Interactive Lecture
Understanding Forces and Constraints
by Dr Arnaldo Delli Carri
A Free Body Diagram (FBD) is a graphical representation used to visualise all the forces acting on a body. It's a fundamental tool in engineering mechanics that helps:
Key Principle: A free body diagram shows a body isolated from its surroundings with all external forces and moments acting upon it.
Internal forces and kinematic quantities such as displacements, velocities, accelerations are NEVER shown on FBDs.
Active forces are externally applied loads that act on a structure or body. These include:
Point Forces: Forces applied at a specific point
Notation: \(\vec{F} = F_x \hat{i} + F_y \hat{j}\) (in 2D)
Point Moments: Couples or torques applied at a point
Notation: \(M\hat{k}\) (anticlockwise positive in 2D)
Uniformly Distributed Load (UDL):
Load intensity: \(w\) (force per unit length) \([\text{ N/m}]\)
Resultant force: \(F = w \cdot L\)
Location: Acts at the centroid of the distribution
Linearly Varying Load:
Load intensity: \(w(x) = w_0 + \frac{w_1 - w_0}{L}x\)
Resultant: \(F = \frac{1}{2}(w_0 + w_1) \cdot L\)
Weight force: \(\vec{W} = m\vec{g}\)
Acts at the centre of mass of the body
Always acts vertically downward
Magnitude: \(W = mg\) where \(g = 9.81 \text{ m/s}^2\)
Reactive forces are the forces that develop at supports or constraints in response to applied loads. You need to identify them correctly before any equilibrium equation makes sense.
This simulation demonstrates a classic equilibrium problem: a ladder leaning against a smooth (frictionless) wall, with its base on rough (friction) ground. A person stands on the ladder. Explore how friction at the ground prevents the ladder from slipping.
Acts perpendicular to the contact surface
Always pushes (never pulls)
Magnitude: \(N \geq 0\)
Static Friction: \(f_s \leq \mu_s N\)
Kinetic Friction: \(f_k = \mu_k N\)
Where \(\mu_s\) and \(\mu_k\) are coefficients of friction
Acts parallel to the contact surface, opposing relative motion
Can only pull (tension), never push
Acts along the line of the cable/rope
Magnitude: \(T \geq 0\)
For massless cables: tension is constant throughout
Hooke's Law: \(F = k \Delta x\)
Where \(k\) is the spring constant (stiffness) and \(\Delta x\) is the extension/compression
Springs can both pull (tension) and push (compression)
Force is proportional to deformation from natural length
This simulation shows a horizontal pinned beam with weight (distributed uniformly along its length), lifted by a cable that wraps around a pulley and goes horizontally to a wall. The cable force must equal half the beam weight to maintain equilibrium.
This simulation demonstrates a traffic light suspended by two springs. It shows how spring forces follow Hooke's Law, and how the springs stretch and settle at whatever angle balances the traffic light's weight against the vertical components of the spring tensions.
For a body in static equilibrium, the following conditions must be satisfied:
Force Equilibrium:
\(\sum F_x = 0\) (Sum of horizontal forces = 0)
\(\sum F_y = 0\) (Sum of vertical forces = 0)
Moment Equilibrium:
\(\sum M_O = 0\) (Sum of moments about any point O = 0)
When a body is in equilibrium under exactly two forces:
When a body is in equilibrium under exactly three forces:
Degree of Static Indeterminacy:
DSI = Number of unknowns - Number of equilibrium equations
If DSI = 0: Statically determinate (can solve using statics alone)
If DSI > 0: Statically indeterminate (requires additional compatibility equations)
If DSI < 0: Mechanism (unstable structure)
A free body diagram is the first step in almost any force or stress calculation. Here is where the technique earns its keep:
| Concept | Equation | Description |
|---|---|---|
| Horizontal force equilibrium | \(\sum F_x = 0\) | \(F_x\) = horizontal forces |
| Vertical force equilibrium | \(\sum F_y = 0\) | \(F_y\) = vertical forces |
| Moment equilibrium | \(\sum M_O = 0\) | \(M_O\) = moments about point O |
| Gravitational force (weight) | \(W = mg\) | \(m\) = mass; \(g = 9.81\) m/s² |
| UDL resultant force | \(F = w \cdot L\) | \(w\) = load intensity; \(L\) = length |
| Static friction (maximum) | \(f_s \leq \mu_s N\) | \(\mu_s\) = coefficient of static friction; \(N\) = normal force |
| Kinetic friction | \(f_k = \mu_k N\) | \(\mu_k\) = coefficient of kinetic friction |
| Spring force (Hooke's Law) | \(F = k \Delta x\) | \(k\) = spring constant; \(\Delta x\) = extension/compression |
| Degree of static indeterminacy | \(\text{DSI} = \text{unknowns} - \text{equations}\) | DSI = 0 (determinate); DSI > 0 (indeterminate) |
A 6-metre beam is simply supported at both ends. It carries a uniformly distributed load of 10 kN/m and a point load of 20 kN at 2 metres from the left support. Draw the free body diagram and calculate the support reactions.
Show Solution
Given:
FBD includes:
Equilibrium equations:
Taking moments about A:
\(\sum M_A = 0\): \(R_B \times 6 - 20 \times 2 - 60 \times 3 = 0\)
\(R_B = \frac{40 + 180}{6} = 36.67 \text{ kN}\)
Vertical equilibrium:
\(\sum F_y = 0\): \(R_A + R_B - 20 - 60 = 0\)
\(R_A = 80 - 36.67 = 43.33 \text{ kN}\)
A 4-metre cantilever beam is fixed at one end and free at the other. It carries a linearly varying load from 0 at the fixed end to 12 kN/m at the free end. Find the reactions at the fixed support.
Show Solution
Given:
Load resultant:
Magnitude: \(F = \frac{1}{2} \times 12 \times 4 = 24 \text{ kN}\)
Location: \(\frac{2}{3} \times 4 = 2.67 \text{ m from fixed end}\)
Reactions at fixed support:
Vertical reaction: \(R_y = 24 \text{ kN (upward)}\)
Horizontal reaction: \(R_x = 0\) (no horizontal loads)
Moment reaction: \(M = 24 \times 2.67 = 64 \text{ kN·m (anticlockwise)}\)
A 100 kg block rests on a 30° incline. The coefficient of static friction is 0.3. Draw the FBD and determine if the block will slide.
Show Solution
Given:
Forces on FBD:
Force components:
Perpendicular to incline: \(N = W \cos 30° = 981 \times 0.866 = 849.5 \text{ N}\)
Parallel to incline: \(W \sin 30° = 981 \times 0.5 = 490.5 \text{ N}\)
Maximum static friction:
\(f_{max} = \mu_s N = 0.3 \times 849.5 = 254.9 \text{ N}\)
Conclusion:
Since \(W \sin 30° = 490.5 \text{ N} > f_{max} = 254.9 \text{ N}\)
The block will slide down the incline.
A uniform 5-metre ladder of mass 20 kg leans against a smooth vertical wall at an angle of 60° from the horizontal ground. A person of mass 80 kg stands at 70% of the ladder's length from the bottom. If the coefficient of static friction at the ground is 0.4, will the ladder slip?
Show Solution
Given:
Forces on FBD:
Equilibrium equations:
Taking moments about the base (anticlockwise positive):
\(\sum M_{base} = 0\): \(N_w L \sin 60° - W_L \frac{L}{2} \cos 60° - W_P (0.7L) \cos 60° = 0\)
\(N_w \times 5 \times 0.866 - 196.2 \times 2.5 \times 0.5 - 784.8 \times 3.5 \times 0.5 = 0\)
\(4.33 N_w - 245.25 - 1373.4 = 0\)
\(N_w = \frac{1618.65}{4.33} = 373.9 \text{ N}\)
Horizontal equilibrium:
\(\sum F_x = 0\): \(f - N_w = 0\) → \(f = 373.9 \text{ N}\)
Vertical equilibrium:
\(\sum F_y = 0\): \(N_g - W_L - W_P = 0\) → \(N_g = 196.2 + 784.8 = 981 \text{ N}\)
Check friction:
\(f_{max} = \mu_s N_g = 0.4 \times 981 = 392.4 \text{ N}\)
Since \(f = 373.9 \text{ N} < f_{max} = 392.4 \text{ N}\)
Conclusion:
The ladder will NOT slip. The required friction force is less than the maximum available.
A 25 kg traffic light is suspended by two cables attached at angles of 30° and 45° from the horizontal. Calculate the tension in each cable.
Show Solution
Given:
Forces on traffic light FBD:
Equilibrium equations:
Horizontal equilibrium:
\(\sum F_x = 0\): \(T_1 \cos 30° - T_2 \cos 45° = 0\)
\(0.866 T_1 = 0.707 T_2\) ... (1)
Vertical equilibrium:
\(\sum F_y = 0\): \(T_1 \sin 30° + T_2 \sin 45° - W = 0\)
\(0.5 T_1 + 0.707 T_2 = 245.25\) ... (2)
From equation (1): \(T_1 = \frac{0.707}{0.866} T_2 = 0.816 T_2\)
Substituting into equation (2):
\(0.5 \times 0.816 T_2 + 0.707 T_2 = 245.25\)
\(0.408 T_2 + 0.707 T_2 = 245.25\)
\(1.115 T_2 = 245.25\)
\(T_2 = 220.0 \text{ N}\)
Therefore: \(T_1 = 0.816 \times 220.0 = 179.5 \text{ N}\)
Answer:
Cable 1 tension: 179.5 N
Cable 2 tension: 220.0 N