Laws of Motion · Lesson 6 of 14
Forces, Free-Body Diagrams and Equilibrium
“Identify common forces, draw only forces on the chosen system and solve particle equilibrium.”
• Describe weight, normal reaction, tension, spring force and fluid contact forces. • Construct a free-body diagram with clear body-to-body force labels. • Solve equilibrium by components. • Explain why a normal force or string tension is not assigned automatically.
List the interactions before writing equations
Choose the body or collection of bodies to be analysed. Replace it in a separate diagram by a simple shape or point. Draw every external force exerted on it by something outside the boundary. Include weight, contact forces, tensions and applied pushes when present. Do not include the forces that the selected body exerts on its surroundings. The term “free-body diagram” means the body has been isolated for analysis, not that it is free of forces.
| Force | Agent and direction | Useful model |
|---|---|---|
| Weight | Earth on body; vertically downward near Earth | W = mg |
| Normal reaction | Surface on body; perpendicular to contact | Unknown until constraints and equations determine it |
| Friction | Contact surface on body; tangent to surface | Opposes relative slip or its tendency |
| Tension | Taut string on attached body; along string away from body | Uniform on an unloaded ideal massless segment |
| Spring force | Spring on body; toward the natural-length position | F = −kx for small deformation |
| Buoyancy | Fluid on immersed body; upward in a static gravitational fluid | Magnitude equals displaced fluid weight |
| Drag | Fluid on body; opposes relative motion through fluid in simple cases | Depends on the flow regime |
Gravity is a non-contact interaction. Ordinary normal, elastic, tension and friction forces arise through contact; their microscopic origin involves electromagnetic interactions between constituents of matter. This chapter uses practical macroscopic force models rather than deriving every contact force microscopically. Charged and magnetic bodies can also have non-contact electromagnetic forces, even though most examples here leave them out.
The four fundamental interaction categories are gravitational, electromagnetic, strong and weak. Ordinary macroscopic mechanics in this chapter mainly uses gravity and forces with an electromagnetic microscopic origin. Strong and weak interactions are important at subatomic scales and are not the contact-force models used for blocks, ropes and tables.
Normal reaction is perpendicular, not always upward
A floor pushes upward on a block, a vertical wall pushes sideways, and an incline pushes along its tilted normal. A non-adhesive contact can push but cannot pull, so its normal-force magnitude must satisfy N ≥ 0. If a calculation requires N < 0, the assumed contact condition has failed. Write the perpendicular-motion equation to determine N; setting N = mg before examining geometry and acceleration is a common mistake.
Strings and springs have different models
An ideal light string has negligible mass and transmits tension along its length. With no extra force applied along a free segment and an ideal smooth pulley, the tension magnitude is the same through that segment. Inextensible means the total length stays fixed and creates a motion constraint. A massless string can still have different tensions on two sides of a knot where an external force is applied. A slack string has zero tension and cannot push.
Equilibrium of a particle
Two forces can balance only if they are equal and opposite. Three concurrent forces in equilibrium can be drawn head to tail as a closed triangle. More forces close a polygon. Resolve the forces into convenient axes rather than adding their magnitudes. A body stationary in a frame that is accelerating need not have zero acceleration in the ground frame; specify the frame before declaring equilibrium.
Problem
A 6 kg mass hangs from a light rope. A horizontal 50 N force is applied at a point on the rope, and the arrangement is stationary. Use g = 10 m/s². Find the upper rope angle to vertical.
- 1.Analyse the mass first. Its lower-string tension T₂ balances weight, so T₂ = 60 N.
- 2.At the pulled point, upper tension T₁ has components balancing the downward 60 N and horizontal 50 N. Thus T₁ cosθ = 60 and T₁ sinθ = 50.
- 3.Dividing gives tanθ = 50/60. Therefore θ = 39.81° to vertical. T₁ = √(60² + 50²) = 78.10 N.
- 4.The rope length affects geometry but not this equilibrium force-angle relation. T₁ and T₂ differ because the sideways force acts at the junction.
A systematic workflow
First draw the physical arrangement and identify the system. Next draw its force diagram. Choose axes and mark acceleration directions. Write component equations and any string or contact constraints. Solve for the unknowns, then check signs and whether assumed contact, tautness or no-slip conditions hold. If you need an internal tension or contact force, analyse one part separately in addition to the combined system.
Equilibrium is a net statement
A floating cork, stationary kite or resting book can each have zero net force while several interactions act. Conversely, a momentarily stationary projectile has nonzero downward acceleration and is not in equilibrium. A snapshot showing zero velocity does not establish zero acceleration; equilibrium requires a force balance, or sufficient information that velocity remains constant over an interval.
Quiz
What belongs on a free-body diagram for a block?
Key Takeaways
• Define the system before deciding whether a force is internal or external. • Normal force is perpendicular to contact and follows from the motion constraint. • Tension pulls along a taut string; a spring supplies a restoring force. • Particle equilibrium requires the vector sum of forces to vanish. • For an extended body, full equilibrium also requires rotational balance.