Common forces in mechanics

Push a stalled car and it does not move — until three or four of you push together and it lurches forward. Let go of a ball and it falls; hang it from a spring and the spring stretches until it can hold the ball still; hang it from a string and the string goes taut with exactly the pull needed. Every one of these everyday moments is the same handful of forces showing up in different disguises: weight, the normal reaction, tension, the spring force, and friction. None of them is a separate law of physics — each is just a name for how one surface, string, or field pushes or pulls on a body, and Newton's three laws tell you what happens once you have identified them correctly. This is the chapter where mechanics stops being abstract vectors on a page and becomes the actual forces you can feel: the seat pushing up on you in a bus that brakes hard, the rope biting into your palms in a tug-of-war, the friction under your shoes that lets you walk at all. Get the forces right — their direction, their size, whether they are fixed or self-adjusting — and Newton's second law, $F_{net} = ma$ , becomes a one-line calculation. Get them wrong (a normal force drawn straight up on a slope, a friction force assumed to equal $\mu N$ even when nothing is sliding) and the whole free-body diagram collapses. Here we build the toolkit properly: inertia and the first law, momentum and the second and third laws, equilibrium of concurrent forces, the common contact and field forces, friction in its static and kinetic forms, and the forces that keep a car from skidding off a curve.

Key takeaways

Newton's first law (inertia): a body stays at rest or in uniform motion unless a net external force acts on it — force is what changes motion, not what sustains it.
Newton's second law: $F_{net} = \dfrac{dp}{dt} = ma$ for constant mass, where momentum $p = mv$ ; impulse $J = F\Delta t$ equals the change in momentum.
Newton's third law: every force has an equal and opposite reaction force, but the two forces of a pair act on two different bodies — they never cancel each other out.
Conservation of momentum: with zero net external force, total momentum is constant — this is how recoil and rocket propulsion work.
The common forces: weight $W = mg$ (field force), normal reaction $N$ , tension $T$ , spring force $F = -kx$ , and friction (all contact forces, all constraint forces except weight).
Friction is self-adjusting up to a limit: static friction $f_s \le \mu_s N$ grows only as large as needed to prevent sliding; once sliding starts, kinetic friction $f_k = \mu_k N$ takes over, and $\mu_k < \mu_s$ .
Key idea: identifying the correct set of contact and field forces on a free-body diagram, and applying Newton's second law along with the self-adjusting nature of the normal force and static friction, is what turns real mechanical situations into solvable equations.

The core insight

Every mechanical problem reduces to naming the actual forces acting on a body — weight, normal reaction, tension, spring force, friction — and applying $F_{net} = ma$ , remembering that normal force and static friction are not fixed numbers but adjust themselves up to a limit to satisfy the constraints of contact and equilibrium.

Where it fits in the syllabus

This topic sits inside CBSE Class 11 Physics, Unit III, "Laws of Motion" — the chapter that follows kinematics and precedes work, energy and power. It is fully in the current 2025-26 prescription with no rationalisation cuts, and it is one of the highest-yield chapters for both board exams and JEE/NEET, because nearly every mechanics numerical (inclines, pulleys, lifts, circular tracks) is built from exactly the forces catalogued here. The chapter assumes you already know vectors and basic kinematics; it does not require calculus beyond the simple rate $dp/dt$ , and it explicitly does not extend into work-energy methods or the torque and rotational dynamics of rigid bodies — those belong to later chapters.

The physical picture

Newton's first law says a body left alone — zero net force — keeps whatever velocity it already has. This built-in reluctance to change velocity is called inertia, and mass is its numerical measure: the more massive an object, the harder it is to speed up, slow down, or turn. The second law connects force to the rate of change of momentum. Momentum is defined as

\vec{p} = m\vec{v},

where: $m$ is mass (kg) and $\vec{v}$ is velocity (m/s), so $p$ has units kg m/s and points along the velocity. Newton's second law states

\vec{F}_{net} = \frac{d\vec{p}}{dt} = m\vec{a} \quad \text{(for constant mass)}.

where: $\vec{F}_{net}$ is the vector sum of every real force acting on the body and $\vec{a}$ is the resulting acceleration. The newton is defined so that $1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2$ — the force that gives a $1\ \text{kg}$ mass an acceleration of $1\ \text{m/s}^2$ . When a force acts only briefly, its total effect is captured by the impulse,

\vec{J} = \vec{F}\,\Delta t = \Delta \vec{p},

where: $\vec{J}$ is impulse (N s, same units as momentum) and $\Delta \vec{p}$ is the change in momentum it produces — this is why a cricketer draws the bat back while catching a ball: stretching $\Delta t$ for a given $\Delta p$ reduces the peak force on the hands. Newton's third law says every force comes as a pair: if body A pushes on body B, then B pushes back on A with equal magnitude and opposite direction, along the same line. The critical, exam-losing detail is that the two forces of a pair act on different bodies, so they can never cancel each other in a single body's free-body diagram — a book resting on a table feels the table's normal force upward, while the reaction to that force acts on the table, not on the book. When several forces act on a single particle and it stays at rest or moves with constant velocity, the forces are in equilibrium: their vector sum is zero. For three concurrent forces this is often solved with Lami's theorem — each force is proportional to the sine of the angle between the other two forces — or simply by resolving along perpendicular axes and setting each sum to zero. Now the cast of common forces you will actually draw on a free-body diagram:

Weight $W = mg$ , the pull of Earth's gravity, always vertically downward — the one field (non-contact) force in this list.
Normal reaction $N$ , the push a surface exerts perpendicular to itself, wherever two surfaces touch. It is a constraint force: it takes whatever value stops the body from passing through the surface, so $N = mg$ only on a flat surface with no vertical acceleration.
Tension $T$ , the pull transmitted along a taut string, always directed along the string away from the body; for an ideal massless string over a frictionless pulley, $T$ is the same throughout.
Spring force, obeying Hooke's law $F = -kx$ , where $k$ is the spring constant (N/m) and $x$ is the displacement from natural length — the minus sign shows the force always restores the spring toward its natural length.
Friction, the contact force parallel to a surface opposing relative sliding, self-adjusting for static contact and fixed at $\mu_k N$ once sliding.

Beyond these everyday forces, physics recognises four fundamental forces — gravitational, electromagnetic, strong nuclear, and weak nuclear — and everything above (normal force, tension, friction, the spring force) is ultimately electromagnetic in origin, arising from the interaction of atoms in contact.

Figure

A free body diagram of a block on an inclined surface showing weight acting straight down split into components along and perpendicular to the slope, normal reaction perpendicular to the surface, and friction acting up the slope opposing sliding

Derivation AAngle of repose and the limit of static friction

The goal of this derivation is to find the steepest angle at which a block can sit on an incline without sliding, and to show that this angle depends only on the coefficient of static friction — not on the mass of the block, the size of the incline, or anything else. This single result, called the angle of repose, is the cleanest place to see weight, normal reaction, and friction working together as one equilibrium problem, which is exactly why it belongs in a chapter on common forces rather than in the friction chapter alone.

Step 1. Set up the free-body diagram. A block of mass $m$ rests on an incline of angle $\theta$ . Three forces act on it: weight $mg$ vertically down, normal reaction $N$ perpendicular to the incline surface, and friction $f$ acting up the slope (since the block tends to slide down). Resolve the weight into components along the incline ( $mg\sin\theta$ , down the slope) and perpendicular to it ( $mg\cos\theta$ , into the surface).

Step 2. Apply equilibrium perpendicular to the incline. Since the block does not accelerate into or out of the surface, $N = mg\cos\theta$ . This is the same result used whenever a block sits on a slope, and it already shows that $N$ is not simply $mg$ — it shrinks as the incline steepens.

Step 3. Apply equilibrium along the incline. For the block to remain at rest, friction must exactly balance the sliding component of weight: $f = mg\sin\theta$ . But static friction cannot exceed its maximum value, $f \le \mu_s N$ . The block is on the verge of sliding exactly when $f$ reaches its ceiling, $f = \mu_s N = \mu_s mg\cos\theta$ .

Step 4. Combine the two equilibrium equations at the point of slipping. Setting $mg\sin\theta = \mu_s mg\cos\theta$ , the mass $m$ and $g$ cancel completely, leaving $\tan\theta = \mu_s$ . This critical angle is called the angle of repose, $\lambda$ , and it satisfies $\tan\lambda = \mu_s$ — identical in form to the angle of friction, since at the point of slipping the resultant of $N$ and $f$ makes angle $\lambda$ with the normal. For any angle less than $\lambda$ the block stays put; for any angle greater than $\lambda$ , gravity's component along the slope exceeds the maximum friction can supply, and the block slides.

Worked examples

Board level - will the block on the incline slide?

A block of mass $5.0\ \text{kg}$ rests on an incline of angle $30^\circ$ with the horizontal. The coefficient of static friction between block and incline is $\mu_s = 0.5$ . Take $g = 9.8\ \text{m/s}^2$ . Determine whether the block remains at rest, and find the friction force actually acting on it.

Step 1. Find the weight and its components. $W = mg = 5.0 \times 9.8 = 49\ \text{N}$ . Along the slope, the sliding component is $W\sin 30^\circ = 49 \times 0.5 = 24.5\ \text{N}$ ; perpendicular to the slope, $W\cos 30^\circ = 49 \times 0.866 = 42.4\ \text{N}$ .

Step 2. Find the normal reaction. Since there is no acceleration perpendicular to the incline, $N = W\cos 30^\circ = 42.4\ \text{N}$ — smaller than the full weight, exactly as expected on a slope.

Step 3. Find the maximum available static friction. $f_{s,\max} = \mu_s N = 0.5 \times 42.4 = 21.2\ \text{N}$ . Compare this with the force friction would need to supply to hold the block at rest: $24.5\ \text{N}$ .

Step 4. Compare and conclude. Since the required friction ( $24.5\ \text{N}$ ) exceeds the maximum available static friction ( $21.2\ \text{N}$ ), static friction cannot hold the block, and it slides down. Once sliding, kinetic friction takes over and (with $\mu_k$ slightly less than $\mu_s$ ) the block accelerates down the slope. Answer: the block slides; the friction actually present at the threshold is capped at $21.2\ \text{N}$ , less than the $24.5\ \text{N}$ needed for equilibrium.

Competitive level - maximum safe speed on a banked curve

A circular curve of radius $r = 100\ \text{m}$ is banked at $\theta = 15^\circ$ . The coefficient of static friction between tyres and road is $\mu_s = 0.3$ . Take $g = 10\ \text{m/s}^2$ and $\tan 15^\circ = 0.268$ . Find (a) the optimum speed at which no friction is needed, and (b) the maximum speed before the car begins to skid outward.

Step 1. Find the optimum (frictionless) speed. On a banked road, if friction is not required, the horizontal component of the normal force alone supplies the centripetal force, giving $\tan\theta = v^2/(rg)$ , so $v_{opt} = \sqrt{rg\tan\theta} = \sqrt{100 \times 10 \times 0.268} = \sqrt{268} \approx 16.4\ \text{m/s}$ .

Step 2. Set up the maximum-speed case, where friction acts down the slope (helping to prevent outward skidding) at its maximum value $\mu_s N$ . The standard result for the maximum speed on a banked road with friction is $v_{\max} = \sqrt{\dfrac{rg(\mu_s + \tan\theta)}{1 - \mu_s\tan\theta}}$ .

Step 3. Substitute the numbers. The numerator inside the root is $rg(\mu_s + \tan\theta) = 100 \times 10 \times (0.3 + 0.268) = 1000 \times 0.568 = 568$ . The denominator is $1 - \mu_s\tan\theta = 1 - (0.3 \times 0.268) = 1 - 0.080 = 0.920$ .

Step 4. Divide and take the square root. $v_{\max}^2 = 568/0.920 = 617.4$ , so $v_{\max} = \sqrt{617.4} \approx 24.8\ \text{m/s}$ , about $89\ \text{km/h}$ . Answer: $v_{opt} \approx 16.4\ \text{m/s}$ with no friction required, and $v_{\max} \approx 24.8\ \text{m/s}$ is the fastest the car can take the curve before friction is exhausted and it skids outward. Timing note: aim for under three minutes. Memorise the two banked-road formulas as a pair — optimum speed drops friction entirely, maximum speed adds $\mu_s$ in the numerator and subtracts $\mu_s\tan\theta$ from the denominator. A quick sanity check: $v_{\max}$ must always come out larger than $v_{opt}$ , since friction only adds to what the banking alone can provide; if your arithmetic gives the opposite, recheck a sign.

Common traps

Treating an action-reaction pair as if it acts on one body and cancels out. The block's weight and the ground's normal reaction on the block are not a Newton's third law pair — they act on the same body and happen to be equal only in special cases. A true third-law pair always acts on two different bodies and never appears together in one body's equation of motion.
Assuming static friction always equals $\mu_s N$ . Static friction is a self-adjusting inequality, $f_s \le \mu_s N$ ; it equals $\mu_s N$ only at the exact point of slipping. Below that, friction supplies exactly whatever is needed to keep the body at rest, nothing more.
Drawing the normal force vertically instead of perpendicular to the actual surface. On an incline $N$ tilts with the slope; only on a horizontal surface does it point straight up.
Adding a "centrifugal force" as if it were real in an inertial (ground) frame. In an inertial frame the only forces on a car turning a curve are gravity, normal reaction, and friction; their resultant supplies the real centripetal force toward the centre. A pseudo-force pointing outward belongs only to a rotating (non-inertial) frame, and this chapter's inertial-frame treatment does not use it.
Forgetting momentum conservation applies only with zero net external force. Recoil and rocket problems work because the internal explosive or thrust forces are equal and opposite internal forces; if an external force (like gravity, over a long time) is significant, momentum is not exactly conserved.

You should be able to…

State Newton's three laws precisely, and correctly identify the two bodies involved in any third-law force pair.
Write $F_{net} = ma$ (from $dp/dt$ ) and impulse $J = F\Delta t = \Delta p$ , and apply them to collisions and time-varying forces.
Draw a correct free-body diagram for a body on an incline, in a lift, or on a banked curve, showing only the forces that actually act on it.
Distinguish contact forces (normal, tension, spring, friction) from the field force of weight, and know that normal force and static friction are self-adjusting constraint quantities, not fixed values.
Apply the static/kinetic friction inequality $f_s \le \mu_s N$ and $f_k = \mu_k N$ , and compute the angle of repose $\tan\lambda = \mu_s$ .
Set up and solve level and banked circular-motion problems using $v_{\max} = \sqrt{\mu_s r g}$ (level road) or the banked-road formulas, staying strictly in the inertial frame.

Ask the tutor: stuck on identifying which forces belong in a free-body diagram, or want another banked-curve numerical worked through step by step — open the AI tutor from this page and ask.

Why does the normal force change on an incline instead of just staying equal to the weight?

The normal force is not a fixed property of a body — it is a constraint force that the surface generates only in the amount needed to stop the body from passing through it. On a flat, non-accelerating surface, that amount happens to equal the full weight, $N = mg$ . Tilt the surface and only the perpendicular component of weight, $mg\cos\theta$ , needs balancing, because the other component, $mg\sin\theta$ , acts along the surface and does not press the block into it. So $N = mg\cos\theta$ shrinks as the incline steepens, until at a vertical wall $N$ would drop to zero.

If static friction can be anything up to μs N, how do I know what value to actually use in a problem?

You never assume static friction equals $\mu_s N$ unless the problem tells you the body is on the verge of sliding. Otherwise, you find the friction force by writing the equilibrium equation for the body and solving for whatever friction is required to keep it at rest — then, as a final check, you compare that required value against the ceiling $\mu_s N$ . If the required friction is less than or equal to $\mu_s N$ , the body stays still and that is the actual friction. If it exceeds $\mu_s N$ , the body cannot stay still, and you switch to kinetic friction $f_k = \mu_k N$ for the resulting motion.

Why is kinetic friction usually less than the maximum static friction?

Once two surfaces are sliding relative to each other, the microscopic high points (asperities) on each surface do not have time to interlock and bond as deeply as they do when the surfaces are momentarily at rest against each other. Static friction can build up stronger microscopic adhesion because the contact points settle in over time. This is why it typically takes a bigger push to start a heavy box moving than to keep it moving once it is already sliding — the coefficient drops from $\mu_s$ to the smaller $\mu_k$ the instant slipping begins.

Is the "centrifugal force" that pushes me outward in a turning car a real force?

Not in the frame of a stationary observer watching from the roadside — which is the inertial frame this chapter uses. In that frame, the only real forces on you are gravity, the seat's normal reaction, and friction, and their net effect is a centripetal force pointing toward the centre of the curve, which is what actually bends your path. The outward push you feel is your own body's inertia — Newton's first law — resisting the change in direction the car is forcing on you; it is not a separate force pulling you outward. If you were analysing the problem from inside the spinning car's own (non-inertial) frame, you would need to invent a fictitious centrifugal force to make the equations balance, but that is a bookkeeping device for that frame, not a real force acting on you.