$W=mg\$ can be replaced by a single vertical force acting through the centre of gravity (CG).
3. Resultant (Net) Torque
- Torque of a single force about point O:
\$\boldsymbol\tau = \vec r \times \vec F\$
where \$\vec r\$ joins O to any point on the line of action of \$\vec F\$.
- Resultant torque:
\$\tau{\text{res}} = \sum{i=1}^{n}(\vec ri \times \vec Fi)\$
- If \$\tau{\text{res}} = 0\$ about one axis, it is automatically zero about any parallel axis that passes through the same point; however, the equilibrium condition requires \$\tau{\text{res}}=0\$ about any axis you choose.
- When the weight is treated as a single force through the CG, its torque about any axis is simply \$W\,d\$, where \$d\$ is the perpendicular distance from the axis to the CG.
4. Principle of Moments (Turning Effects)
The principle of moments is a direct statement of \$\displaystyle\sum\tau = 0\$:
For a body in equilibrium, the sum of clockwise moments about any axis equals the sum of anticlockwise moments about the same axis.
For a single force:
\$\tau = F\,d\,\sin\theta\$
- \$F\$ – magnitude of the force.
- \$d\$ – distance from the axis to any point on the line of action.
- \$\theta\$ – angle between \$\vec r\$ and \$\vec F\$ (often \$\sin\theta = 1\$ when \$d\$ is taken as the perpendicular distance).
Couple
A pair of equal and opposite forces whose lines of action are parallel but not collinear. The net force of a couple is zero, yet it produces a non‑zero torque:
\$\tau_{\text{couple}} = F\,d\$
Example: Two forces \$F\$ act on a door, one at the hinge (parallel to the door) and one at the far edge (perpendicular). Their resultant torque is \$F\,d\$, where \$d\$ is the separation of the lines of action.
5. Centre of Gravity (CG)
- Definition: the point at which the total weight of a body may be considered to act.
- For a uniform, symmetric object the CG coincides with the geometric centre (e.g., \$L/2\$ for a uniform beam of length \$L\$).
- When analysing equilibrium, replace the distributed weight by a single force \$W=mg\$ acting through the CG – this simplifies both force and torque equations.
6. Vector‑Triangle Method for Coplanar Forces
If three (or more) forces lie in a single plane and the body is in equilibrium, they can be represented by a closed vector polygon.
- Draw each force as an arrow with length proportional to its magnitude and direction as given.
- Place the arrows tip‑to‑tail in any order.
- If the body is in equilibrium, the final tip meets the starting point, forming a closed triangle (or polygon).
- The missing side of the triangle gives the magnitude and direction of the unknown force.
7. Sign Conventions for Torques
| Direction | Symbol | Typical Use in Cambridge Exams |
|---|---|---|
| Clockwise | − (negative) | Most textbook examples adopt this convention. |
| Anticlockwise | + (positive) | Often taken as positive. |
Choose a convention at the start of a problem and apply it consistently. The equilibrium condition is satisfied when the algebraic sum (including signs) is zero.
8. Illustrative Example – Uniform Beam on Two Supports
Consider a uniform beam of length \$L\$ supported at its ends \$A\$ and \$B\$, with a weight \$W\$ acting at its centre (the CG).
Step 1 – Force equilibrium (vertical direction):
\$RA + RB - W = 0 \qquad (1)\$
Step 2 – Choose an axis through \$A\$ to eliminate \$R_A\$ from the torque equation:
\$\sum\tauA = 0 \;\Rightarrow\; RB\,L - W\frac{L}{2}=0 \qquad (2)\$
From (2) \$RB = \dfrac{W}{2}\$; substituting into (1) gives \$RA = \dfrac{W}{2}\$.
Both \$\displaystyle\sum\vec F = 0\$ and \$\displaystyle\sum\tau = 0\$ are satisfied – the beam is in static equilibrium.
9. Systematic Problem‑Solving Strategy
| Step | Action | Key Check |
|---|---|---|
| 1 | Draw a clear free‑body diagram (FBD). | All external forces, their magnitudes (if known), directions, and points of application are shown. |
| 2 | Identify the centre of gravity and replace the weight by \$W=mg\$ acting through it. | Weight is a single vertical force through the CG. |
| 3 | Choose a convenient axis for torque calculations. | Prefer an axis that passes through one or more unknown reactions to eliminate them from the torque equation. |
| 4 | Write the equilibrium equations: \$\displaystyle\sum\vec F = 0\$ (separate into horizontal & vertical components) \$\displaystyle\sum\tau = 0\$ (about the chosen axis). | Apply the sign convention consistently. |
| 5 | Solve the simultaneous equations for the unknown forces. | Check that each solution satisfies both the force and torque equations. |
| 6 | Verify units, physical plausibility and that no reaction is contradictory (e.g., a “negative” support force unless tension is specified). | Values should be realistic for the geometry and constraints. |
10. Common Mistakes & How to Avoid Them
- Incorrect torque sign: Decide at the start whether clockwise is positive or negative and stick to it. What‑if: Re‑evaluate the sign if the final sum is not zero.
- Omitting force components: Resolve any angled force into horizontal and vertical components before writing \$\sum\vec F = 0\$. What‑if: A missing vertical component will give an impossible reaction force.
- Poor choice of axis: Selecting an axis that does not pass through an unknown reaction often leaves extra variables in the torque equation. What‑if: Try a different axis to simplify.
- Assuming rigidity automatically: Deformation introduces internal forces; the equilibrium conditions apply only to the rigid‑body approximation. What‑if: Check whether the body can be considered rigid for the problem’s scale.
- Forgetting the centre of gravity: Treat the weight as a single force through the CG, not at an arbitrary point. What‑if: Re‑locate the weight to the CG and recompute torques.
- Neglecting the vector‑triangle method: When three coplanar forces are in equilibrium, they must form a closed triangle; using this can simplify finding an unknown force. What‑if: Sketch the triangle; the missing side gives the unknown.
11. Practice Questions
Ladder against a smooth wall
A uniform ladder 4 m long and weighing 200 N leans against a smooth vertical wall, making an angle of \$30^\circ\$ with the ground. The foot of the ladder is 1.5 m from the wall. Determine the horizontal and vertical components of the force exerted by the ground on the ladder, assuming static equilibrium.
Three‑string support for a rectangular plate
A rectangular plate of mass \$5\,\$kg is held horizontally by three vertical strings attached at corners \$A\$, \$B\$, and \$C\$ (as shown). The distances between the strings are known. Find the tension in each string when the plate is in equilibrium.
Torque wrench problem
A torque wrench applies a force of \$30\,\$N at the end of a \$0.40\,\$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?
Vector‑triangle application
A sign is suspended by three tension members as shown. Two tensions are known: \$T1 = 120\,\$N acting at \$30^\circ\$ above the horizontal to the left, and \$T2 = 150\,\$N acting vertically upwards. Determine the magnitude and direction of the third tension \$T_3\$ using the vector‑triangle method.
Couple on a door
Two forces of magnitude \$F\$ act on a door 1.2 m wide. One force is applied at the edge, perpendicular to the plane of the door; the other is applied 0.30 m from the hinge, parallel to the door surface, producing a couple. If the door is in rotational equilibrium, find the required magnitude of \$F\$ when the perpendicular force is \$40\,\$N.
Summary
- Static equilibrium requires both \$\displaystyle\sum\vec F = 0\$ (no net translation) and \$\displaystyle\sum\tau = 0\$ (no net rotation) about any axis.
- The principle of moments (clockwise = anticlockwise) is a direct consequence of \$\displaystyle\sum\tau = 0\$.
- Replace the distributed weight by a single force \$W=mg\$ acting through the centre of gravity to simplify calculations.
- For three or more coplanar forces, the vector‑triangle method provides a quick visual way to find unknown forces.
- Adopt a consistent sign convention for torques, draw a clear FBD, choose a convenient axis, and always verify both equilibrium equations after solving.