Premed · Premed · Physics 1
Lecture 13: Static Equilibrium
Physics I — Mechanics & Thermodynamics
Learning Objectives
By the end of this lecture, students will be able to:
- State the conditions for static equilibrium (translational and rotational)
- Solve for unknown forces and torques in systems in static equilibrium
- Strategically choose pivot points to simplify torque calculations
- Analyze common equilibrium scenarios: beams, ladders, and suspended signs
- Determine the center of gravity and its role in stability
Lecture Content
I. Conditions for Static Equilibrium
An object is in static equilibrium when it is at rest and remains at rest. Two conditions must be satisfied simultaneously. First, translational equilibrium requires that the net external force is zero: Sum F = 0, which in component form gives Sum F_x = 0 and Sum F_y = 0. Second, rotational equilibrium requires that the net external torque about any point is zero: Sum tau = 0. Together, these conditions provide three independent equations in two dimensions (two force equations and one torque equation). Three-dimensional problems yield six equations but are beyond the typical introductory scope.
II. Center of Gravity
The center of gravity (CG) is the point where the total gravitational force effectively acts. In a uniform gravitational field, where g is constant throughout the object, the center of gravity coincides with the center of mass. Its location is given by x_cg = (Sum m_i x_i) / M = x_cm. For symmetric objects of uniform density, the CG is at the geometric center. For composite objects, the CG of each part is found separately, and the overall CG is determined by taking a mass-weighted average. An object supported at its center of gravity will balance without any tendency to rotate.
III. Strategy for Solving Equilibrium Problems
The standard approach to equilibrium problems begins with drawing a clear diagram of the object and identifying all forces. Next, construct a free-body diagram with all forces at their correct points of application: weight acts at the center of gravity, normal forces act at contact points, and tension acts along the rope or cable at the attachment point. Choose a coordinate system, then choose a pivot point for torque calculations.
The pivot point should be chosen strategically: placing it where an unknown force acts eliminates that force from the torque equation, since a force applied at the pivot produces zero torque. Apply Sum F_x = 0, Sum F_y = 0, and Sum tau = 0, then solve the resulting system. A useful check is that for a system truly in equilibrium, the net torque is zero about every point, so you can calculate torques about any convenient location.
IV. Beams and Supports
Horizontal beams supported at one or two points are classic equilibrium problems. For a simply supported beam with a load, the support reactions are the upward normal forces at the support points. The weight of the beam acts at its center, and any additional loads act at their given positions. Sum F_y = 0 provides one equation relating the support forces, while Sum tau = 0 about one support gives the other support force directly.
For a cantilever beam fixed at one end, the wall exerts both a force (with vertical and horizontal components) and a torque on the beam. This results in three unknowns at the wall: F_x, F_y, and tau_wall.
<image>A horizontal beam of length L and mass M supported by a pivot at the left end and a cable attached to the right end at angle theta. A person of mass m stands at a distance d from the left end. The free-body diagram shows: weight Mg at L/2, weight mg at distance d, tension T in the cable at the right end (decomposed into components), and pivot force components (H horizontal, V vertical) at the left end. Torque equation about the pivot is written out, showing how the pivot forces vanish when the pivot is chosen as the axis.</image>
V. Ladder Problems
A ladder leaning against a wall is a classic two-dimensional equilibrium problem. The typical assumptions are that the wall is frictionless (exerting only a horizontal normal force), the floor has friction (exerting both normal and friction forces), and the ladder has weight acting at its center.
The forces on the ladder are its weight (mg) at the center, the normal force from the floor (N_floor, upward), friction from the floor (f, horizontal, toward the wall), and the normal force from the wall (N_wall, horizontal, away from the wall). The equilibrium equations are Sum F_x = 0, giving f = N_wall; Sum F_y = 0, giving N_floor = mg plus any person's weight; and Sum tau = 0 about the base, which determines N_wall. The ladder slips when the required friction exceeds mu_s N_floor.
VI. Hanging Signs and Suspended Objects
A sign hanging from a beam attached to a wall by a hinge and supported by a cable is another standard equilibrium configuration. The hinge exerts a force with horizontal and vertical components, the cable exerts tension along its length, and the sign's weight acts at the point of suspension.
Choosing the hinge as the pivot eliminates the hinge forces from the torque equation, making it straightforward to solve for the cable tension first. The hinge forces are then found from the force equations. When multiple cables or supports are present, all three equilibrium equations may need to be used simultaneously.
VII. Stability and Tipping
An object is stable if, when slightly displaced, it tends to return to equilibrium. An object tips when the vertical line through its center of gravity falls outside the base of support. A wider base and a lower center of gravity produce greater stability.
Whether an object tips or slides depends on the relative magnitudes of the forces required for each. Tipping occurs when the torque about the pivot edge becomes unbalanced, while sliding occurs when the applied force exceeds mu_s N. If the force required to tip is less than the force required to slide, the object tips first, and vice versa.
These principles have direct applications in biomechanics. When standing, the center of gravity must remain over the feet. Carrying a heavy load on one side shifts the center of gravity, and the body instinctively leans in the opposite direction to compensate.
<image>Panel A: A tall, narrow box on a surface being pushed by a horizontal force at height h. The center of gravity is marked. The box will tip about the bottom-right edge when the torque from the applied force exceeds the torque from gravity about that edge. The tipping condition equation is shown. Panel B: The same box being pushed, but now sliding instead of tipping (lower push point, higher friction). Panel C: A triangle of stability showing a wide-based object (stable, CG line well within the base) vs. a narrow-based object (easily tipped, CG line near the base edge).</image>
