Static Equilibrium —Lecture
Summary
Based on SlideShare: 'Lecture 4
static_equilibrium' (for classroom
use)
2.
Equilibrium (Overview)
• Netforce is zero: ΣF = 0
• Net torque is zero: Στ = 0
• No linear or angular acceleration when both
hold
3.
Static vs DynamicEquilibrium
• Static: object at rest (v = 0), a = 0, α = 0
• Dynamic: motion at constant speed and/or
constant angular speed (a = 0, α = 0)
4.
Free Body Diagram(FBD)
‑
• Isolate the body and draw all forces (W, N, T,
friction, pushes)
• Choose x–y axes to simplify components
• Write ΣFx = 0 and ΣFy = 0
• Write Στ = 0 about a convenient pivot
5.
Example: Two equal-anglestrings
holding mass m
ΣFy: 2T sinθ = mg → T = mg/(2 sinθ)
ΣFx: components cancel by symmetry
6.
Example: One horizontal,one at
30°
ΣFy: T_ang·sin30° = mg → T_ang = mg/sin30° = 2mg
ΣFx: T_horiz = T_ang·cos30°
7.
Torque τ
• τ= r × F; magnitude τ = r F sinθ
• Moment arm r = r sinθ so τ = F r
⊥ ⊥
• Pick pivots that eliminate unknown forces
8.
Beam pinned atA, rope at B
making angle θ
Load W at distance x from A, rope tension T at end B
Στ_A = 0 → T sinθ · L = W · x T = (W x) / (L sinθ)
⇒
Then use ΣFx = 0 and ΣFy = 0 for hinge reactions
9.
Worked numbers (W= 120 N at
midpoint, θ = 37°)
Στ_A: T sin37° · L = W · (L/2) T = (W/2)/sin37°
⇒
With sin37° ≈ 0.6 → T ≈ 100 N
R_Ax = T cos37° ≈ 80 N; R_Ay = W − T sin37° ≈ 60 N
Checklist & Pitfalls
•FBD → ΣF → Στ → solve
• Use perpendicular component for torque (sin
vs cos!)
• Weight acts at the center of mass
• Define clockwise/CCW signs consistently
12.
Practice Problems
• 1)2 kg mass held by two strings at 30° and 60°
→ find T1, T2
• 2) Beam L=2 m, W at 0.5 m, cable θ=45° →
find T, R_Ax, R_Ay
• 3) Ladder length 5 m against wall with
frictionless wall → find forces
13.
Source
• Original deckon SlideShare: 'Lecture 4
static_equilibrium' (for reference)
• This PPT is a didactic summary for classroom
use.