Equilibrium of a Particle
When a load is lifted at constant velocity or simply hangs at rest, it is in equilibrium. This chapter develops equilibrium for a particle and applies it to find the forces carried by cables, cords, and springs that support suspended loads.
Learning objectives
- Identify when dimensions and rotation can be neglected in a particle model.
- Draw a complete free-body diagram for a joint, cable, spring, or smooth contact.
- Resolve forces and solve the independent two- or three-dimensional equilibrium equations.
- Use a spring law or cable capacity as an additional physical constraint.
- Verify force directions, units, and feasibility of the calculated support forces.
The single most common source of lost marks in this chapter is a missing or sloppy free-body diagram. Draw the FBD first, every time — even for problems that look trivial. If you can't draw it, you don't yet understand the problem.
Recognition strategy
Only two collinear forces act on the instrument: weight and spring force. Particle equilibrium determines the spring force, and Hooke's law then converts that force to extension.
Given information
- \(m=12.0\,\text{kg}\), \(g=9.81\,\text{m/s}^2\), and \(k=2.40\,\text{kN/m}=2400\,\text{N/m}\); the instrument is at rest.
Model and interpretation
Take upward as positive and treat the spring as linear over this small extension. The spring pulls upward on the instrument; its extension is measured from free length, not from an arbitrary installed dimension.
Verification
Units: Weight uses \(\text{kg}\cdot\text{m/s}^2=\text{N}\), and dividing N by \(\text{N/m}\) leaves metres, so both the force and extension have the correct dimensions.
Signs and direction: The solved positive \(F_s\) is upward under the selected convention, opposing the downward weight as a supporting spring must; the extension itself is reported as a positive elongation.
Plausibility: A 2.40 kN/m spring needs only about 0.05 m to generate 0.118 kN, and \(2400(0.0491)=117.8\,\text{N}\) confirms the rounded result.
Engineering interpretation
The static extension establishes clearance and preload geometry, but it is not the full service displacement. Vibration, travel stops, fatigue life, and the spring's actual allowable deflection must be checked before specifying a production support.
Recognition strategy
All three forces meet at the ring, so it is a two-dimensional particle problem with two unknown tensions. Resolve each cable using its own signed direction rather than using a resultant triangle.
Given information
- Downward applied load is 900 N; cable A is up-left at 30°; cable B is up-right at 45°; use +x right and +y up.
Model and interpretation
Ideal cables pull away from the ring along their axes. Cable mass, stretch, and hardware dimensions are neglected, making force balance the applicable static model.
Verification
Units: Trigonometric factors are dimensionless, so every term in both equilibrium equations remains in newtons and may legitimately be summed with the 900 N applied load.
Signs and direction: The left cable has negative x and positive y components while the right cable has positive x and positive y components; the solved positive tensions match the assumed pulling directions.
Plausibility: Horizontal components are \(659\cos30^\circ=571\,\text{N}\) and \(807\cos45^\circ=571\,\text{N}\), while vertical components total \(329+571=900\,\text{N}\).
Engineering interpretation
The steeper-looking load path is not automatically the most highly loaded one; coupled horizontal balance matters. Selection of actual slings also requires rated capacity, sling-angle rules, fittings, dynamic lift effects, and inspection requirements.
Recognition strategy
The equally spaced geometry and centered payload create three equal tensions whose horizontal components cancel by symmetry. Capacity is an inequality, so the minimum drop occurs when each cable reaches its allowable tension.
Given information
- \(W=1.20\,\text{kN}\), anchor radius \(r=0.800\,\text{m}\), three identical cables, and allowable tension \(T_{\max}=0.600\,\text{kN}\).
Model and interpretation
Treat the payload connection as a particle centered beneath the anchor circle. Each cable length is \(\sqrt{h^2+r^2}\), and only its vertical fraction \(h/\sqrt{h^2+r^2}\) supports weight.
Verification
Units: The vertical direction ratio is metres divided by metres and is dimensionless, so \(3(0.600\,\text{kN})[h/\sqrt{h^2+r^2}]\) correctly produces the 1.20 kN load.
Signs and direction: Each cable pulls upward toward its anchor, giving positive vertical components, while the three radial horizontal components cancel only because the anchors and payload are truly symmetric.
Plausibility: At \(h=0.716\,\text{m}\), the vertical direction cosine is \(0.667\), and \(3(0.600)(0.667)=1.20\,\text{kN}\); a shallower drop would indeed exceed capacity.
Engineering interpretation
This geometry reaches the stated static cable limit, leaving no allowance for installation error or dynamic handling. A real overhead fixture must use approved rigging hardware and include code-required factors, unequal-load effects, and professional review.
3.1 Condition for the Equilibrium of a Particle
Static Equilibrium
A particle is in equilibrium when it stays at rest (if originally at rest) or moves with constant velocity (if originally in motion). The term "static equilibrium" usually refers to a particle at rest.
Equilibrium requires that the resultant of all forces acting on the particle be zero. This is a statement of Newton's first law and is both a necessary and a sufficient condition.
If the vector sum of all forces acting on a particle is zero, the particle remains at rest or continues to move at constant velocity. This is both necessary and sufficient for equilibrium.
Sufficiency follows from Newton's second law, \( \sum\mathbf{F}=m\mathbf{a} \). If the forces satisfy Eq. 3–1, then \( m\mathbf{a}=\mathbf{0} \), so the acceleration \( \mathbf{a}=\mathbf{0} \) and the particle moves at constant velocity or stays at rest.
"Constant velocity" counts as equilibrium too — a crate being hoisted at steady speed is just as much an equilibrium problem as one hanging still. Watch for the phrase "constant velocity" in problem statements; it's a signal to apply \(\sum\mathbf{F}=\mathbf{0}\).
3.2 The Free-Body Diagram
Free-Body Diagram (FBD)
A free-body diagram is a sketch of the particle imagined as isolated and "free" from its surroundings, showing every known and unknown force that acts on it. It is the essential first step in any equilibrium analysis.
Before building an FBD, it helps to recognize three supports that appear repeatedly in particle problems: springs, cables/pulleys, and smooth contacts.
Springs
Linearly Elastic Spring
A linearly elastic spring (or cord) changes length in direct proportion to the force applied to it. Its stiffness, or spring constant \(k\), measures this elasticity. The deformation \(s = l - l_0\) is measured from the unloaded length \(l_0\).
A positive \(s\) (elongation) means the force pulls on the spring; a negative \(s\) (shortening) means it pushes. For example, a spring with \(l_0 = 0.8\ \text{m}\) and \(k = 500\ \text{N/m}\) stretched to \(1\ \text{m}\) has \(s = 0.2\ \text{m}\), so \(F = ks = 500(0.2) = 100\ \text{N}\).
Cables and Pulleys
Ideal Cable
Unless stated otherwise, cables are taken to be weightless and inextensible. A cable can carry only tension — a pulling force directed along the cable. A continuous cable passing over a frictionless pulley keeps a constant tension \(T\) throughout its length, for any wrap angle.
Smooth Contact
Smooth (Frictionless) Surface
When an object rests on a smooth surface, the surface reaction is a normal force \(N\) directed perpendicular to the surface at the point of contact.
Procedure for Drawing a Free-Body Diagram
1 · Draw the Outlined Shape
Imagine the particle isolated or cut free from its surroundings. Remove all supports and sketch the particle's outline.
2 · Show All Forces
Indicate every force acting on the particle — active forces that tend to cause motion, and reactive forces from supports or constraints that resist motion.
3 · Identify Each Force
Label known forces with their magnitudes and directions. Use letters for the magnitudes and directions of unknown forces.
Students often draw the sphere's weight directly on the knot. It isn't there. Each body in a connected system gets its own FBD, and forces only transfer through the members that physically touch that body.
3.3 Coplanar Force Systems
Coplanar Equilibrium
When all forces on a particle lie in a single plane (the \(x\!-\!y\) plane), each force is resolved into \(\mathbf{i}\) and \(\mathbf{j}\) components. Equilibrium requires both component sums to vanish independently.
These two equations solve for at most two unknowns. Each component carries an algebraic sign matching its arrowhead direction along the axis. If an unknown force's sense is assumed and the solution comes out negative, the true sense is simply opposite to the assumption — the magnitude is still valid.
Procedure for Analysis
Free-Body Diagram
Set up \(x,y\) axes in any convenient orientation. Label all known and unknown magnitudes and directions. The sense of an unknown force may be assumed.
Equations of Equilibrium
Apply \(\sum F_x = 0\) and \(\sum F_y = 0\). Components along a positive axis are positive; along a negative axis, negative. If a spring is involved with more than two unknowns, add \(F = ks\) to relate spring force to deformation. A negative result reverses the assumed sense.
When a spring appears, the chain is almost always: solve equilibrium → get spring force → use \(F=ks\) for the stretch → use geometry for the final length. Don't try to jump straight to geometry before you know the forces.
3.4 Three-Dimensional Force Systems
Spatial Equilibrium
For a particle under a three-dimensional force system, the resultant must vanish: \(\sum\mathbf{F}=\mathbf{0}\). Resolving into \(\mathbf{i},\mathbf{j},\mathbf{k}\) components yields three independent scalar equations.
The algebraic sum of force components along each coordinate axis must be zero. These three equations solve for at most three unknowns — typically coordinate direction angles or force magnitudes.
Procedure for Analysis
Free-Body Diagram
Establish \(x,y,z\) axes in any convenient orientation, label all known and unknown magnitudes and directions, and assume senses for unknowns as needed.
Equations of Equilibrium
Use the scalar equations when components are easy to read off. When the geometry is hard to visualize, write each force as a Cartesian vector, substitute into \(\sum\mathbf{F}=\mathbf{0}\), and set the \(\mathbf{i},\mathbf{j},\mathbf{k}\) components each to zero. A negative result reverses the assumed sense.
My recommended order every time: (1) write each force as magnitude × unit vector, (2) sum into three component equations, (3) look for the equation with the fewest unknowns and solve there first, then back-substitute. Examples 7 and 8 both reward this "easiest equation first" habit.
Chapter Summary
Particle Equilibrium
A particle at rest or moving at constant velocity is in equilibrium: all forces form a zero resultant, \(\sum\mathbf{F}=\mathbf{0}\). Always begin with a free-body diagram showing every force with its known or unknown magnitude and direction.
Two Dimensions
Apply \(\sum F_x = 0\) and \(\sum F_y = 0\). Relate spring force to deformation with \(F = ks\). A continuous cable over a frictionless pulley keeps constant tension throughout.
Three Dimensions
Apply \(\sum F_x = 0\), \(\sum F_y = 0\), \(\sum F_z = 0\). For awkward geometry, express each force as a Cartesian vector, sum to zero, and set each component to zero.
Physical meaning
Equilibrium means the vector polygon of all forces closes. A component equation equals zero because there is no unbalanced tendency to accelerate along that axis; it does not mean that individual forces vanish.
Problem-solving strategy
- Isolate one explicit particle or ring, rather than the entire connected assembly.
- Replace every attachment with its physical action: tension pulls away, a smooth contact acts normal, and a spring acts along its axis.
- Select axes that simplify the most forces and write each force with signed components.
- Count unknown scalar quantities before selecting the two or three available force equations.
- Use an unused equation, capacity limit, or geometry relation to audit the solution.
Before moving on
- Can you defend every arrow on a particle FBD?
- Can you solve a joint with two unknown tensions without mixing angle references?
- Can you test a capacity-controlled configuration against every cable limit?