Welcome to the Principle of Moments
Have you ever wondered why door handles are placed as far away from the hinges as possible, or why it is so much easier to loosen a tight nut with a long spanner rather than a short one? The answer lies in the turning effect of forces, known in physics as moments.
In this chapter of AS 1 Forces, Energy and Electricity, we will break down how forces cause rotation, what makes objects balance perfectly, and how to solve equilibrium problems step-by-step. Don't worry if this seems a bit daunting at first—once you master a few straightforward rules, solving moment problems becomes very predictable and satisfying!
1. What is a Moment?
A force can cause an object to accelerate in a straight line, but if the object is fixed at a point (a pivot or fulcrum), the force can cause it to rotate. The measure of this turning effect is called the moment of a force.
Definition: The moment of a force about a point is defined as the product of the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force.
The mathematical formula is:
\(\text{Moment} = F \times d\)
Where:
- \(F\) = magnitude of the applied force, measured in newtons (\(\text{N}\))
- \(d\) = perpendicular distance from the pivot to the line of action of the force, measured in metres (\(\text{m}\))
Unit of a Moment: The SI unit is the newton-metre (\(\text{N m}\)). Note: Be careful never to write this as joules (\(\text{J}\)), even though the base units look similar. Joules are reserved for energy and work, while \(\text{N m}\) is used for moments!
Direction: A moment always acts in a specific rotational direction: either clockwise or anticlockwise.
What if the force is at an angle?
If the force is applied at an angle \(\theta\) to the beam or lever, only the component of the force perpendicular to the lever causes a turning effect. In that case:
\(\text{Moment} = F \times d \times \sin\theta\)
Alternatively, you can find the perpendicular distance from the pivot to the line of action of the force using trigonometry: \(d_{\perp} = d \sin\theta\).
Key Takeaway: To get the biggest turning effect for the smallest effort, apply a large force at the greatest possible perpendicular distance from the pivot.
2. Centre of Gravity and Centre of Mass
Every single particle in an extended object experiences a gravitational pull towards the Earth. Rather than calculating thousands of tiny downward forces, we can simplify our calculations by finding one single point where all the weight seems to act.
Definition: The centre of gravity of an object is the single point through which the entire weight of the object may be considered to act.
Uniform vs. Non-Uniform Objects
- A uniform object has an evenly distributed mass and a regular shape. For example, the centre of gravity of a uniform metre rule is located exactly at its geometric centre—the \(50.0\text{ cm}\) mark.
- A non-uniform object has an uneven mass distribution (like a baseball bat or a tapered rod), so its centre of gravity is shifted towards the heavier end.
Did you know? In a uniform gravitational field (such as near the Earth's surface), the centre of gravity is located at the exact same location as the centre of mass.
Key Takeaway: When drawing force diagrams for a uniform beam, always draw the downward weight force \(W = mg\) acting directly from the geometric midpoint of the beam.
3. Couples and Torque of a Couple
Sometimes forces act in pairs to produce pure rotation without moving the object linearly in any direction. Think about turning a car's steering wheel with both hands, or twisting the cap off a bottle.
Definition of a Couple: A couple consists of a pair of equal and opposite parallel forces whose lines of action do not coincide.
Because the two forces are equal in magnitude and opposite in direction, the resultant linear force is zero (\(F - F = 0\)). This means a couple produces rotation without any translational (linear) acceleration.
Torque of a Couple
The turning effect produced by a couple is called its torque.
Definition: The torque of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces.
The formula is:
\(\text{Torque of a couple} = F \times d\)
Where:
- \(F\) = magnitude of one of the forces (\(\text{N}\))
- \(d\) = perpendicular distance between the two parallel forces (\(\text{m}\))
Key Takeaway: Unlike a single moment, which depends on where you choose the pivot, the torque of a couple has the exact same value about any point in its plane.
4. The Principle of Moments
When an object is balanced and not rotating, it is in rotational equilibrium.
The Principle of Moments states:
For an object in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.
In mathematical shorthand:
\(\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}\)
Or written another way:
\(\sum M = 0\)
5. The Conditions for Complete Equilibrium
For any rigid body to be in total, complete equilibrium (meaning it does not accelerate linearly AND does not rotate), it must satisfy two fundamental conditions:
Condition 1: Translational Equilibrium (No Resultant Force)
The vector sum of all external forces acting on the body must be zero.
\(\sum F = 0\)
In 1D/2D problems, this simply means:
\(\sum \text{Upward Forces} = \sum \text{Downward Forces}\)
\(\sum \text{Leftward Forces} = \sum \text{Rightward Forces}\)
Condition 2: Rotational Equilibrium (No Resultant Moment)
The algebraic sum of the moments of all forces about any chosen pivot must be zero.
\(\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}\)
Quick Review: An object can have zero resultant force and still rotate (like a spinning wheel experiencing a couple), or zero resultant moment and still accelerate linearly. Both conditions must be satisfied for total equilibrium!
6. Step-by-Step Guide to Solving Moments Problems
Don't panic when faced with a complex beam or bridge question. Follow this foolproof 4-step method:
Step 1: Draw a clear Free-Body Diagram
Draw the object horizontally. Add all forces acting on the object with clear arrows. Don't forget:
- The weight of the beam acting downwards at its centre of gravity.
- Any added loads/masses acting downwards.
- Normal contact forces (reaction forces \(R_1\), \(R_2\)) or tension forces acting upwards at the supports.
Step 2: Choose a smart pivot point
You can take moments about any point you like. A clever trick is to place your pivot directly on top of the line of action of an unknown force you don't care about or want to eliminate. Since \(d = 0\), its moment will be zero (\(F \times 0 = 0\))!
Step 3: Apply the Principle of Moments
Identify which forces cause clockwise rotation and which cause anticlockwise rotation about your chosen pivot. Set them equal:
\(\sum (F_{\text{cw}} \times d_{\text{cw}}) = \sum (F_{\text{acw}} \times d_{\text{acw}})\)
Solve for the unknown variable.
Step 4: Use the force balance to find remaining unknowns
If you need to find a second unknown support force, use \(\sum F_{\text{up}} = \sum F_{\text{down}}\). This is usually much faster than taking moments a second time!
7. Worked Example
Problem: A uniform wooden plank of length \(4.0\text{ m}\) and mass \(20\text{ kg}\) rests horizontally on two supports, \(A\) and \(B\). Support \(A\) is placed at the left end (\(0\text{ m}\)), and support \(B\) is placed \(1.0\text{ m}\) from the right end (at \(3.0\text{ m}\)). A person of mass \(60\text{ kg}\) stands \(0.5\text{ m}\) from the left end. Take \(g = 9.81\text{ N kg}^{-1}\). Calculate the upward reaction force provided by support \(B\).
Solution:
1. Identify the forces acting on the plank:
- Weight of the person: \(W_p = mg = 60 \times 9.81 = 588.6\text{ N}\) acting downwards at \(0.5\text{ m}\) from \(A\).
- Weight of the uniform plank: \(W_{\text{plank}} = mg = 20 \times 9.81 = 196.2\text{ N}\) acting downwards at the midpoint (\(2.0\text{ m}\) from \(A\)).
- Reaction force at \(A\): \(R_A\) acting upwards at \(0\text{ m}\).
- Reaction force at \(B\): \(R_B\) acting upwards at \(3.0\text{ m}\) from \(A\).
2. Choose a pivot:
Take moments about Support A (this eliminates \(R_A\) because its distance is \(0\text{ m}\)).
3. Calculate Clockwise and Anticlockwise moments about A:
- Clockwise moments (forces pulling downwards):
Due to person: \(588.6\text{ N} \times 0.5\text{ m} = 294.3\text{ N m}\)
Due to plank's weight: \(196.2\text{ N} \times 2.0\text{ m} = 392.4\text{ N m}\)
Total clockwise moment = \(294.3 + 392.4 = 686.7\text{ N m}\)
- Anticlockwise moments (forces pushing upwards):
Due to \(R_B\): \(R_B \times 3.0\text{ m}\)
4. Apply the Principle of Moments:
\(\sum \text{Anticlockwise Moments} = \sum \text{Clockwise Moments}\)
\(R_B \times 3.0 = 686.7\)
\(R_B = \frac{686.7}{3.0} = 228.9\text{ N} \approx 229\text{ N}\)
Bonus: If asked to find \(R_A\), simply use the force condition:
\(\sum F_{\text{up}} = \sum F_{\text{down}}\)
\(R_A + R_B = W_p + W_{\text{plank}}\)
\(R_A + 228.9 = 588.6 + 196.2\)
\(R_A = 784.8 - 228.9 = 555.9\text{ N} \approx 556\text{ N}\)
8. Common Mistakes to Avoid
1. Forgetting the beam's own weight: Unless the question explicitly states that the beam is "light" or "of negligible mass", you must include the weight of the beam acting at its centre of gravity.
2. Measuring distances from the wrong place: Always measure the distance from the pivot to each force, NOT from the end of the ruler or plank (unless the end is the pivot!).
3. Forgetting unit conversions: Distances in questions are often given in centimetres (\(\text{cm}\)) or millimetres (\(\text{mm}\)). Convert them to metres (\(\text{m}\)) right at the start to ensure your moments are in \(\text{N m}\).
4. Confusing mass with weight: Remember that mass (in \(\text{kg}\)) must be multiplied by \(g = 9.81\text{ N kg}^{-1}\) (or the value given on your CCEA data sheet) to get the force of weight in newtons (\(\text{N}\)).
5. Forgetting that a couple's torque uses the full distance between forces: When calculating the torque of a couple, multiply one force by the total distance between the lines of action of both forces, not half the distance.
Summary Checklist
- \(\text{Moment} = F \times d_{\perp}\) (measured in \(\text{N m}\)).
- Centre of gravity: The point where all the weight of an object appears to act.
- Couple: Two equal, opposite, parallel forces with different lines of action (\(\text{Torque} = F \times d\)).
- Principle of Moments: \(\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}\) in equilibrium.
- Conditions for Complete Equilibrium: \(\sum F = 0\) and \(\sum M = 0\).