Chapter 1 · Toolkit 3 of 3Basics for textbook sections 1.6–1.12

Forces, Newton's Laws & Motion Under Constant Force

An electric charge feels a force, and a force makes things accelerate, turn or stay balanced. Mechanics has the same story. Coulomb's law looks almost exactly like Newton's law of gravitation and obeys Newton's third law. An electron in a uniform field moves like a ball thrown sideways under gravity. A dipole in a field feels a torque (ടോർക്ക് (ബലാഘൂർണം)), just like a spanner. This page revises the Class 11 mechanics you need and points to where each idea returns in this chapter.

📖 What you need

1. Force and Newton's laws

A force is a push or a pull. It is a vector (സദിശം), measured in newtons (N). Newton's three laws of motion:

  1. First law (inertia): a body stays at rest or keeps moving in a straight line at constant speed unless a net force acts on it.
  2. Second law: the net force equals mass × acceleration, F = ma, and the acceleration is along the net force. A given force gives a light body a big acceleration and a heavy body a small one: a = F/m.
  3. Third law: forces come in pairs. If body 1 pushes body 2 with F21, body 2 pushes body 1 with F12 = −F21: equal in size, opposite in direction, and acting on different bodies.

The electric force between two charges obeys the third law (section 1.6). An electron and a proton attract each other with equal forces. Yet the electron, almost 2000 times lighter, accelerates almost 2000 times more.

🧪 Action and reaction – two charged pucks push apart

The force arrows are always the same length: F12 = −F21, whatever the masses. The accelerations are F/m₁ and F/m₂, so the lighter puck races away faster. To keep things simple the force is held constant here; real electric forces weaken as the pucks separate. The motion runs for 2 s in real time on frictionless ice.

Worked example 1 · F = ma

Q. A net force of 10 N acts on a 2 kg trolley. Find its acceleration and its speed after 3 s if it starts from rest.

a = F/m. Then use v = u + at with u = 0.

🎬 Exam layout, step by step
Answer: a = 10 ÷ 2 = 5 m/s²; v = 0 + 5 × 3 = 15 m/s.

2. Gravitation: the first inverse square law

Newton's law of gravitation: any two masses attract each other with a force

F = G m₁m₂ ÷ r²     G = 6.67 × 10−11 N m² kg−2

Double the distance and the force becomes one quarter; triple it and it becomes one ninth. This is the inverse square law (വിപരീത വർഗ നിയമം). Coulomb's law, F = k q₁q₂/r², has exactly the same form, with charges in place of masses and k = 9 × 10⁹ N m² C−2 in place of G. Three big differences: gravity only attracts, but electric forces can attract or repel. Gravity cannot be blocked or cancelled, but positive and negative charges can cancel. And the electric force is enormously stronger. Between an electron and a proton it is about 10³⁹ times the gravitational force (Example 1.4).

🧪 The inverse square law

Two students of 50 kg standing 1 m apart attract each other with less than a millionth of a newton. That is why we never notice gravity between everyday objects. Move r from 1 m to 2 m: the force falls to ¼. Move it to 3 m: it falls to 1/9.

Worked example 2 · Gravity between two students

Q. Two students, each of mass 50 kg, stand 1 m apart. Find the gravitational force between them. What would it be at 2 m?

F = G m₁m₂/r². For 2 m, use the inverse square law: twice the distance, one quarter of the force.

🎬 Exam layout, step by step
Answer: F = 6.67 × 10−11 × 50 × 50 ÷ 1² ≈ 1.67 × 10−7 N; at 2 m it is ¼ of that, ≈ 4.2 × 10−8 N.

3. Motion under a constant force

When the force on a body is constant, its acceleration a = F/m is constant too. Starting with velocity u along the direction of the force, after time t:

v = u + at   ·   s = ut + ½at²   ·   v² = u² + 2as

Projectile: a ball thrown horizontally with speed vₓ keeps that horizontal speed, since no horizontal force acts. Meanwhile it falls with acceleration g: in time t it goes x = vₓt sideways and y = ½gt² down. Eliminating t gives a parabola:

y = g x² ÷ (2vₓ²)

An electron entering the space between two charged plates is in a uniform electric field (ഏകീകൃത വൈദ്യുത മണ്ഡലം). It feels a constant force eE, so its acceleration is a = eE/m. It traces exactly the same parabola with g replaced by eE/m (exercise 1.33). The cathode-ray tube and the ink-jet printer use this.

🧪 Same parabola, two worlds

Ball: g = 9.8 m/s², shown in real time. Electron: e/m = 1.76 × 10¹¹ C/kg, the plates are 5 cm long and 1 cm apart, and the speed is in units of 10⁷ m/s. The electron crosses the plates in a few nanoseconds, so its flight is shown hugely slowed down. The electron bends towards the positive plate, opposite to the field, because its charge is negative.

Worked example 3 · A ball rolls off a table

Q. A ball rolls off a table with a horizontal speed of 5 m/s and lands 0.4 s later. Find (a) the height of the table and (b) how far from the table it lands. (g = 9.8 m/s²)

Vertically: it starts with zero vertical speed, so y = ½gt². Horizontally: constant speed, so x = vₓt.

🎬 Exam layout, step by step
Answer: (a) y = ½ × 9.8 × 0.4² = 0.784 m; (b) x = 5 × 0.4 = 2 m.

4. Torque and couples

A force applied away from a pivot can turn a body. The turning effect is the torque (ടോർക്ക് (ബലാഘൂർണം)):

τ = r × F   ·   τ = rF sin θ   (unit N m)

Here r is the distance from the pivot to the point where the force acts, and θ is the angle between r and F. You get the biggest turning effect by pushing at right angles (θ = 90°) and far from the pivot. Pushing straight along the spanner (θ = 0°) does nothing. That is why door handles are placed far from the hinges.

A couple is a pair of equal and opposite forces acting along different lines. The net force is zero, so the body does not move off, but it turns. An electric dipole in a uniform field is exactly a couple: +q feels qE one way and −q feels qE the other way. The torque on it is τ = pE sin θ (section 1.12).

🧪 Spanner and couple

The torque is largest at θ = 90° and zero at 0° and 180°, when the force points along the spanner, straight through the pivot. The couple on the right has the same F and arm, and the same sin θ dependence. That is the dipole of section 1.12 in disguise.

Worked example 4 · Tightening a nut

Q. A mechanic pulls a spanner with a force of 20 N, 25 cm from the nut. Find the torque when the force is (a) at 90° to the spanner and (b) at 30° to it.

τ = rF sin θ with r = 0.25 m. sin 90° = 1 and sin 30° = 0.5.

🎬 Exam layout, step by step
Answer: (a) τ = 0.25 × 20 × 1 = 5 N m; (b) τ = 0.25 × 20 × 0.5 = 2.5 N m.

5. Equilibrium: stable and unstable

A body is in equilibrium (സന്തുലനം (സ്ഥിര/അസ്ഥിര)) when the net force on it (and the net torque) is zero. But what happens after a small push?

Exercise 1.32 asks whether a charge can rest in stable equilibrium in an electrostatic field. Gauss's law answers no. A charge midway between two equal charges is in equilibrium, but that equilibrium is unstable. The same goes for a dipole pointing against the field.

🧪 Nudge the marble

The green arrow is the net force on the marble after the nudge. It points back to the middle in the bowl (a restoring force: stable) and away from the top on the hill (unstable). On the flat table it is zero (neutral).

⭐ Key ideas

6. Check yourself

Next: Electric charge – the chapter begins ▶