Forces, Free-Body Diagrams, and Constraints
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable resolving vectors into orthogonal components and applying Newton's second law, \(\vec{F} = m\vec{a}\), to a single particle. Familiarity with basic trigonometry for right-angled triangles is also assumed.
Why this matters
Free-body diagrams are the essential bridge between physical intuition and mathematical calculation.
In Newtonian mechanics, the most common source of error is not algebraic but conceptual: students often miss forces or invent fictitious ones. This lecture establishes the rigorous procedure for identifying all real forces acting on a body and distinguishing them from constraint forces that arise from physical limitations. By mastering this technique, you gain the ability to decompose complex mechanical systems into solvable vector equations, laying the groundwork for analyzing everything from simple blocks on inclines to complex rigid-body dynamics.
The Concept of a Free-Body Diagram
A free-body diagram (FBD) is a visual representation of a single object isolated from its environment. To construct one, you must first define the system of interest—typically a single particle or rigid body. Once defined, you "cut" the object free from all contacts and fields. Every interaction that was previously hidden by contact or distance must now be represented as a vector arrow drawn on the isolated object (for a particle model, from a single point). These arrows represent real forces: gravity, tension, normal reaction, and friction. Crucially, you must exclude any force that acts on a different object. For example, if a block pushes on a wall, the force of the block on the wall is not included in the FBD of the block; only the reaction force of the wall on the block is included. This isolation prevents the common error of double-counting action-reaction pairs within a single system.
Classifying Forces: Contact and Field
Forces in classical mechanics are broadly categorized into contact forces and field forces. Contact forces arise from direct physical interaction, such as the normal force \(N\) and friction \(f\). Field forces, like gravity, act at a distance. In the context of constraints, we often deal with idealized contact forces. The normal force is always perpendicular to the surface of contact and points away from the surface. It is a reactive force; it adjusts its magnitude to prevent interpenetration of bodies, up to the limit of the material's strength. Friction, conversely, acts parallel to the surface and opposes relative motion or the tendency of motion. It is crucial to distinguish between static friction, which adjusts its magnitude to prevent slipping, and kinetic friction, which has a constant magnitude proportional to the normal force.
Constraints and the Normal Force
Constraints are physical restrictions that limit the degrees of freedom of a system. A surface acting as a constraint imposes a condition that the object cannot pass through it. Mathematically, this is enforced by the normal force. Consider a block on an inclined plane at angle \(\theta\). The constraint is that the block remains on the plane. The gravitational force \(mg\) acts vertically downwards. We resolve this into components parallel and perpendicular to the plane. The perpendicular component is \(mg\cos\theta\). Since there is no acceleration perpendicular to the plane (the constraint holds), the normal force must balance this component exactly. Thus, \(N = mg\cos\theta\). This result is a direct consequence of the constraint condition and Newton's second law applied in the direction normal to the surface.
Friction and the Coulomb Model
Friction is modeled using the Coulomb approximation. Static friction \(f_s\) satisfies \(f_s \le \mu_s N\), where \(\mu_s\) is the coefficient of static friction. It takes the value necessary to prevent motion, up to its maximum limit. If the applied force exceeds this limit, the object slips, and kinetic friction \(f_k = \mu_k N\) takes over, where \(\mu_k\) is typically less than \(\mu_s\). On an inclined plane, the condition for the block to remain at rest is that the component of gravity down the slope, \(mg\sin\theta\), does not exceed the maximum static friction, \(\mu_s mg\cos\theta\). This leads to the critical angle condition \(\tan\theta \le \mu_s\). If this condition is violated, the block accelerates down the plane with \(a = g(\sin\theta - \mu_k\cos\theta)\). For a block sliding over a fixed surface, kinetic friction does negative work and dissipates mechanical energy as heat.
The derivations
Projectile Motion from Constant-Acceleration Kinematics
For a particle of constant mass \(m\) moving under a constant acceleration \(\vec{a}\), the position and velocity are linear and quadratic functions of time: \(\vec{v}(t)=\vec{v}_0+\vec{a}t\) and \(\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}t^2\). When this acceleration is the uniform gravitational field \(\vec{a}=-g\,\hat{y}\) near the Earth's surface and air resistance is neglected, a projectile launched with speed \(v_0\) at angle \(\theta\) above the horizontal traces the parabola \(y(x)=x\tan\theta-\dfrac{g\,x^2}{2v_0^2\cos^2\theta}\), with time of flight \(T=\dfrac{2v_0\sin\theta}{g}\), maximum height \(H=\dfrac{v_0^2\sin^2\theta}{2g}\), and range on level ground \(R=\dfrac{v_0^2\sin 2\theta}{g}\).
Friction: the Coulomb Model and the Inclined Plane
For a rigid block of mass \(m\) resting on a rigid plane inclined at angle \(\theta\) to the horizontal, the normal reaction is \(N = mg\cos\theta\). The block remains at rest if and only if \(\tan\theta \le \mu_s\), where \(\mu_s\) is the coefficient of static friction. If \(\tan\theta > \mu_s\), the block slides down with acceleration \(a = g(\sin\theta - \mu_k\cos\theta)\), where \(\mu_k\) is the coefficient of kinetic friction. If the block is pushed up the incline, its acceleration while moving upward is \(a = -g(\sin\theta + \mu_k\cos\theta)\). The work done by kinetic friction over a sliding distance \(d\) is \(W_f = -\mu_k mg\cos\theta\, d\), which is dissipated as thermal energy.
A common misconception is that friction always opposes the velocity of the object. In reality, friction opposes the relative motion (or tendency of relative motion) between the surfaces in contact. For example, in a car accelerating forward, the static friction on the driving wheels acts in the forward direction, providing the force that accelerates the car. The ground pushes the tyre forward, while by Newton's third law the tyre pushes the ground backward. Confusing the direction of friction with the direction of velocity leads to incorrect free-body diagrams in rolling and driving problems. C
Worked example
Problem. A block of mass \(m = 5.0\ \mathrm{kg}\) rests on a rough inclined plane with an angle of inclination \(\theta = 30^\circ\). The coefficients of friction are \(\mu_s = 0.5\) and \(\mu_k = 0.4\). Determine whether the block remains at rest. If it slides, calculate its acceleration. Take \(g = 9.8\ \mathrm{m/s^2}\).
- Check the static friction limit. Calculate the component of gravity down the slope: \(F_{g,\parallel} = mg\sin\theta = 5.0 \times 9.8 \times \sin(30^\circ) = 24.5\ \mathrm{N}\).
- Calculate the maximum static friction: \(f_{s,\text{max}} = \mu_s N = \mu_s mg\cos\theta = 0.5 \times 5.0 \times 9.8 \times \cos(30^\circ) \approx 21.2\ \mathrm{N}\).
- Compare the forces. Since \(F_{g,\parallel} > f_{s,\text{max}}\) (\(24.5 > 21.2\)), the block will slide.
- Calculate the acceleration using kinetic friction. The net force down the slope is \(F_{\text{net}} = mg\sin\theta - \mu_k mg\cos\theta\).
- Substitute values: \(a = g(\sin\theta - \mu_k\cos\theta) = 9.8(\sin(30^\circ) - 0.4\cos(30^\circ))\).
- Compute the result: \(a = 9.8(0.5 - 0.4 \times 0.866) = 9.8(0.5 - 0.3464) = 9.8(0.1536) \approx 1.51\ \mathrm{m/s^2}\).
Units check. The term \(g\) has units \(\mathrm{m/s^2}\). The trigonometric functions and coefficients are dimensionless. The result is in \(\mathrm{m/s^2}\), which is consistent with acceleration.
Three quick tests
- If the angle of the incline is increased, how does the normal force change?
Answer
It decreases, as \(N = mg\cos\theta\) and \(\cos\theta\) decreases for \(\theta \in (0, 90^\circ)\). - Can static friction do work on a sliding object?
Answer
No, by definition, static friction acts when there is no relative motion at the contact point. If the object is sliding, kinetic friction acts. - In a free-body diagram, should you include the force exerted by the object on the surface?
Answer
No, only forces acting on the object are included. The force on the surface is part of the free-body diagram of the surface, not the object.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-420 · D-422 · Threads: force, energy, matter