Statics Of Rigid Bodies Simulations

A collection of interactive 3D visualizations and simulations to help you master concepts in statics of rigid bodies.

Force Systems - Theory & Concepts - Force Components

A comprehensive overview of Force Systems explaining core principles.

Rectangular Components of a Force

Results

Magnitude (F):

100.0 N

Angle (θ):

45.0°

x-component (Fx):

70.7 N

y-component (Fy):

70.7 N

Magnitude (F)100 N

Angle (θ)45°

F_x = F \cos(\theta)

F_y = F \sin(\theta)

Force Systems - Theory & Concepts - Moment

A comprehensive overview of Force Systems explaining core principles.

Moment of a Force (M = r × F)

Moment Result

500.0 N·m

Counter-Clockwise (+)

$M = F_y \times d = (F \times \sin \theta) \times d$

$M = (100 \times \sin(90^\circ)) \times 5$

$M = 100.0 \times 5 = 500.0 \text{ N} \cdot \text{m}$

Force (F)100 N

Distance (r)5 m

Force Angle (θ)90°

Force Systems - Theory & Concepts - Equivalent Force System

A comprehensive overview of Force Systems explaining core principles.

Equivalent Force-Couple System

Force Magnitude (F)100 N

Distance from O to P (d)2 m

Force Systems - Theory & Concepts - Vector Resolution3 D

A comprehensive overview of Force Systems explaining core principles.

3D Vector Configuration

Adjust the magnitude and two coordinate direction angles. The third angle ( $\\gamma$ ) is calculated automatically to satisfy the identity $\\cos^2\\alpha + \\cos^2\\beta + \\cos^2\\gamma = 1$ .

Magnitude (F)100 N

Angle α (from x-axis)60°

Angle β (from y-axis)60°

Calculated Angle γ (from z-axis):

\gamma \approx 45.0^\circ

Rectangular Components

X - Component (

F_x

)+50.0 N

Y - Component (

F_y

)+50.0 N

Z - Component (

F_z

)+70.7 N

Cartesian Vector Formulation

\mathbf{F} = \{50.0\mathbf{i} + 50.0\mathbf{j} + 70.7\mathbf{k}\}\text{ N}

Equilibrium of Particles - Theory & Concepts - Particle Equilibrium

A comprehensive overview of Equilibrium of Particles explaining core principles.

2D Particle Equilibrium (Concurrent Forces)

Calculated Tensions

Tension 1 (T1):

89.7 N

Tension 2 (T2):

73.2 N

$\Sigma F_x = 0 \implies T_2 \cos(\theta_2) - T_1 \cos(\theta_1) = 0$

$\Sigma F_y = 0 \implies T_1 \sin(\theta_1) + T_2 \sin(\theta_2) - W = 0$

Weight (W)100 N

Angle 1 (θ1)45°

Angle 2 (θ2)30°

Equilibrium of Rigid Bodies - Theory & Concepts - Beam Equilibrium

A comprehensive overview of Equilibrium of Rigid Bodies explaining core principles.

Interactive Physics Simulation

Simply Supported Beam Equilibrium Simulator

Study the conditions of static translational and rotational equilibrium on a simply supported beam. Move the point load P to observe reaction force shifts.

Load Position (X)4.0 m

Load Magnitude (P)600 N

Static Equilibrium Equations

Rotational Equilibrium ($\Sigma M_A = 0$):

-(600·4.0) + (B_y·10) = 0 \implies B_y = 240.0 N

Translational Equilibrium ($\Sigma F_y = 0$):

A_y + B_y - 600 = 0 \implies A_y = 360.0 N

When the load moves closer to support A, reaction $A_y$ increases while $B_y$ decreases, ensuring the seesaw moment remains perfectly balanced.

Reaction Support Ay

360.0 N

Reaction Support By

240.0 N

Internal Forces in Structural Members - Theory & Concepts

A comprehensive overview of Internal Forces in Structural Members explaining core principles.

Interactive Physics Simulation

Shear Force & Bending Moment Diagram Simulator

Apply a point load P at any position along a simply supported beam. Observe the resulting reaction forces, step shear diagram V, and parabolic bending moment diagram M.

Point Load Position (x)5.0 m

Load Magnitude (P)15 kN

Shear & Moment Calculus Relationships

Shear is the spatial derivative of Bending Moment:

V(x) = \frac{dM(x)}{dx}

Change in Moment is the area under Shear curve:

M(x) = \int V(x) \, dx

Notice that the maximum bending moment occurs exactly at the point where the shear diagram crosses zero ($x = 5\text{ m}$).

Shear Diagram V (kN)

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Bending Moment Diagram M (kN·m)

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Reaction R1 (Left)

7.50 kN

Reaction R2 (Right)

7.50 kN

Max Bending Moment

37.50 kN·m

Dry Friction - Theory & Concepts

A comprehensive overview of Dry Friction explaining core principles.

Interactive Physics Simulation

Friction on an Inclined Plane Simulator

Study Coulomb's friction laws. Increase the angle of inclination to find the critical slip angle where static friction yields to sliding kinetic motion.

✓ Static Equilibrium

Angle of Inclination (θ)15 deg

Block Weight (W)120 N

Static Coeff (μs)0.50

Kinetic Coeff (μk)0.40

Inclined Friction Calculus

Critical Slip Angle:

\theta_{crit} = \tan^{-1}(\mu_s)

Active Slip Angle:15° ≤ 26.6°

If the incline angle $\theta$ is smaller than $\theta_{\text{crit}}$, the parallel gravity pull ($W_x$) matches static friction ($f_s$). If it exceeds it, kinetic friction ($f_k$) governs.

Clamping Normal Force (N)

115.9 N

Parallel Gravity Pull (Wx)

31.1 N

Max Static Limit (μs·N)

58.0 N

Active Friction Force (f)

31.1 N

Distributed Forces, Centroids, and Centers of Gravity - Theory & Concepts

A comprehensive overview of Distributed Forces, Centroids, and Centers of Gravity explaining core principles.

Composite Centroid Simulation

Flange Height ($h_1$)40 mm

Web Width ($w_2$)40 mm

Area 1 (Flange):8000 mm²

Area 2 (Web):6400 mm²

Total Area:14400 mm²

Notice how the centroid (Ȳ, red dot) shifts towards the part with the larger area.
Ȳ = (Σ y_i * A_i) / Σ A_i

Moments of Inertia - Theory & Concepts - Moment Of Inertia

A comprehensive overview of Moments of Inertia explaining core principles.

Parallel-Axis Theorem

Adjust the dimensions and distance of the rectangle to see how the moment of inertia changes relative to the reference axis.

Width (b): 10 cm

Height (h): 20 cm

Distance (d): 15 cm

I_bar (bh³/12):6666.7 cm⁴

Area (bh):200.0 cm²

Ad² term:45000.0 cm⁴

Total I:51666.7 cm⁴

Cables and Arches - Theory & Concepts

A comprehensive overview of Cables and Arches explaining core principles.

Interactive Physics Simulation

Suspension Cable Force & Shape Simulator

Compare structural cable configurations. Analyze how concentrated point loads or uniform distributed loads yield parabolic vs. catenary curves.

Cable Load Profile Type

Uniform Load (w)12 kN/m

Midspan Sag Distance (h)5.0 m

Cable Mechanics Governing Equations

Minimum Cable Tension (at lowest point):

T_0 = \\frac{w \\cdot L^2}{8 \\cdot h}

Maximum Cable Tension (at support anchors):

T_{max} = \\sqrt{T_0^2 + \\left(\\frac{w \\cdot L}{2}\\right)^2}

Span L: 20 meters

Minimum Cable Tension (T0)

120.0 kN

Maximum Anchor Tension (Tmax)

169.7 kN

Three-Hinged Arches - Theory & Concepts - Three Hinged Arch

A comprehensive overview of Three-Hinged Arches explaining core principles.

Three-Hinged Arch Analysis

Load Position (

x

): 5.0 m

Load Magnitude (

P

): 100 kN

Real-time Reactions:

$A_y$ : 75.0 kN

$B_y$ : 25.0 kN

Thrust ( $H$ ): 50.0 kN