Static Equilibrium Conditions

• Static equilibrium requires two conditions to be met.

Condition for Equilibrium

• Net external force on an object must be zero, so the object's center of mass isn't accelerating.
• Written as a formula: $\sum \vec{F} = 0$
• In component form:
• $\sum F_x = 0$
• $\sum F_y = 0$
• $\sum F_z = 0$
• Net external torque on the object about any axis must be zero, so the object has no angular acceleration.
• Written as a formula: $\sum \vec{\tau} = 0$

Center of Gravity

• The center of gravity is where the weight of the object can be considered to act when calculating the torque due to gravity.
• For symmetrical objects with uniform density it's located at the geometric center.

Problem-Solving Strategy for Static Equilibrium Problems

• Choose the object to which the equilibrium conditions will be applied.
• Draw a free-body diagram showing all external forces. Indicate the point of application of each force.
• Choose a convenient coordinate system and resolve all forces into x and y components.
• Apply the first condition of equilibrium, $\sum F_x = 0$ and $\sum F_y = 0$.
• Choose a convenient axis of rotation and calculate the torque due to each force about that axis.
• Apply the second condition of equilibrium, $\sum \tau = 0$.
• Solve the resulting system of equations to find the unknown quantities.
• Verify that the answer is reasonable.

Example Problem: Beam Supported by Ropes

• A uniform beam of length $L$ and weight $W$ is supported by two ropes.
• The first rope is at the left end, the second is at $L/4$ from the right end.
• Problem: Find the tensions in the two ropes

Solution

• Object: the beam.
• Free-body diagram shows:
• Weight $W$ acting downward at the beam's center (distance $L/2$ from either end).
• Tension $T_1$ in the first rope, upward at the left end.
• Tension $T_2$ in the second rope, upward at a distance $L/4$ from the right end.
• Coordinate system: x-axis is horizontal, y-axis is vertical.
• First condition of equilibrium:
• $\sum F_x = 0$ (no horizontal forces)
• $\sum F_y = T_1 + T_2 - W = 0$
• Axis of rotation: the left end of the beam to eliminate the torque due to $T_1$.
• Second condition of equilibrium:
• $\sum \tau = 0$
• $-W(L/2) + T_2(3L/4) = 0$
• Solving the equations:
• Torque equation yields : $T_2 = \frac{2}{3}W$
• Force equation yields : $T_1 = W - T_2 = W - \frac{2}{3}W = \frac{1}{3}W$
• $T_1 = \frac{1}{3}W$
• $T_2 = \frac{2}{3}W$

Explanation of static equilibrium conditions: the net external force must be zero to prevent the object's center of mass from accelerating. The net external torque on the object about any axis must be zero, ensuring no angular acceleration.