JEE Mains Physics · Laws of Motion

Equilibrium of Forces

A body at rest or moving at constant velocity has zero net force, so the forces along any two perpendicular directions each add to zero.

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Why this matters

Seventeen PYQs, fourteen of them multiple choice, and one from 2026. Ten balance the strings, chains and forces that meet at a point: two strings from a ceiling, a rope pulled sideways at its middle, a hanging chain. Seven find a normal reaction or a support force: a roller pushed at an angle, a block held on a smooth incline, and three rigid bodies (a ladder, a hinged rod, a bar on a shoulder) that need moments as well.

Concept 1 of 2: Strings, chains and forces at a point

Pick the point or body that is at rest and resolve every force on it along two perpendicular lines, usually horizontal and vertical. The horizontal balance fixes how the tensions compare; the vertical balance fixes how big they are. A rope or chain can be cut anywhere: each piece is in equilibrium on its own.

Definition

At rest: $\sum F_x = 0$ and $\sum F_y = 0$ . Three forces in equilibrium form a closed triangle, head to tail.
Two strings from a ceiling at $\theta_1$ and $\theta_2$ to the horizontal: $T_1\cos\theta_1 = T_2\cos\theta_2$ and $T_1\sin\theta_1 + T_2\sin\theta_2 = mg$ . The steeper string carries more.
A horizontal force F at a rope's midpoint: the lower half carries only mg, so the upper half makes $\tan\theta = F/mg$ with the vertical.
A chain of mass m hanging between two supports at $\theta$ to the horizontal: each half's weight is held by $T\sin\theta = mg/2$ ; the tension at the lowest point is the horizontal part, $T_0 = T\cos\theta = \tfrac{mg}{2}\cot\theta$ .
A chain holding a load: the support carries the chain's weight plus the load.
The force that restores equilibrium is minus the sum of the others.

Equilibrium at a point

\sum F_x = 0,\quad \sum F_y = 0, \qquad T_0 = \frac{mg}{2}\cot\theta

Worked example

A 6 kg lamp hangs from two strings that make

30^{\circ}

and

60^{\circ}

with the ceiling. Find both tensions. (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q15Moderate

Example 1 · Laws of Motion · Equilibrium of Forces

A flexible chain of mass

m

hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is

30^{\circ}

. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is

\_\_\_\_

The lower half of the rope carries only the weight

With a sideways force at the midpoint, the rope below that point still hangs straight and holds mg. Only the upper half tilts, and its angle comes from tan θ = F/mg.

The lowest-point tension is not half the weight

At the lowest point the chain is horizontal, so its tension is the horizontal part of the support tension, (mg/2) cot θ. Half the weight is the vertical part at each support.

The steeper string carries more

A string close to vertical does most of the lifting. If an answer gives the larger tension to the flatter string, the cosines have been swapped.

Drill 9 more on strings, chains and forces at a point

Concept 2 of 2: Normal reactions, smooth inclines and supports

A surface pushes back with whatever force is needed to stop the body sinking into it, always perpendicular to the surface. So the normal reaction is not mg as soon as anything else presses or pulls. On a smooth incline the body stays still only if some force cancels the part of its weight along the slope.

Definition

N is perpendicular to the surface and takes whatever value the other forces need.
A force F at $\theta$ to a level floor: pushing down, $N = mg + F\sin\theta$ ; pulling up, $N = mg - F\sin\theta$ .
Smooth incline, block held by a horizontal force: $F\cos\theta = mg\sin\theta$ , so $F = mg\tan\theta$ and $N = mg/\cos\theta$ .
Smooth incline with a counter-mass hanging over a pulley, at rest: $m_2 g = m_1 g\sin\theta$ and $N = m_1 g\cos\theta$ .
Constant velocity is equilibrium: on a smooth slope the applied force cancels $mg\sin\theta$ .
A ladder, rod or bar also needs zero turning effect: take moments about a point where an unknown force acts. A smooth wall or shoulder pushes perpendicular to its own surface.

Normal reaction

N = mg \pm F\sin\theta, \qquad F_{\text{horizontal}} = mg\tan\theta

Worked example

A 50 kg roller on level ground is acted on by a 300 N force at

37^{\circ}

to the horizontal (

\sin 37^{\circ} = 0.6

). Find the normal reaction when the force (a) pulls upward and (b) pushes downward. (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q8Moderate

Example 2 · Laws of Motion · Equilibrium of Forces

As shown in figure, a

70\text{ }kg

garden roller is pushed with a force of

\overrightarrow{F}= 200\text{ }N

at an angle of

30^{\circ}

with horizontal. The normal reaction on the roller is (Given

g = 10\text{ }m{\text{ }s}^{- 2}

)

N is not always mg

Any force with a vertical part changes the normal reaction. A push down at an angle adds F sin θ; a pull up takes it away. Writing N = mg here gives one of the wrong options.

Constant velocity means zero net force

A body moving at steady speed is in equilibrium, exactly as if it were at rest. The applied force only cancels the other forces; it does not exceed them.

A smooth contact pushes perpendicular to the surface

A bar on a smooth shoulder or a ladder on a smooth wall feels a force at right angles to the contact surface, not straight up. Taking it as vertical changes the answer.

Drill 6 more on normal reactions, smooth inclines and supports

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Strings, chains and forces at a point
Equilibrium at a point
$\sum F_x = 0,\quad \sum F_y = 0, \qquad T_0 = \frac{mg}{2}\cot\theta$
Normal reactions, smooth inclines and supports
Normal reaction
$N = mg \pm F\sin\theta, \qquad F_{\text{horizontal}} = mg\tan\theta$

Watch out for (6)

The lower half of the rope carries only the weight
→ Strings, chains and forces at a point
The lowest-point tension is not half the weight
→ Strings, chains and forces at a point
The steeper string carries more
→ Strings, chains and forces at a point
N is not always mg
→ Normal reactions, smooth inclines and supports
Constant velocity means zero net force
→ Normal reactions, smooth inclines and supports
A smooth contact pushes perpendicular to the surface
→ Normal reactions, smooth inclines and supports

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