Common Forces in Mechanics — Weight, Normal, Tension, Spring (NCERT 4.9)

1. The Working Force Catalogue

Every free-body diagram in school mechanics is built from a short catalogue: weight (gravity's pull), the normal reaction (a surface's perpendicular push), tension (a string's pull along its length), the spring force (a restoring pull or push), and — so important it gets its own page — friction, the tangential part of contact forces. The laws of the last pages tell you how forces govern motion; this page tells you what the forces are, which direction each one points, and — critically — which of them adjusts itself to circumstances. Two of the catalogue entries are fixed by nature (weight is mg, spring force is kx), and two are response forces whose magnitudes the situation decides (normal and static friction). That fixed-versus-responsive distinction is the conceptual spine of every FBD question NTA asks.

Approach the catalogue as a drawing discipline, not a memory list: for each force know (i) its source body, (ii) its direction rule, (iii) its magnitude law, and (iv) whether it can switch off. Weight: source Earth, always down, mg, never switches off. Normal: source contact surface, perpendicular to surface, "whatever is needed", can vanish (contact lost). Tension: source string, along string away from body, "whatever is needed" (set by the second law), vanishes when the string goes slack. Friction: source surface, along surface opposing relative sliding, up to μ_sN, vanishes without pressing (its full treatment lives on the friction page).

2. Complete Theory: Each Force, Ruled and Quantified

Weight. W = mg, Earth's gravitational pull, directed vertically down, acting (for bookkeeping) at the body's centre of gravity. Its value is set by location — g shrinks with altitude and latitude, the business of Gravitation — but in any single problem it is a fixed, known vector. The third-law partner of a body's weight is its equal pull on Earth, not the normal force on it (a distinction policed on the third-law page).

Normal reaction — five scenarios NTA cycles. The normal is the surface's perpendicular response to being pressed; it equals whatever the perpendicular axis's second-law equation demands. (1) Horizontal surface, no other vertical forces: N = mg. (2) Incline of angle θ: N = mg cos θ — less than mg, because the surface only answers the perpendicular component. (3) Force F pushing down at an angle θ below horizontal on a horizontal floor: N = mg + F sin θ — pressing increases N. (4) Force F pulling up at an angle θ: N = mg − F sin θ — pulling decreases N; if F sin θ = mg, N hits zero and the body lifts off. (5) Lift accelerating vertically with a: N = m(g + a) upward-accelerating, N = m(g − a) downward-accelerating, N = 0 in free fall (the apparent-weight family, tabulated on the hub). Scenario 3 matters doubly: since friction scales with N, pressing a block harder also makes it harder to slide — a two-layer question NTA loves.

Tension and the ideal string–pulley system. A string under tension pulls each end along itself, away from the body it holds — strings can never push. For a light (massless) string, Newton's second law applied to any segment demands equal pulls at both ends; over a frictionless, massless pulley the same argument makes T identical throughout — the pulley bends the force, it does not shrink it. Two connected bodies then share one tension and one speed (string length is constant), the pair of facts that makes every table-block-pulley system solvable; the worked systems sit on the solving-problems page. When a string is cut, its tension vanishes instantly — a favourite trick in "find the acceleration just after cutting" questions, where the surviving forces must suddenly accelerate the body.

Spring force (Hooke's law). Within its elastic limit a spring fights displacement from natural length with F = kx, pointing back toward natural length (vector form F = −kx). The constant k (N m⁻¹) measures stiffness; it is a property of the spring, not the situation. Equilibrium hanging gives kx = mg — the standard bridge between the catalogue and the second law. Springs differ from strings in one exam-relevant way: a spring's force cannot change instantaneously (the extension needs time to change), so "just after cutting" questions that would zero a string's tension leave a spring's force intact. Springs store energy too — the ½kx² account belongs to Work, Energy and Power.

Contact forces, decomposed. The total contact interaction between surface and body is one force; we habitually resolve it into a perpendicular part (the normal) and a tangential part (friction). This picture explains both response forces at once: press harder and the whole contact force grows — its perpendicular part (more N) and with it the friction ceiling (μ_sN). Nothing in mechanics is "just N" or "just friction"; they are the two projections of one push.

3. Visualising the Catalogue on One Block

4. Solved Examples

5. Practice Questions

Q1. A 60 kg woman stands on a weighing scale in a lift. Find the reading (in newtons and in "kg") when the lift (i) ascends at constant 5 m/s, (ii) accelerates up at 2 m s⁻², (iii) accelerates down at 2 m s⁻², (iv) free-falls (g = 10 m s⁻²).
ANSWER: (i) N = mg = 600 N → 60 kg (constant velocity changes nothing); (ii) N = m(g + a) = 60 × 12 = 720 N → 72 kg; (iii) N = m(g − a) = 60 × 8 = 480 N → 48 kg; (iv) N = 0 → 0 kg — weightless, though mg still acts. Only acceleration writes the reading; velocity is invisible to the scale.

Q2 (MCQ). A light string passes over a frictionless pulley connecting a 3 kg and a 5 kg mass. The tension is: (a) 50 N (b) 30 N (c) between 30 and 50 N, same on both sides (d) different on the two sides
ANSWER: (c) — one ideal string means one T; the second-law system on the solving-problems page gives T = 2m₁m₂g/(m₁+m₂) = 2 × 15 × 10/8 = 37.5 N, which lies between the two weights (30 and 50 N). Options (a) and (d) break the ideal-string rule.

Q3. A 10 kg block on a floor is pulled by a 50 N rope at 53° above the horizontal. Find N and the horizontal acceleration if the floor is frictionless (g = 10 m s⁻²).
ANSWER: Vertical: N = mg − F sin 53° = 100 − 50 × 0.8 = 60 N. Horizontal: a = F cos 53°/m = 50 × 0.6/10 = 3 m s⁻². The pull carries 40 N of the weight — raise the angle to 53°→ higher and N keeps shrinking; at F sin θ = 100 N the block would lift off with N = 0.

6. Key Formulas & Takeaways

Scenario formulas: N = mg; mg cos θ; mg + F sin θ; mg − F sin θ; m(g + a); m(g − a); 0. Each is just the vertical (or perpendicular) second-law equation solved for N — never memorise without the axis sketch that generates it.

7. Frequently Asked Questions