Laws of Motion · Part 1

• A bus starts suddenly: your feet move with the floor but your upper body tends to stay at rest, so you lean backward.
• A bus brakes: your upper body tends to keep moving, so you lurch forward.
• Beating a carpet: the carpet moves, the dust tends to stay at rest and falls out.
• A larger mass has more inertia: it needs more force for the same acceleration.

Newton’s first law holds only in an inertial frame, one that is not accelerating. The ground is a good enough inertial frame for most NEET problems.

\(\vec F_{\rm net} = d\vec p/dt\) is the general law. For a body whose mass changes it gives

• Sand falling on a moving belt at rate dm/dt: the extra force to keep the belt at constant speed v is \(v\,dm/dt\).
• A machine gun firing n bullets per second, each of mass m at speed v: the average recoil force is \(nmv\).
• If momentum is given as a function of time, differentiate it: the slope of a p–t graph is the force.

A genuine action–reaction pair always:

1. acts on two different bodies;
2. is of the same type (both gravitational, both contact, both tension);
3. is equal in size and opposite in direction, at every instant, whether the bodies accelerate or not.

The weight of a book and the normal force on it act on the same body and are of different types, so they are not a pair. The horse–cart puzzle is resolved the same way: the cart’s pull on the horse acts on the horse. What moves the horse forward is the ground’s push on its hooves.

1. Choose one body (or one system) and draw it alone.
2. Draw the weight mg from its centre, straight down.
3. Go round the boundary. At each contact add a normal force (perpendicular to the surface) and, if needed, friction (along the surface). At each string add a tension pulling away from the body.
4. Add applied forces. Do not add “force of motion”, “centripetal force” or ma as extra forces.
5. Pick axes, ideally along the acceleration, and write \(\sum F = ma\) along each.

• Pull F at angle θ above the horizontal on a level floor: \(N = mg - F\sin\theta\).
• Push F at angle θ below the horizontal: \(N = mg + F\sin\theta\).
• Smooth incline: \(N = mg\cos\theta\). Smooth incline with a horizontal force holding the block at rest: \(F = mg\tan\theta\) and \(N = mg/\cos\theta\).
• A block pressed against a wall by a horizontal force F: N = F, unrelated to mg.

Two blocks \(m_1\) and \(m_2\) on a smooth floor, pushed by F applied to \(m_1\):

1. System: \(a = F/(m_1+m_2)\).
2. Only the contact force accelerates \(m_2\): \(N = m_2a = \dfrac{m_2F}{m_1+m_2}\).

Pushing from the other side changes the contact force to \(m_1F/(m_1+m_2)\). The contact force equals the force needed to accelerate everything ahead of the contact.

• Massless string: the net force on any piece must be zero (zero mass × finite acceleration), so the tension is the same at both ends of every piece.
• Hanging rope of mass M with a load m at the bottom: at distance x from the bottom, \(T = (m + Mx/L)g\). So the top carries \((m + M)g\).
• Rope of mass M pulling a block m on a smooth floor with force F: a = F/(M + m). The tension at the block end is ma, which is less than F. A massless rope would transmit the full F.

• Series: \(1/k = 1/k_1 + 1/k_2\). Parallel: \(k = k_1 + k_2\).
• Cutting a spring into n equal pieces: each has constant nk (k ∝ 1/length).
• A spring force cannot change instantly, because the extension needs time to change. A string tension can change instantly.
• So just after a string is cut, any body still attached to a spring feels the old spring force. This gives non-obvious instantaneous accelerations, such as g upward.

Where several strings meet at a light knot, the knot has no mass, so the forces on it must add to zero. Resolve along two axes, or use Lami’s theorem from the Vectors chapter when exactly three strings meet. The tension in each string is then fixed by geometry and the hanging weight.

1. Treat all connected bodies as one system moving with a common acceleration a. Internal tensions cancel. \(a = \dfrac{\text{net external driving force}}{\text{total mass}}\).
2. Isolate the simplest single body and apply \(\sum F = ma\) to find the tension.
3. Check with another body.

Block \(m_1\) on a smooth table, pulled by a hanging \(m_2\): \(a = \dfrac{m_2g}{m_1+m_2}\) and \(T = \dfrac{m_1m_2g}{m_1+m_2}\). In a chain of blocks pulled by F, the tension at any joint equals the force needed to accelerate everything behind that joint.

Write the total string length in terms of the positions and differentiate twice.

• A load hanging from a movable pulley, with the string fixed at one end and pulled at the other: if the free end moves by x, the load moves by x/2. So \(a_{\rm load} = a_{\rm end}/2\).
• The movable pulley has two segments pulling up, so \(2T - Mg = Ma_{\rm load}\).
• Energy check: the tension does equal and opposite work at the two ends, so T times the end’s displacement must equal 2T times the load’s displacement.

• Atwood with \(m_1 = m_2\): a = 0 and T = mg.
• \(m_2 \gg m_1\): a → g and \(T \to 2m_1g\).
• Tension always lies between \(m_1g\) and \(m_2g\).
• The clamp of a fixed pulley carries the resultant of the two tensions: 2T when both strings hang vertically, \(T\sqrt2\) when they are at right angles. It is not \((m_1+m_2)g\) while the masses accelerate.
• Atwood machine in a lift accelerating up at \(a_0\): replace g by \(g + a_0\) everywhere.

Take up as positive. For a person of mass m on a scale, \(N - mg = ma\), so \(N = m(g + a)\).

• At rest or moving at constant velocity (up or down): a = 0, N = mg.
• Moving up and speeding up, or moving down and slowing down: a upward, N = m(g + a) > mg.
• Moving up and slowing down, or moving down and speeding up: a downward, N = m(g − a) < mg.
• Free fall (cable broken): a = g downward, N = 0 (weightlessness).
• Downward acceleration greater than g: the person would press against the ceiling instead.

The scale reading depends on the direction of the acceleration, never on the direction of the velocity.

Everything inside a lift with upward acceleration \(a_0\) behaves as if gravity were \(g_{\rm eff} = g + a_0\) (or \(g - a_0\) for downward acceleration).

• A pendulum’s period becomes \(T = 2\pi\sqrt{L/g_{\rm eff}}\).
• A ball dropped from height h inside the lift reaches the floor in \(\sqrt{2h/g_{\rm eff}}\).
• The cable tension for a lift of total mass M (cage plus load) is \(M(g + a_0)\).
• A beam (pan) balance still balances, because both pans feel the same \(g_{\rm eff}\). A spring balance changes its reading.

A 2 kg box sits on the smooth floor of a truck that accelerates forward at 3 m/s².

• Ground frame: no horizontal real force acts on the box, so it stays at rest. The truck slides forward beneath it.
• Truck frame: the box feels a pseudo force 2 × 3 = 6 N backward, so it accelerates backward at 3 m/s² relative to the truck.

Both descriptions agree on what happens. If the floor is rough enough, static friction (6 N forward) keeps the box moving with the truck; Part 2 treats the limit of this.

In a frame with horizontal acceleration \(a_0\), real gravity and the pseudo force combine into an effective gravity \(g_{\rm eff} = \sqrt{g^2 + a_0^2}\), tilted backward at \(\tan\theta = a_0/g\).

• A plumb line or pendulum hangs along \(g_{\rm eff}\).
• The water surface in a tank on an accelerating truck tilts at the same angle, rising at the back.
• A pendulum’s period in the car becomes \(2\pi\sqrt{L/g_{\rm eff}}\), a little shorter than at rest.

Let the wedge be pushed horizontally at \(a_0\) so that its sloping face moves into the block. In the wedge frame the pseudo force \(ma_0\) acts on the block, away from the face.

• Perpendicular to the slope: \(N = m(g\cos\theta + a_0\sin\theta)\).
• Along the slope (down positive): \(a_{\rm rel} = g\sin\theta - a_0\cos\theta\).
• No sliding when \(a_0 = g\tan\theta\); then \(N = mg/\cos\theta\).
• For \(a_0 > g\tan\theta\) the block slides up the slope.

Impulse is the area under the F–t graph. For a triangle use ½ × base × height; for a rectangle, base × height. Areas below the time axis count as negative.

The same impulse can be delivered as a large force for a short time or a small force for a long time. Average force = impulse ÷ contact time.

• A fielder pulls the hands back while catching: the ball’s momentum change is fixed, but the stopping time grows, so the force on the hands falls.
• Jumping onto sand or bending the knees on landing, air bags, packing material, shock absorbers: all increase Δt.
• A ball that bounces back suffers a larger Δp (up to twice) than one that stops, so it needs a larger impulse.

• Recoil: a gun of mass M fires a bullet m at v. Total momentum stays zero: \(MV = mv\), so \(V = mv/M\), opposite to the bullet.
• Explosion at rest: fragment momenta add to zero. For two pieces they are equal and opposite. For three pieces, any one is opposite to the vector sum of the other two.
• Ball hitting a wall at an angle: only the component normal to the wall reverses, so \(|\Delta p| = 2mv\cos\theta\) with θ measured from the normal.

Car on a road banked at θ, moving in a horizontal circle of radius r.

1. Without friction: vertical \(N\cos\theta = mg\); horizontal (toward the centre) \(N\sin\theta = mv^2/r\). Divide: \(\tan\theta = v^2/rg\). Note \(N = mg/\cos\theta\), larger than mg.
2. With friction μ, at the maximum speed friction acts down the slope: \(N\cos\theta - \mu N\sin\theta = mg\) and \(N\sin\theta + \mu N\cos\theta = mv^2/r\). Dividing:

At the minimum speed friction acts up the slope. On a level road (θ = 0), \(v_{\max} = \sqrt{\mu rg}\).

• A cyclist leans at θ from the vertical with \(\tan\theta = v^2/rg\), the same as the banking angle.
• Conical pendulum (string L at θ to the vertical): \(T\cos\theta = mg\), \(T\sin\theta = mv^2/r\), period \(2\pi\sqrt{L\cos\theta/g}\).
• Car over the top of a convex bridge (hump) of radius r: \(mg - N = mv^2/r\), so N falls as v rises. Contact is lost at \(v = \sqrt{gr}\).
• Car at the bottom of a dip: \(N - mg = mv^2/r\), so the passengers feel heavier.

Mass on a string of length r in a vertical circle:

• Top: \(T + mg = mv^2/r\). The string stays taut only if \(v_{\rm top} \ge \sqrt{gr}\).
• Bottom: \(T - mg = mv^2/r\). Energy conservation gives \(v_{\rm bottom}^2 = v_{\rm top}^2 + 4gr\), so the minimum speed at the bottom is \(\sqrt{5gr}\).
• With the minimum speeds, the tension is 0 at the top and 6mg at the bottom.