Source: Experimental Physics and Engineering Lab, Texas A&M University
Tags: particle statics, free body diagram, FBD, tension, Newton's second law, static equilibrium, energy conservation, kinetic energy, potential energy, cable snap velocity, engineering mechanics
Particle statics problems treat objects as point masses in equilibrium, where the sum of all forces equals zero. These notes cover how to draw free body diagrams, apply Newton's second law in component form, and use energy conservation to find velocities. The first two problems walk through a cable-snap energy problem and a two-rope tension system with an applied force.
Particle idealisation
Treating an object as a point mass with no physical dimensions. All forces act through a single point. This simplifies analysis because you don't need to worry about moments or rotation.
Free body diagram (FBD)
A sketch of an isolated body showing every external force acting on it, with directions and labels. The single most important step in any statics problem.
Static equilibrium
The condition where the net force on a particle is zero in every direction: ΣFx = 0 and ΣFy = 0. The particle has zero acceleration.
Newton's second law (ΣF = ma)
The governing equation. In statics, a = 0, so it reduces to ΣF = 0. In dynamics problems (like a snapping cable), a ≠ 0 and energy methods may be more convenient.
Kinetic energy (Ke)
Energy of motion: Ke = ½mv². Measured in joules (SI) or foot-pounds (imperial).
Gravitational potential energy (Ug)
Energy due to height: Ug = mgh. The reference height (h = 0) can be chosen for convenience.
Conservation of energy (ΔEk = ΔEg)
When only gravity does work (no friction, no other external work), a loss in potential energy equals a gain in kinetic energy, and vice versa.
Tension
The pulling force transmitted through a cable, rope, or string. Always acts along the line of the cable, directed away from the object.
Isolate the particle (cut it free from all supports, cables, surfaces).
Draw the particle as a dot.
Add every force: weight (straight down), tension in each cable (along the cable, away from the particle), normal forces, applied forces.
Label magnitudes and angles. Angles should reference a known axis (usually horizontal or vertical).
Choose a coordinate system. Align one axis with as many forces as possible to reduce the number of components you need to resolve.
Any force F at angle θ from the horizontal splits into:
Fx = F cos θ (horizontal component)
Fy = F sin θ (vertical component)
Sign convention matters. Pick positive directions (e.g. right = +x, up = +y) and stick with them throughout the problem.
Write ΣFx = 0 and ΣFy = 0.
This gives two equations, so you can solve for up to two unknowns.
If more unknowns exist, you need additional equations (geometry, a third equilibrium equation from a second particle, etc.).
A ball on a cable swings through an arc.
Given: θ = 30°, cable length ΔL = 4.5 m.
Find: the ball's velocity if the cable snaps (at the lowest point of the swing).
The ball starts at angle θ = 30° from the lowest point.
Height above the lowest point: h = ΔL – ΔL cos 30° = 4.5 – 4.5 cos 30°
Alternatively from the solution: h = 4.5 sin 20° = 2.25 ... but the working uses h = 2.25 (read from the triangle geometry in the diagram).
Corrected from the solution page: h ≈ 2.25 m (vertical drop), horizontal distance x = 4.5 cos 30° ≈ 3.89 m.
At the top of the swing the ball is momentarily at rest (Ke = 0). At the bottom, all potential energy has converted to kinetic energy.
ΔEk = ΔEg
Ug + Ke(initial) = Ke(final)
mgh + 0 = ½mv²
Mass cancels:
gh = ½v²
9.8 × 2.25 = ½v²
v ≈ 6.64 m/s
When the only force doing work is gravity, mass cancels and the velocity depends only on the height dropped. This is a classic energy-method result.
A particle is held in place by two ropes, AB and AC, with an applied force FR.
Given: α = 25°, |FR| = 85 lbs, and a 10° angle appears in the geometry.
Rope AB makes 25° with the vertical, rope AC makes 40° with the horizontal.
Find: tension in rope AB and tension in rope AC.
Four forces act on the particle at point A:
TAB (tension along rope AB)
TAC (tension along rope AC)
FR = 85 lbs (applied force at 10° below horizontal)
W (weight, if relevant, though this problem focuses on the applied force geometry)
X-direction (ΣFx = 0):
TAB cos 25 + TAC cos 40 = 85 cos 10
0.9063 TAB + 0.7660 TAC = 83.71 ... (i)
Y-direction (ΣFy = 0):
TAB sin 25 – TAC sin 40 = –85 sin 10 = –14.76
0.4226 TAB – 0.6428 TAC = –14.76 ... (ii)
From equation (ii): TAB = (–14.76 + 0.6428 TAC) / 0.4226
Substitute into equation (i) and solve for TAC:
TAC ≈ 53.79 lbs
Back-substitute to get TAB:
TAB ≈ 46.89 lbs
For concurrent-force problems, the method is always the same: draw the FBD, resolve every force into x and y components, write two equilibrium equations, solve the resulting system of two equations in two unknowns.
Formula | Meaning |
|---|---|
ΣF = ma | Newton's second law (= 0 in statics) |
Ke = ½mv² | Kinetic energy |
Ug = mgh | Gravitational potential energy |
ΔEk = ΔEg | Energy conservation (gravity only) |
Fx = F cos θ | Horizontal component of force |
Fy = F sin θ | Vertical component of force |
⚠️ Free body diagrams are the foundation of every statics and dynamics problem. If the FBD is wrong, everything downstream is wrong. Draw one for every single problem.
⚠️ When using energy conservation, make sure the only work being done is by gravity. If there is friction or an external applied force doing work, you need the full work-energy theorem instead.
⚠️ Sign errors in components are the most common mistake. Write out your sign convention before you start resolving forces.
⚠️ In the cable-snap problem, mass cancels. If the exam gives you mass, it may be a distractor, or you may need it for a follow-up part (e.g. finding the tension before the cable snaps).
Q: A 2 kg ball hangs from a 3 m cable and is released from 40° to the vertical. What is its speed at the lowest point?
A: h = 3 – 3 cos 40° = 3 – 2.298 = 0.702 m. Using gh = ½v², v = √(2 × 9.8 × 0.702) ≈ 3.71 m/s. Mass does not affect the answer.
Q: Why does mass cancel in the energy conservation equation for a falling or swinging object?
A: Because both kinetic energy (½mv²) and gravitational potential energy (mgh) are proportional to mass. When you set them equal, m divides out from both sides.
Q: A particle is in static equilibrium with three forces acting on it. You know two of the forces. How many scalar equations do you need to find the third force?
A: Two: ΣFx = 0 and ΣFy = 0. Each gives one equation, and the unknown force has two unknowns (magnitude and direction), or two component unknowns.
Q: In Question 2, what would happen to the tensions if the applied force FR increased while all angles stayed the same?
A: Both TAB and TAC would increase proportionally, since the equilibrium equations are linear in the tensions and in FR.
particle statics, free body diagram, FBD, tension in cables, Newton's second law, static equilibrium, concurrent forces, force components, energy conservation, kinetic energy, potential energy, cable snap problem, two-rope tension, ENGR 214, PHYS 214, Texas A&M, engineering mechanics, force resolution, trigonometric components, gravitational PE, velocity from height