Comprehensive Physics Study Notes: Equilibrium, Moments, and Young Modulus (AS Level 9702)

System Parameters and Setup Specifications

• System Description: A uniform square picture suspended in static equilibrium from a fixed pin located at point P via a supporting wire.
• Physical Quantities Given:
  • Weight of the picture: $W = 19\,\text{N}$
  • Tension in the wire: $T = 13\,\text{N}$
  • Diameter of the wire: $d = 0.80\,\text{mm} = 8.0 \times 10^{-4}\,\text{m}$
  • Young modulus of the wire material: $E = 9.3 \times 10^{10}\,\text{Pa}$
• Geometric and Physical Features:
  • The top edge of the square picture is perfectly horizontal.
  • The centre of gravity of the picture is positioned vertically directly below the pivot point P.
  • The wire obeys Hooke's Law throughout the deformation.
  • The wire forms an angle $\theta$ relative to the horizontal plane.

Rotational Equilibrium and Moment Definitions

• Moment of a Force Definition:
  • The moment of a force about a specified point is defined as the product of the magnitude of the force and the perpendicular distance from the line of action of the force to the pivot point.
  • Mathematical expression: $\text{Moment} = F \times d_{\perp}$
• Moment Analysis for Weight:
  • The centre of gravity of the picture lies directly on the vertical line passing through point P.
  • Because the line of action of the weight vector passes directly through point P, the perpendicular distance $d_{\perp}$ from point P to the line of action of the weight is zero ($d_{\perp} = 0\,\text{m}$).
  • Consequently, the moment produced by the weight about point P is zero ($\text{Moment} = 19\,\text{N} \times 0\,\text{m} = 0\,\text{N\,m}$).

Resolution of Forces and Angle Calculation

• Vertical Force Equilibrium:
  • For the system to remain in static equilibrium, the net vertical force acting on the picture must equal zero ($\sum F_y = 0$).
  • The vertical upward support is provided by the two symmetric sections of the wire attached to the picture.
  • Each wire segment exerts a vertical upward force component given by $T \sin(\theta)$.
  • The sum of the upward vertical components of tension from both sides of the wire balances the downward weight of the picture.
• Calculation of Angle $\theta$:
  • Equilibrium equation:         $2 \times T \sin(\theta) = W$
  • Substituting values $T = 13\,\text{N}$ and $W = 19\,\text{N}$:         $2 \times 13\,\text{N} \times \sin(\theta) = 19\,\text{N}$
  • $26 \times \sin(\theta) = 19$
  • $\sin(\theta) = \frac{19}{26} \approx 0.730769$
  • $\theta = \arcsin(0.730769) \approx 46.95^\circ$
  • Rounding to two significant figures yields $\theta = 47^\circ$ (or $47.0^\circ$ to three significant figures).

Material Deformations and Stress-Strain Analysis

• Cross-Sectional Area of the Wire:
  • Diameter $d = 0.80\,\text{mm} = 8.0 \times 10^{-4}\,\text{m}$
  • Radius $r = \frac{d}{2} = 0.40\,\text{mm} = 4.0 \times 10^{-4}\,\text{m}$
  • Cross-sectional area $A$:         $A = \pi \times r^2 = \pi \times (4.0 \times 10^{-4}\,\text{m})^2 = 1.6\pi \times 10^{-7}\,\text{m}^2 \approx 5.0265 \times 10^{-7}\,\text{m}^2$
• Tensile Stress Calculation:
  • Tensile stress ($\sigma$) is defined as force per unit cross-sectional area:         $\sigma = \frac{T}{A}$
  • Substituting tension $T = 13\,\text{N}$:         $\sigma = \frac{13\,\text{N}}{5.0265 \times 10^{-7}\,\text{m}^2} \approx 2.5862 \times 10^7\,\text{Pa}$
• Tensile Strain Calculation:
  • Young modulus ($E$) is defined as the ratio of tensile stress ($\sigma$) to tensile strain ($\varepsilon$):         $E = \frac{\sigma}{\varepsilon}$
  • Rearranging for strain ($\varepsilon$):         $\varepsilon = \frac{\sigma}{E}$
  • Substituting Young modulus $E = 9.3 \times 10^{10}\,\text{Pa}$ and calculated stress $\sigma = 2.5862 \times 10^7\,\text{Pa}$:         $\varepsilon = \frac{2.5862 \times 10^7\,\text{Pa}}{9.3 \times 10^{10}\,\text{Pa}} \approx 2.7809 \times 10^{-4}$
  • Rounded to two significant figures, the strain $\varepsilon$ in the wire is $2.8 \times 10^{-4}$ (dimensionless).