Static Equilibrium, Elasticity, and Fracture: Study Notes

Chapter 9: Static Equilibrium; Elasticity and Fracture

9-1 The Conditions for Equilibrium

Static equilibrium is a fundamental concept in physics describing objects that remain at rest under the action of multiple forces. For an object to be in equilibrium, both the net force and the net torque acting on it must be zero.

• Definition of Equilibrium: An object is in equilibrium if it is at rest and the sum of all forces and torques acting on it is zero.

• Normal Force and Gravity: For a book resting on a table, the upward normal force balances the downward force of gravity.

• First Condition (Translational Equilibrium): The vector sum of all forces along each coordinate axis must be zero: $\sum \vec{F} = 0$

• Second Condition (Rotational Equilibrium): The sum of all torques about any axis must be zero. The choice of axis is arbitrary: $\sum \tau = 0$

• Important Note: If an object is in equilibrium, the sum of torques around any axis is zero.

Example: A hanging object supported by two ropes at different angles must have the sum of forces in both x and y directions equal to zero for equilibrium.

9-2 Solving Statics Problems

Solving statics problems involves systematic steps to ensure all forces and torques are accounted for. This process is essential for analyzing structures, machines, and everyday objects.

• Step 1: Choose one object at a time and draw a free-body diagram showing all forces acting on it and their points of application.

• Step 2: Select a coordinate system and resolve all forces into their components (usually x and y).

• Step 3: Write the equilibrium equations for the forces: $\sum F_x = 0$ $\sum F_y = 0$

• Step 4: Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation: $\sum \tau = 0$ Choosing a clever axis can simplify calculations, especially if it passes through unknown forces.

• Step 5: Solve the resulting system of equations for the unknowns.

Example: For a uniform beam hinged to a wall and held by a cable, draw the free-body diagram, resolve forces, and apply equilibrium conditions to find the tension in the cable and the forces at the hinge.

Example Problem: Hinged Beam and Cable

A uniform beam of length 2.20 m and mass 25 kg is mounted by a hinge on a wall. The beam is held horizontally by a cable at 30°, and supports a sign of mass 28 kg at its end. The goal is to determine the components of the hinge force and the tension in the supporting cable.

• Free-Body Diagram: Identify all forces: weight of beam ($mg$), weight of sign ($Mg$), tension in cable ($F_T$), and hinge reaction forces ($F_{Hx}$, $F_{Hy}$).

• Equilibrium Equations:

  • Sum of forces in x-direction: $\sum F_x = 0$

  • Sum of forces in y-direction: $\sum F_y = 0$

  • Sum of torques about hinge: $\sum \tau_{hinge} = 0$

• Application: Use geometry and trigonometry to resolve the tension into components and set up equations for equilibrium.

Example Calculation: The tension in the cable can be found by balancing torques about the hinge, considering the perpendicular distances from the hinge to the points of force application.

Ranking Problems: Tension and Vertical Components

Several problems involve ranking the magnitude of tension forces or their vertical components for objects suspended by ropes at different angles.

• Vertical Component of Tension: For identical masses suspended by pairs of ropes, the sum of the vertical components of tension must balance the weight of the mass.

• Ranking: The vertical component is maximized when the ropes are more vertical; as the angle increases from the vertical, the vertical component decreases.

• Individual Rope Tension: The tension in each rope depends on the angle and the distribution of the load.

Example: For three identical balls suspended by ropes at different angles, rank the cases according to the sum of vertical components or the tension in a specific rope.

Mobile Balance Problem

In problems involving mobiles (objects suspended from horizontal bars), the principle of static equilibrium is used to determine the required mass for balance.

• Torque Balance: For the mobile to remain horizontal, the sum of torques about the suspension point must be zero.

• Application: If one toy has a known mass and is suspended farther from the pivot, the mass of another toy closer to the pivot can be found using the equation: $m_1 d_1 = m_2 d_2$ where $m$ is mass and $d$ is distance from the pivot.

Example: If toy #4 has mass 10 g and is suspended four times farther from the pivot than toy #2, toy #2 must have mass 40 g to balance the mobile.

Summary Table: Conditions for Equilibrium

Condition	Equation	Description
Translational Equilibrium	$\sum \vec{F} = 0$	Net force on object is zero; object does not accelerate.
Rotational Equilibrium	$\sum \tau = 0$	Net torque on object is zero; object does not rotate.

Additional info: These notes focus on the principles of static equilibrium, including the conditions for equilibrium, problem-solving strategies, and applications to real-world systems such as beams, cables, and mobiles. Elasticity and fracture are mentioned in the chapter title but not covered in the provided materials.