Concept Card: Force, Inertia & Newton’s First Law of Motion
1. Physical Definition and Nature of Force:
A force is an external vector interaction (push or pull) that changes or tends to change an object’s state of rest or of uniform rectilinear motion, or causes deformation.
Forces in nature are classified fundamentally into four categories:
- Gravitational Force: Universal, always attractive, long-range, weakest fundamental force ($\sim 10^{-38}$ relative to strong nuclear).
- Electromagnetic Force: Operates between charged particles; attractive or repulsive; long-range. Dominates atomic structures, contact interactions (normal reaction, friction, string tension, elasticity).
- Strong Nuclear Force: Short-range ($\sim 10^{-15}\text{ m}$); binds protons and neutrons within the nucleus; charge-independent.
- Weak Nuclear Force: Short-range ($\sim 10^{-18}\text{ m}$); governs radioactive beta decay and neutrino interactions.
2. The Concept and Types of Inertia:
Inertia is the inherent resistance of any material body to any change in its velocity (state of rest or uniform straight-line motion).
Quantitative Measure: Mass ($m$) is the scalar measure of translational inertia. Greater mass implies greater inertia.
- Inertia of Rest: Tendency of a body to remain stationary (e.g., rider falling backward when a horse accelerates forward; dust falling from a beaten carpet; a coin dropping into a glass when the card below it is flicked).
- Inertia of Motion: Tendency of a body to persist in uniform motion (e.g., passengers lurching forward upon sudden braking of a vehicle; an athlete running before a long jump).
- Inertia of Direction: Tendency of a body to maintain its direction of motion (e.g., mud thrown off a rotating bicycle wheel tangentially; passengers leaning outwards when a car negotiates a curved bend).
3. Newton’s First Law of Motion (Galileo’s Principle of Inertia):
Formal Statement: Every material body perseveres in its state of rest, or of uniform motion in a straight line, unless compelled to change that state by an unbalanced external force.
Mathematical condition for equilibrium:
$$\sum \vec{F}_{\text{ext}} = \vec{0} \iff \vec{a} = \vec{0} \iff \vec{v} = \text{constant}$$
Component representation:
$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$
Significance:
1. Provides the qualitative definition of force: Force is the causal agency of acceleration ($\vec{a} \neq \vec{0}$).
2. Defines inertial frames of reference.
4. Reference Frames: Inertial vs. Non-Inertial:
- Inertial Frame: A non-accelerating reference frame ($\vec{a}_{\text{frame}} = \vec{0}$) in which Newton’s laws hold true without modification. Any frame in uniform linear translation relative to an inertial frame is also inertial.
- Non-Inertial Frame: An accelerated or rotating frame ($\vec{a}_{\text{frame}} \neq \vec{0}$). To apply Newton’s laws in a non-inertial frame, one must introduce an apparent pseudo-force (inertial force):
$\vec{F}_{\text{pseudo}} = -m\vec{a}_{\text{frame}}$
directed opposite to the acceleration of the frame.
5. Translational Equilibrium & Lami’s Theorem:
A particle is in translational equilibrium if the net external force acting on it is zero ($\vec{F}_{\text{net}} = \vec{0}$).
Lami’s Theorem: If three coplanar concurrent forces $\vec{F}_1, \vec{F}_2, \vec{F}_3$ maintain a particle in static equilibrium:
$$\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$$
where $\alpha$ is the angle between $\vec{F}_2$ and $\vec{F}_3$, $\beta$ is between $\vec{F}_1$ and $\vec{F}_3$, and $\gamma$ is between $\vec{F}_1$ and $\vec{F}_2$.
Solved Examples
Example 1 (Dynamic Equilibrium & Net Force Evaluation):
State the net force acting on each of the following objects:
(a) A raindrop falling with a constant terminal speed of $9\text{ m/s}$.
(b) A cork of mass $10\text{ g}$ floating on water.
(c) A kite held stationary in the sky against strong wind.
(d) A commercial airplane flying at a constant cruising speed of $900\text{ km/h}$ at a steady altitude of $10\text{ km}$.
Solution:
(a) Constant velocity $\implies \vec{a} = 0 \implies \vec{F}_{\text{net}} = \vec{0}$ (downward gravity balances upward air resistance).
(b) The floating cork is at rest $\implies \vec{a} = 0 \implies \vec{F}_{\text{net}} = \vec{0}$ (gravity is balanced by buoyant force).
(c) Stationary kite $\implies \vec{a} = 0 \implies \vec{F}_{\text{net}} = \vec{0}$ (string tension, aerodynamic lift, and weight sum to zero).
(d) Aircraft maintains constant speed in a straight line at fixed altitude $\implies \vec{a} = 0 \implies \vec{F}_{\text{net}} = \vec{0}$ (thrust balances drag; aerodynamic lift balances weight).
Example 2 (Concurrent Force Equilibrium & Lami’s Theorem):
A small sphere of mass $m = 6.0\text{ kg}$ is suspended by a light cord of length $2.0\text{ m}$ from a ceiling. A horizontal force $F$ is applied to the sphere until the cord makes an angle of $\theta = 30^\circ$ with the vertical in static equilibrium. Taking $g = 10\text{ m/s}^2$:
(a) Resolve forces along horizontal and vertical axes to determine string tension $T$ and force $F$.
(b) Verify the results using Lami’s Theorem.
Solution:
(a) Resolving forces:
Vertical equilibrium: $T\cos(30^\circ) = mg \implies T\left(\frac{\sqrt{3}}{2}\right) = (6.0)(10) = 60\text{ N} \implies T = \frac{120}{\sqrt{3}} = 40\sqrt{3}\text{ N} \approx 69.28\text{ N}$.
Horizontal equilibrium: $F = T\sin(30^\circ) = (40\sqrt{3})(0.5) = 20\sqrt{3}\text{ N} \approx 34.64\text{ N}$.
(b) Using Lami’s Theorem:
Angle between $F$ and $mg$ is $90^\circ$.
Angle between $T$ and $mg$ is $180^\circ – 30^\circ = 150^\circ$.
Angle between $T$ and $F$ is $90^\circ + 30^\circ = 120^\circ$.
$\frac{F}{\sin(150^\circ)} = \frac{mg}{\sin(120^\circ)} = \frac{T}{\sin(90^\circ)}$
$T = \frac{60}{\sin(120^\circ)} = \frac{60}{\sqrt{3}/2} = 40\sqrt{3}\text{ N}$.
$F = mg\frac{\sin(150^\circ)}{\sin(120^\circ)} = 60\frac{0.5}{\sqrt{3}/2} = \frac{60}{\sqrt{3}} = 20\sqrt{3}\text{ N}$ (Identical result!).
Example 3 (Smooth Wedge Acceleration in Inertial vs. Non-Inertial Frames):
A block of mass $m$ rests on a smooth inclined wedge of angle $\theta$. The wedge is placed on a frictionless horizontal floor.
(a) Find the horizontal acceleration $a_0$ that must be given to the wedge so that the block remains stationary relative to the wedge.
(b) Analyze the problem from both the ground frame (inertial) and the wedge frame (non-inertial).
(c) Find the normal force exerted by the wedge on the block during this motion.
Solution:
(a) & (b)
– Ground Frame (Inertial): The block accelerates horizontally with $a_0$ ($\vec{a} = a_0\hat{i}$).
Vertical: $N\cos\theta – mg = 0 \implies N = \frac{mg}{\cos\theta}$.
Horizontal: $N\sin\theta = ma_0 \implies \left(\frac{mg}{\cos\theta}\right)\sin\theta = ma_0 \implies a_0 = g\tan\theta$.
– Wedge Frame (Non-Inertial): Wedge accelerates right with $a_0$, introducing pseudo-force $ma_0$ to the left on the block.
Balancing forces along the incline: $ma_0\cos\theta = mg\sin\theta \implies a_0 = g\tan\theta$.
(c) Normal force: $N = mg\cos\theta + ma_0\sin\theta = mg\cos\theta + m(g\tan\theta)\sin\theta = mg\left[\cos\theta + \frac{\sin^2\theta}{\cos\theta}\right] = \frac{mg}{\cos\theta}$.
Example 4 (Simple Pendulum in an Accelerating Carriage):
A simple pendulum of bob mass $m$ is suspended from the ceiling of a train accelerating horizontally with uniform acceleration $a$.
(a) Find the equilibrium inclination angle $\theta$ of the string with the vertical in the train frame.
(b) Calculate the tension $T$ in the string at this equilibrium orientation.
(c) If the string is cut, describe the trajectory of the bob as observed inside the train and on the platform.
Solution:
(a) In the non-inertial frame of the train, pseudo-force $F_p = ma$ acts horizontally opposite to acceleration.
$T\sin\theta = ma, \quad T\cos\theta = mg \implies \tan\theta = \frac{a}{g} \implies \theta = \arctan\left(\frac{a}{g}\right)$ tilted backward.
(b) $T = \sqrt{(mg)^2 + (ma)^2} = m\sqrt{g^2 + a^2} = mg_{\text{eff}}$.
(c) When string is cut:
– Observer inside train: Bob is acted on by effective gravity $\vec{g}_{\text{eff}} = -a\hat{i} – g\hat{j}$; moves in a straight line backward and downward along angle $\theta = \arctan(a/g)$.
– Observer on ground: Bob has forward initial horizontal velocity $v_{\text{train}}$ and experiences solely downward gravity $mg$; executes a standard parabolic trajectory.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A particle is observed from two reference frames $S_1$ and $S_2$. The frame $S_1$ is inertial while $S_2$ is non-inertial. An external force $\vec{F}$ acts on the particle. Which of the following statements is correct regarding the observed acceleration of the particle?
(A) The acceleration is zero in both frames.
(B) The acceleration is non-zero in $S_1$ but must be zero in $S_2$.
(C) The acceleration cannot be zero in both frames simultaneously.
(D) The acceleration is identical in both frames.
Problem 2 (JEE Main – Single Correct):
A block of mass $5\text{ kg}$ is kept on a smooth horizontal table. If three forces $F_1 = 10\text{ N}$ due East, $F_2 = 10\text{ N}$ due North, and $F_3 = 10\sqrt{2}\text{ N}$ due South-West act simultaneously on the block, the net acceleration of the block is:
(A) $2\text{ m/s}^2$ South-West
(B) $2\sqrt{2}\text{ m/s}^2$ North-East
(C) Zero
(D) $4\text{ m/s}^2$ North
Problem 3 (JEE Main – Single Correct):
A body of mass $2\text{ kg}$ moves with a constant velocity $\vec{v} = (3\hat{i} + 4\hat{j})\text{ m/s}$. The net force acting on the body is:
(A) $10\text{ N}$
(B) $5\text{ N}$
(C) Zero
(D) $25\text{ N}$
Problem 4 (JEE Main – Single Correct):
An object of mass $m$ suspended by a string from the roof of an elevator experiences zero tension in the string. The motion of the elevator could be:
(A) Moving upwards with uniform speed
(B) Moving downwards with uniform speed
(C) Falling freely under gravity
(D) Accelerating upwards with acceleration $g$
Problem 5 (JEE Main – Single Correct):
A mass $m$ is suspended by two strings making angles $45^\circ$ and $45^\circ$ with the horizontal ceiling. The tension in each string is:
(A) $mg$
(B) $\frac{mg}{\sqrt{2}}$
(C) $\sqrt{2}mg$
(D) $\frac{mg}{2}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding Newton’s First Law and Inertia?
(A) Inertia depends solely on the mass of a body, not on its shape or state of motion.
(B) Newton’s First Law is independent of Newton’s Second Law and establishes the definition of inertial frames.
(C) In an inertial frame, a body cannot experience any acceleration without a non-zero net external force.
(D) A reference frame attached to the surface of the Earth is strictly an inertial frame of reference.
Problem 7 (JEE Advanced – One or More Correct):
A person is standing in an elevator. Under which of the following conditions does the normal contact force exerted by the floor on the person equal the person’s true weight ($N = mg$)?
(A) The elevator is at rest.
(B) The elevator moves upwards with uniform velocity.
(C) The elevator moves downwards with uniform velocity.
(D) The elevator accelerates upwards with uniform acceleration.
Problem 8 (JEE Advanced – One or More Correct):
A block of mass $m$ is placed on a smooth wedge of angle $\theta$ which accelerates horizontally with acceleration $a$. Which of the following statements is/are correct?
(A) The block remains at rest relative to the wedge if $a = g\tan\theta$.
(B) The normal force between the block and wedge is $\frac{mg}{\cos\theta}$ when $a = g\tan\theta$.
(C) If $a > g\tan\theta$, the block slides up the inclined plane.
(D) If $a < g\tan\theta$, the block slides down the inclined plane.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
Three coplanar forces of magnitudes $F_1 = 30\text{ N}, F_2 = 40\text{ N}$, and $F_3 = F$ maintain a particle in static equilibrium. If the angle between $\vec{F}_1$ and $\vec{F}_2$ is $90^\circ$, find the magnitude of force $F$ in Newtons.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A plumb line hangs from the roof of a railroad car. When the car accelerates along a level track with uniform acceleration $a = 7.5\text{ m/s}^2$, the plumb line inclines at an angle $\theta$ to the vertical. Taking $g = 10\text{ m/s}^2$, evaluate $4 \times \tan\theta$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (C) | 4. (C) | 5. (B) | 6. (A, B, C) | 7. (A, B, C) | 8. (A, B, C, D) | 9. 50 | 10. 3
Solution 1:
In inertial frame $S_1$, $\vec{a}_1 = \vec{F}/m$. In frame $S_2$ accelerating with $\vec{a}_0 \neq \vec{0}$, $\vec{a}_2 = \vec{a}_1 – \vec{a}_0$. Since $\vec{a}_0 \neq \vec{0}$, both $\vec{a}_1$ and $\vec{a}_2$ cannot be simultaneously zero.
Correct Answer: (C)
Solution 2:
$\vec{F}_1 = 10\hat{i}\text{ N}, \vec{F}_2 = 10\hat{j}\text{ N}$.
South-West direction: $\vec{F}_3 = 10\sqrt{2}\left(\frac{-\hat{i} – \hat{j}}{\sqrt{2}}\right) = -10\hat{i} – 10\hat{j}\text{ N}$.
Net force: $\vec{F}_{\text{net}} = (10 – 10)\hat{i} + (10 – 10)\hat{j} = \vec{0} \implies \vec{a} = \vec{0}$.
Correct Answer: (C)
Solution 3:
Constant velocity $\implies \vec{a} = \vec{0} \implies \vec{F}_{\text{net}} = m\vec{a} = \vec{0}$.
Correct Answer: (C)
Solution 4:
In a freely falling elevator, effective acceleration is $g_{\text{eff}} = g – g = 0$, so string tension is zero ($T = m(g – a) = 0$).
Correct Answer: (C)
Solution 5:
Resolving vertically: $2T\sin(45^\circ) = mg \implies 2T\left(\frac{1}{\sqrt{2}}\right) = mg \implies \sqrt{2}T = mg \implies T = \frac{mg}{\sqrt{2}}$.
Correct Answer: (B)
Solution 6:
– (A) Mass is the sole scalar measure of translational inertia (True).
– (B) Newton’s 1st law defines the inertial frame and the concept of force qualitatively (True).
– (C) In an inertial frame, $\sum \vec{F} = 0 \iff \vec{a} = 0$ (True).
– (D) Earth rotates and orbits, so it is strictly a non-inertial frame (False).
Correct Answer: (A, B, C)
Solution 7:
When the elevator moves with uniform velocity (or at rest), $\vec{a} = 0$, giving $N = mg$ in cases A, B, and C.
Correct Answer: (A, B, C)
Solution 8:
In the frame of the wedge, forces along the incline are $mg\sin\theta$ downward and $ma\cos\theta$ upward.
At $a = g\tan\theta$, $ma\cos\theta = mg\sin\theta \implies$ equilibrium (A is True).
Normal force $N = mg\cos\theta + ma\sin\theta = \frac{mg}{\cos\theta}$ (B is True).
If $a > g\tan\theta$, $ma\cos\theta > mg\sin\theta$, block slides up (C is True).
If $a < g\tan\theta$, block slides down (D is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
For three concurrent forces in equilibrium: $\vec{F}_3 = -(\vec{F}_1 + \vec{F}_2)$.
Since $\vec{F}_1 \perp \vec{F}_2$, $|\vec{F}_3| = \sqrt{F_1^2 + F_2^2} = \sqrt{30^2 + 40^2} = 50\text{ N}$.
Correct Answer: 50
Solution 10:
In the 50
Solution 10:
In the accelerating car’s frame: $\tan\theta = \frac{a}{g} = \frac{7.5}{10} = 0.75$.
Then $4 \times \tan\theta = 4 \times 0.75 = 3$.
Correct Answer: 3