in class 10-28

Equilibrium and Torque

• When in static equilibrium, angular acceleration ($\alpha$) is zero; thus, the sum of torques equals zero.

• Free body diagrams are expanded to include torques when analyzing systems in equilibrium.

Key Concepts

• Static Equilibrium Conditions:

  • Sum of forces ($\Sigma F = 0$)

  • Sum of torques ($\Sigma \tau = 0$)

• Torque Calculation:

  • Individual torque ($\tau = r \cdot f_{\perpendicular}$)

  • $r$ = distance from pivot point

  • $f_{\perpendicular}$ = perpendicular component of force

• Choice of Pivot Points:

  • Pivot point can be chosen freely; affects torque calculation but not overall equilibrium assessment.

Extended Free Body Diagrams

• Represent shapes (e.g., rods) to depict forces acting at specific points.

• Important to indicate locations and angles for accurate torque calculations.

Example Problem: Lifting a Heavy Bucket

• Bucket of mass 910 kg (weight $W = mg$).

• Identify forces acting on the beam and calculate the necessary lift force to maintain equilibrium.

Example Problem: Two Normal Forces at a Support

• Analyze normal forces when distributing weight across two support points of a board.

• Assign pivot and calculate torques for both ends of the board.

Ladder Problem

• For a ladder leaning against a wall:

  • Forces includes weight, normal forces at contact points, and friction.

  • Apply conditions for static equilibrium to find the coefficient of static friction needed to prevent slipping.

• General Equation:

  • Coefficient of static friction $\mu_s \geq \frac{1}{2} \tan(\phi)$ (where $\phi$ is the angle).

Important Concepts

• Rigid body assumptions: No flexing or bending.

• Dynamic vs. Static Analysis: Emphasis on static scenarios makes calculations simpler due to zero angular acceleration conditions.