Statics

Statics studies forces on bodies at rest or moving at constant velocity (equilibrium). It is the foundation of structural analysis and machine design.

Key Principle: Equilibrium

For a body to be in equilibrium:

ΣF = 0   (sum of all forces equals zero)
ΣM = 0   (sum of all moments equals zero)

No acceleration means all forces and moments balance.

Moments and Torque

What is a Moment?

A moment (or torque) is the rotational effect of a force. It depends on:

1. Magnitude of the force
2. Distance from the pivot point (the moment arm)

M = F × d

Where:
- M = moment (N⋅m or lb⋅ft)
- F = force (N or lb)
- d = perpendicular distance from pivot to line of action (m or ft)

Sign Convention

Counterclockwise -> positive moment
Clockwise        -> negative moment

Example: Opening a Door

         <- F = 50 N
        |
    [===|========]   door (width = 0.8 m)
        |
        v hinge

Moment at hinge:

M = F × d
M = 50 N × 0.8 m = 40 N⋅m (counterclockwise)

Pushing farther from the hinge produces more torque. That is why doorknobs sit far from hinges.

Moment of a Force About a Point

        ^ F
        |
        |
    ----*--------   Point O
         d_perp

M_O = F × d_perp

Where d_perp is the perpendicular distance from point O to the line of action of F.

Principle of Moments (Varignon's Theorem)

The moment of a resultant force equals the sum of moments of its components.

M_O = F_x × d_y + F_y × d_x

This becomes useful for complex force systems.

Types of Supports and Reactions

Understanding support reactions is the heart of structural analysis.

Support Types

2D Equilibrium Equations

For planar (2D) problems, three equations are available:

ΣFx = 0   (horizontal forces)
ΣFy = 0   (vertical forces)
ΣM  = 0   (moments about any point)

That means a 2D static problem can solve for up to 3 unknowns.

Beam Analysis

Beams are structural elements that resist bending.

Simply Supported Beam

   Point Load P
        v
    ====*====
    ^       ^
    R_A     R_B

    |<--a-->|<--b-->|
    |<------L------>|

Given: beam length L, load P at distance 'a' from left. Find: reactions R_A and R_B.

Solution

ΣFy = 0:  R_A + R_B - P = 0           ... (1)
ΣM_A = 0: R_B × L - P × a = 0         ... (2)

From equation (2):
R_B = (P × a) / L

From equation (1):
R_A = P - R_B = P - (P × a)/L = P(L - a)/L = (P × b)/L

Check: sum of moments about B should also be zero.

Cantilever Beam

        Load P
          v
    |=====*=====
    ^M_A         (free end)
    | R_A

    |<----L---->|

Given: beam length L, load P at free end. Find: reaction force R_A and moment M_A at fixed support.

Solution

ΣFy = 0:  R_A - P = 0
          R_A = P

ΣM_A = 0: M_A - P × L = 0
          M_A = P × L

Uniformly Distributed Load (UDL)

    w (N/m)  load per unit length
    vvvvvvvvvvvv
    ============
    ^         ^
    R_A       R_B
    |<---L--->|

Total load: W = w × L (acts at the center of the beam).

For symmetric support:
R_A = R_B = W/2 = (w × L)/2

Example Problem: Beam with Multiple Loads

    100 N       200 N
      v           v
    ==*====*======
    ^            ^
    R_A          R_B

    |<-2m->|<-3m->|

Find: reactions R_A and R_B.

Solution

ΣFy = 0:
R_A + R_B - 100 - 200 = 0
R_A + R_B = 300          ... (1)

ΣM_A = 0 (taking moments about point A):
R_B × 5 - 100 × 2 - 200 × 5 = 0
5 R_B = 200 + 1000 = 1200
R_B = 240 N

From equation (1):
R_A = 300 - 240 = 60 N

Verification (check ΣM_B = 0)

R_A × 5 - 100 × 3 - 200 × 0 = 0
60 × 5 - 300 = 0  (checks)

Trusses

A truss is a structure built from slender members (bars) connected at joints (pins). Common in bridges, roofs, and towers.

Assumptions for Truss Analysis

1. All members connect by frictionless pins.
2. Loads apply only at joints.
3. Members are two-force members (force along the member only).
4. The weight of members is negligible.

Member Forces

• Tension (T): member is being pulled apart.
• Compression (C): member is being pushed together.

Method of Joints

Apply equilibrium at each joint.

Example: Simple Truss

        2000 N
          v
        B *
       /|\
      / | \
     /  |  \
  A *   |   * C
    ^   |   ^
   R_A  |  R_C

  |<-2m>|<-2m>|
  Height = 2m

Step 1: Find Reactions

By symmetry, R_A = R_C = 1000 N.

Step 2: Analyze Joint A

At joint A:
- R_A = 1000 N (upward)
- F_AB (along member AB, unknown)
- F_AC (along member AC, horizontal, unknown)

Angles: θ = 45° (for member AB)

ΣFx = 0:  F_AC - F_AB cos(45°) = 0
ΣFy = 0:  R_A + F_AB sin(45°) = 0

From ΣFy:
1000 + F_AB × 0.707 = 0
F_AB = -1414 N (negative means compression)

From ΣFx:
F_AC = -1414 × 0.707 = -1000 N (tension)

Continue this process at each joint to find all member forces.

Method of Sections

Cut through the truss and analyze a section as a free body. Useful when a specific member force is needed without working through every joint.

      Cut line
    - - - - - -
         v P
       B *
      /|  \
     / |   \
  A *  |    * C
    ^  |    ^

Analyze the left or right section. Apply ΣFx, ΣFy, ΣM = 0.

Friction

Friction is the force that resists sliding between surfaces.

Static vs Kinetic Friction

Usually μ_s > μ_k (it is harder to start moving than to keep moving).

Friction Force

F_friction ≤ μ_s × N

Where:
- F = friction force
- μ_s = coefficient of static friction
- N = normal force

Example: Block on Incline

       [====]  m = 10 kg
      /
     /  θ = 30°
    /________

μ_s = 0.5

Will the block slide?

Forces parallel to plane:
- Component of weight: W_parallel = mg sin(θ) = 10 × 9.81 × sin(30°) = 49.05 N
- Maximum friction: F_max = μ_s × N = μ_s × mg cos(θ)
  F_max = 0.5 × 10 × 9.81 × cos(30°) = 42.5 N

Since W_parallel > F_max (49.05 > 42.5), the block slides.

Centroids and Centers of Gravity

The centroid is the geometric center of a shape. The center of gravity is where the total weight acts.

Simple Shapes

Composite Shapes

For complex shapes, divide into simple parts:

x_bar = (Σ A_i × x_i) / (Σ A_i)
y_bar = (Σ A_i × y_i) / (Σ A_i)

Where:
- A_i = area of part i
- x_i, y_i = centroid of part i

Example: L-Shaped Beam

    [====2====]   Rectangle 1: 4m × 1m
    [         ]
    [ 1       ]   Rectangle 2: 1m × 3m
    [         ]
    [=========]

Find centroid from bottom-left corner:

Rectangle 1: A_1 = 4 × 1 = 4 m², centroid at (2, 3.5)
Rectangle 2: A_2 = 1 × 3 = 3 m², centroid at (0.5, 1.5)

x_bar = (4 × 2 + 3 × 0.5) / (4 + 3) = (8 + 1.5) / 7 = 1.36 m
y_bar = (4 × 3.5 + 3 × 1.5) / 7 = (14 + 4.5) / 7 = 2.64 m

Practice Problems

Problem 1: Beam Reactions

A 6 m beam is supported at both ends. A 500 N load is applied 2 m from the left end. Find the reactions.

<details> <summary>Solution</summary>

ΣM_A = 0:  R_B × 6 - 500 × 2 = 0
           R_B = 1000/6 = 166.7 N

ΣFy = 0:   R_A + 166.7 - 500 = 0
           R_A = 333.3 N

</details>

Problem 2: Moment Calculation

A 100 N force is applied perpendicular to a wrench 0.3 m from the bolt. What is the moment on the bolt?

<details> <summary>Solution</summary>

M = F × d = 100 N × 0.3 m = 30 N⋅m

</details>

Problem 3: Friction

A 50 kg box sits on a floor (μ_s = 0.4). What horizontal force is needed to start it moving?

<details> <summary>Solution</summary>

N = mg = 50 × 9.81 = 490.5 N
F_max = μ_s × N = 0.4 × 490.5 = 196.2 N

At least 196.2 N is required to overcome static friction.

</details>

Key Takeaways

• Equilibrium: ΣF = 0 and ΣM = 0.
• Moments: M = F × d (force times perpendicular distance).
• Support reactions: different supports provide different constraints.
• Beams: sum forces and moments to find reactions.
• Trusses: method of joints or sections for member forces.
• Friction: F ≤ μ × N; static friction prevents motion.

Next Steps

Continue to 03-dynamics.md to add motion to the analysis.