Turning effects of forces

1. Overview

The turning effect of a force, known as a moment, occurs when a force causes an object to rotate about a fixed point called a pivot or fulcrum. While translational dynamics focuses on the resultant force causing linear acceleration, rotational mechanics focuses on the positioning and orientation of forces. For an object to be in a state of static equilibrium, it must satisfy two conditions: the resultant force must be zero (no translation) and the resultant moment must be zero (no rotation). This topic establishes the mathematical framework for calculating these effects using the concepts of Centre of Gravity, Moments, and Couples.

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Key Definitions

• Centre of Gravity: The point at which the entire weight of an object is considered to act.
• Moment of a Force: The product of the force and the perpendicular distance from the pivot to the line of action of the force.
• Couple: A pair of forces of equal magnitude acting in opposite directions along parallel lines of action.
• Torque of a Couple: The product of one of the forces and the perpendicular distance between the lines of action of the two forces.
• Principle of Moments: For an object in rotational equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about that same point.

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Content

3.1 The Centre of Gravity (CG)

The weight of an object is not concentrated at a single atom but is distributed across its entire mass. However, for the purpose of solving mechanics problems, we treat the weight as a single vector acting from the Centre of Gravity.

• Uniform vs. Non-Uniform Objects:
  • In a uniform object, the mass is distributed evenly. The CG coincides with the geometric centre (centroid). For example, the CG of a uniform 1.0 m ruler is at the 50.0 cm mark.
  • In a non-uniform object, the CG shifts toward the denser or heavier end.
• CG Outside the Object: The CG does not have to be located within the physical material of the object. For a wedding ring or a hollow sphere, the CG is located at the empty geometric centre.
• Stability and Toppling:
  • An object is stable if its CG is low and it has a wide base.
  • The Toppling Rule: An object will topple if the vertical line acting downwards from its CG falls outside the edge of its base. At this point, the weight creates a resultant moment about the edge of the base, causing rotation.
• Experimental Determination (Plumb Line Method):
  1. Suspend the irregular lamina (flat shape) from a hole near its edge using a pin.
  2. Hang a plumb line (a string with a weight) from the same pin.
  3. Draw a line on the lamina following the path of the plumb line string.
  4. Repeat the process by suspending the lamina from a different hole.
  5. The point where the two lines intersect is the Centre of Gravity.

3.2 The Moment of a Force

The moment measures the "turning strength" of a force. It depends on both the magnitude of the force and how far from the pivot it is applied.

The Equation: $M = F \times d$ (This equation is not on the formula sheet and must be memorised)

Where:

• $M$ = Moment (N m)
• $F$ = Force applied (N)
• $d$ = Perpendicular distance from the pivot to the line of action of the force (m)

The "Line of Action" Concept: The line of action is an imaginary infinite line extending in both directions along the force vector. The distance $d$ used in the formula must always be the shortest distance (the 90° drop) from the pivot to this line.

Calculating Moments at an Angle: If a force $F$ is applied at an angle $\theta$ to a beam of length $L$:

1. Component Method: Resolve the force into a component perpendicular to the beam ($F \sin\theta$). The moment is $M = (F \sin\theta) \times L$.
2. Distance Method: Find the perpendicular distance from the pivot to the line of action ($d = L \sin\theta$). The moment is $M = F \times (L \sin\theta)$. Both methods result in the same formula: $M = FL \sin\theta$.

3.3 Couples

A couple is a specific arrangement of forces that produces rotation only.

Conditions for a Couple:

1. Two forces of equal magnitude.
2. Forces act in opposite directions.
3. Forces are parallel (they must not act along the same line).

Resultant Force of a Couple: Because the two forces are equal and opposite, the resultant force ($\sum F$) is zero. According to Newton’s Second Law ($F = ma$), if the resultant force is zero, there is no linear acceleration. Therefore, a couple cannot move an object from point A to point B; it can only make it spin.

3.4 Torque of a Couple

The turning effect of a couple is specifically called torque. Unlike the moment of a single force, which changes depending on where you place the pivot, the torque of a couple is constant regardless of the pivot's location.

The Equation: $T = F \times d$ (This equation is not on the formula sheet and must be memorised)

Where:

• $T$ = Torque of the couple (N m)
• $F$ = Magnitude of one of the forces (N)
• $d$ = Perpendicular separation between the two lines of action (m)

Derivation of Torque Independence: Consider a rod of length $d$ with a force $F$ at each end acting in opposite directions. Let's take moments about a point $P$ located a distance $x$ from one end.

• Moment of Force 1 = $F \times x$ (Clockwise)
• Moment of Force 2 = $F \times (d - x)$ (Clockwise)
• Total Torque $T = Fx + F(d - x) = Fx + Fd - Fx = Fd$ The variable $x$ cancels out, proving that the torque is the same about any point.

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4. Worked Examples

Worked Example 1 — Equilibrium of a Non-Uniform Beam

A non-uniform horizontal beam $AB$ of length 5.0 m is supported by a pivot at its centre. The beam is kept in equilibrium by a 120 N weight placed 0.5 m from end $A$ and a 40 N weight placed at end $B$. Calculate the weight of the beam and state where it acts.

Step 1: Identify the pivot and forces.

• Pivot is at 2.5 m (the centre).
• Force 1 ($W_1$): 120 N at 0.5 m from $A$. Distance from pivot = $2.5 - 0.5 = 2.0$ m (Anticlockwise).
• Force 2 ($W_2$): 40 N at 5.0 m from $A$. Distance from pivot = $5.0 - 2.5 = 2.5$ m (Clockwise).
• Weight of beam ($W_b$): Acts at the CG. Since the beam is non-uniform, we don't know the CG. However, the question asks for the weight. Correction: In this specific setup, if the beam is balanced at the centre, the CG must be on one side to balance the moments. Let's assume the CG is at distance $x$ from the pivot.

Step 2: Apply Principle of Moments. Sum of Clockwise Moments = Sum of Anticlockwise Moments $$(40 \text{ N} \times 2.5 \text{ m}) + (W_b \times x) = (120 \text{ N} \times 2.0 \text{ m})$$ $$100 + W_b x = 240$$ $$W_b x = 140 \text{ N m}$$ If the weight of the beam is known to be 70 N, then $x = 140 / 70 = 2.0$ m from the pivot.

Answer: The weight of the beam creates a moment of 140 N m to balance the system.

Worked Example 2 — Torque on a Circular Valve

A technician applies a couple to a circular steam valve of radius 20 cm by pulling with a force of 85 N on one side and pushing with 85 N on the opposite side. Calculate the torque.

Step 1: Identify the force and perpendicular separation.

• Force $F = 85$ N
• Radius $r = 20$ cm = 0.20 m
• The distance $d$ between the two forces is the diameter: $d = 2 \times 0.20 = 0.40$ m.

Step 2: Apply the torque equation. $$T = F \times d$$ $$T = 85 \text{ N} \times 0.40 \text{ m}$$ $$T = 34 \text{ N m}$$

Answer: 34 N m

Worked Example 3 — Forces at an Angle

A uniform sign of weight 150 N is supported by a horizontal 2.0 m rod of negligible mass. The rod is hinged to a wall at point $P$ and supported by a cable attached to the end of the rod at an angle of 35° to the horizontal. Calculate the tension $T$ in the cable.

Step 1: Identify moments about the hinge $P$.

• Weight of sign ($W$): 150 N acts at the end of the rod (2.0 m from $P$).
• Moment (Clockwise) = $150 \text{ N} \times 2.0 \text{ m} = 300 \text{ N m}$.
• Tension ($T$): Acts at 2.0 m from $P$ at an angle of 35°.
• Perpendicular component of Tension = $T \sin(35^\circ)$.
• Moment (Anticlockwise) = $(T \sin 35^\circ) \times 2.0 \text{ m}$.

Step 2: Equate moments for equilibrium. $$300 = T \sin(35^\circ) \times 2.0$$ $$300 = T \times 0.5736 \times 2.0$$ $$300 = 1.147 \times T$$ $$T = 300 / 1.147 = 261.5 \text{ N}$$

Answer: 260 N (to 2 s.f.)

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Key Equations

Quantity	Equation	Symbols	SI Units	Data Sheet?
Moment of a Force	$M = Fd$	$d$ = perp. distance to pivot	N m	No
Torque of a Couple	$T = Fd$	$d$ = perp. separation of forces	N m	No
Weight	$W = mg$	$g = 9.81 \text{ m s}^{-2}$	N	No
Equilibrium (Forces)	$\sum F = 0$	Sum of all vector forces	N	No
Equilibrium (Moments)	$\sum M = 0$	Clockwise = Anticlockwise	N m	No

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Common Mistakes to Avoid

• ❌ Wrong: Using the distance along the beam ($L$) instead of the perpendicular distance ($d$) when the force is at an angle.
  • ✅ Right: Always use $d = L \sin\theta$ or resolve the force into its perpendicular component.
• ❌ Wrong: Doubling the force when calculating the torque of a couple (e.g., $T = 2F \times d$).
  • ✅ Right: The formula $T = Fd$ already accounts for both forces; $F$ is the magnitude of one of the forces.
• ❌ Wrong: Forgetting to include the weight of a "uniform beam" in the calculation.
  • ✅ Right: If the word "uniform" appears, draw a weight vector $W$ acting at the exact midpoint of the beam.
• ❌ Wrong: Giving units for moments as Joules (J).
  • ✅ Right: While the dimensions are the same ($1 \text{ J} = 1 \text{ N m}$), energy is a scalar and moments are vectors. Always use N m for moments and torques.
• ❌ Wrong: Using mass (kg) instead of weight (N) in moment equations.
  • ✅ Right: Multiply mass by $g$ (9.81) before calculating the moment.

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Exam Tips

1. The "Pivot Choice" Strategy: In equilibrium problems, you can take moments about any point. Choose a point where an unknown force acts (like a hinge or a support). Since the distance $d$ to that force will be zero, its moment becomes zero, eliminating one variable from your equation.
2. Standard Form: For very large or small moments, use standard form (e.g., $4.5 \times 10^4$ N m).
3. Significant Figures: Always provide your final answer to the same number of significant figures as the least precise data point given in the question (usually 2 or 3 s.f.).
4. Directional Clarity: When solving complex problems, start by writing: "Taking moments clockwise about the pivot..." This shows the examiner your frame of reference.
5. Couple Identification: If asked to explain why a system is a couple, you must explicitly state: "The forces are equal in magnitude, opposite in direction, and parallel." Mentioning only two of these three will often lose the mark.
6. Read the "Negligible" Keyword: If a question says a rod has "negligible mass," do not include its weight in your moment diagram. If it doesn't say this, you must find or calculate its weight.