Physics of Torque and Equilibrium Study Notes

Learning Objectives

At the end of this study guide, you should be able to:

  • Define center of gravity.

  • Determine the center of gravity of a uniform symmetry shape.

  • Define torque.

  • Solve problems regarding torque.

  • Explain conditions of equilibrium for rigid bodies.

  • Solve problems regarding the conditions of equilibrium for rigid bodies.

Centre of Gravity (CoG)

  • Definition: The centre of gravity of a body is defined as that point at which the weight of the body acts.

  • Symmetry Rule: The centre of gravity of a symmetrical body is located along the axis of symmetry.

  • Alternative Terminology: Centre of gravity is also sometimes referred to as the centre of mass.

  • Conceptual Interpretation:

    • It is the point at which the whole weight of the body can be taken as acting through.

    • It is the point from which the object will balance.

  • CoG for Regular and Symmetrical Shapes:

    • Uniform Rod: The centre of gravity is at the middle point of the rod, where the weight of the uniform rod is average.

    • Circular Disc: The centre of gravity is located at the geometric centre of the disc.

    • Cylinder: The centre of gravity is at the midpoint of its geometric axis.

    • Parallelogram Lamina: The centre of gravity is at the point of intersection of the two diagonals (e.g., intersection of lines AC and BD).

    • Cuboid: The centre of gravity is at the geometric centre where the diagonals meet.

    • Triangle: The centre of gravity (and geometrical centre) is found by drawing a line from each corner to the opposite line's midpoint (the intersection of medians).

  • External Centre of Gravity:

    • For some objects, the CoG is not located on the physical material of the actual object.

    • Examples:

      • A stool.

      • A ring (CoG is in the empty center).

      • Two uniform perpendicular planks forming an L-shape.

Stability

  • Definition of Stability: Stability is directly affected by the position of the Centre of Gravity (CoG). An object becomes unstable when its CoG falls outside of the object’s base.

  • Methods to Increase Stability:

    1. Wider Base: Increasing the surface area of the base support.

    2. Lower Centre of Gravity: Placing the majority of the weight closer to the ground.

  • Real-World Examples of Stability:

    • The Traffic Cone: Highly stable due to its low centre of gravity and exceptionally wide base.

    • Modern Racing Car: Designed for maximum stability at high speeds using a wide wheel base and a very low centre of gravity.

Moment of Force (Torque)

  • Definition: The moment or torque of a force is a measure of the tendency of the force to rotate the body upon which it acts about an axis.

  • Fundamental Formula:     Moment=F×d\text{Moment} = F \times d

    • Where FF is the force applied.

    • Where dd is the perpendicular distance from the pivot point (axis of rotation) to the line of action of the force.

  • Torque (τ\tau) vs. Force (FF):

    • Linear forces cause accelerations based on Newton's Second Law: ∑F=m×a\sum F = m \times a.

    • Torques cause angular accelerations: ∑τext,o=Io×αo\sum \tau_{\text{ext,o}} = I_o \times \alpha_o

    • (Note: IoI_o is the Moment of Inertia and αo\alpha_o is the Angular Acceleration).

  • Detailed Torque Equations:

    • τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

    • Magnitude of torque: τ=r×F×sin⁡(θ)\tau = r \times F \times \sin(\theta)

    • Where rr is the length of the position vector (distance from pivot) and θ\theta is the angle between the position vector and the force vector.

  • Units: The SI unit for torque is newton-metre (NmNm). While this is dimensionally the same as the units for energy (Joules), torque and energy are not equivalent concepts.

  • Lever Arm (Moment Arm):

    • The lever arm (dd) is the shortest (perpendicular) distance from the axis of rotation to a line drawn along the direction of the force.

    • Calculation: d=r×sin⁡(θ)d = r \times \sin(\theta).

    • General Torque calculation using lever arm: τ=F×d\tau = F \times d.

  • Direction: The direction of the torque can be determined using the right-hand rule. For example, if a force causes a counter-clockwise rotation, the torque vector points out of the screen/page.

Principle of Moments and Equilibrium

  • Rotational Equilibrium: A body is in rotational equilibrium when it has no tendency to change its speed of rotation.

  • The Principle of Moments: For a body to be in rotational equilibrium, the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about that same point.

  • Conditions for Static Equilibrium of Rigid Bodies:

    1. Translational Equilibrium: The sum of forces must be zero (The sum of forces downwards = the sum of forces upwards): ∑F=0\sum F = 0.

    2. Rotational Equilibrium: The sum of clockwise moments = the sum of anticlockwise moments: ∑M=0\sum M = 0.

  • Simply-Supported Beams:

    • Beam: A structure loaded by force acting transversely (sideways) to its length, causing the beam to bend.

    • Simply-supported beam: The simplest arrangement where the beam is supported at each end, and loads are distributed along its length.

Worked Examples: Calculating Moments/Torque

  • Calculation with an Angle (Bolt and Wrench):

    • Scenario: A force of 20 N20\,N applied 24 cm24\,cm from a bolt at an angle of 30∘30^{\circ}.

    • Formula: M=F×l×cos⁡(θ)M = F \times l \times \cos(\theta)

    • Calculation: 20 N×0.24 m×cos⁡(30∘)=4.2 Nm20\,N \times 0.24\,m \times \cos(30^{\circ}) = 4.2\,Nm

  • Container on a Crane:

    • Scenario: A crane with arm length 20 m20\,m inclined at 30∘30^{\circ} with the vertical carries a container of mass 2000 kg2000\,kg.

    • Calculation: τ=r×(m×g)×sin⁡(30∘)\tau = r \times (m \times g) \times \sin(30^{\circ})

    • τ=20 m×(2000 kg×9.81 m/s2)×0.5=196,200 Nm\tau = 20\,m \times (2000\,kg \times 9.81\,m/s^{2}) \times 0.5 = 196,200\,Nm

  • Salvaging a Ship:

    • Scenario: A force of 500 kN500\,kN applied at point A (100 m100\,m from contact point) at an angle. The relevant angle for calculation is 100∘100^{\circ}.

    • Calculated Torque: τ=100 m×500,000 N×sin⁡(100∘)=49,240,388 Nm≈49.24 MNm\tau = 100\,m \times 500,000\,N \times \sin(100^{\circ}) = 49,240,388\,Nm \approx 49.24\,MNm

Worked Examples: Equilibrium and Reaction Forces

  • Balancing a Light Rod (Single Pivot):

    • Weights: 1.2 N1.2\,N at 0.25 m0.25\,m and a force FF at 0.35 m0.35\,m on one side; 2.5 N2.5\,N at 0.4 m0.4\,m on the other side.

    • Equation: (1.2 N×0.25 m)+F(0.35 m)=2.5 N×0.4 m(1.2\,N \times 0.25\,m) + F(0.35\,m) = 2.5\,N \times 0.4\,m

    • 0.3+0.35F=1→0.35F=0.70.3 + 0.35F = 1 \rightarrow 0.35F = 0.7

    • Result: F=2 NF = 2\,N

  • Reaction Forces (RA and RB) on a Beam:

    • Total downward forces: 20+30+20+22.5=92.5 kN20 + 30 + 20 + 22.5 = 92.5\,kN

    • Sum of forces equation: RA+RB−92.5=0RA + RB - 92.5 = 0

    • Taking moments about A to find RB:

      • (20 kN×1 m)+(30 kN×2 m)+(20 kN×3 m)+(22.5 kN×3.75 m)=RB×4.5 m(20\,kN \times 1\,m) + (30\,kN \times 2\,m) + (20\,kN \times 3\,m) + (22.5\,kN \times 3.75\,m) = RB \times 4.5\,m

      • 224.375 kNm=RB×4.5 m→RB=49.86 kN224.375\,kNm = RB \times 4.5\,m \rightarrow RB = 49.86\,kN

    • Solve for RA: RA+49.86=92.5→RA=42.64 kNRA + 49.86 = 92.5 \rightarrow RA = 42.64\,kN

Questions & Discussion

  • Kite Problem: A kite hangs freely from a branch. Which point is the centre of gravity?

    • Analysis: For an object to hang freely in equilibrium, the CoG must be directly below the point of suspension. If the kite is attached to the branch at point B, and point D is directly below B, then point D is the centre of gravity.

  • Box Problem: A box is balanced at the edge of a table.

    • Analysis: For the box to remain balanced and not tip over, its centre of gravity must be supported by the base (the table). The point of balance suggests the CoG is located exactly above the edge of the table or within the portion of the box resting on the table.