The Timpson Lyceum · University Physics I (2025)

Applying Newton's Laws – Lecture 5 (University Physics I)

Overview

This lecture is almost entirely about applying Newton's laws rather than developing new theory: it works through a long string of example problems and then introduces a few new force models. The first half drills the two regimes of Newton's laws — equilibrium ($\sum \vec{F} = 0$) for objects at rest or moving at constant velocity, and dynamics ($\sum \vec{F} = m\vec{a}$) for accelerating objects — using hanging ropes, hanging weights, inclined ramps, an ice boat, an elevator, a connected food tray, and a pulley-coupled glider. The second half adds three new force models: friction (static and kinetic), fluid resistance (drag), and the forces behind uniform circular motion. The unifying skill is drawing a correct free-body diagram, choosing convenient $x$–$y$ axes, and resolving each force into components. Sanity-checking the final symbolic result against limiting cases (angle $\to 0$, mass $\to 0$, etc.) is emphasized throughout.

Key Concepts

The two problem types

• Equilibrium: if an object is at rest or moving at constant velocity, $\vec{a}=0$, so $\sum F_x = 0$ and $\sum F_y = 0$. Forces cancel.
• Dynamics: if the object accelerates, $\sum \vec{F} = m\vec{a}$. Apply this component-by-component.
• Weight: $w = mg$, with $g = 9.8\ \text{m/s}^2$, always directed straight down.

Ropes, tension, and weight distribution

• For a massless rope the tension is the same at both ends. For a rope of weight $w_R$ holding an object of weight $w$, the top of the rope must support both: $T_{\text{top}} = w + w_R$, while $T_{\text{bottom}} = w$. (Like a human chain — the top link carries everyone below it.)

Inclined-plane / ramp problems

• Choose axes along and perpendicular to the incline.
• The weight splits into a component along the ramp, $w\sin\theta$ (tends to slide the object down), and one into the ramp, $w\cos\theta$ (balanced by the normal force).
• For a car held on a ramp by a cable: $N = w\cos\alpha$ and $T = w\sin\alpha$.
• For a block pulled up a frictionless ramp by a hanging counterweight at constant speed, the tension equals the slid block's gravity component, so the counterweight need only be $w_2 = w_1\sin\theta$ — far less than the load itself.

Newton's second law in motion

• Ice boat: find acceleration from kinematics ($v = v_0 + at$), then $F = ma$. Adding a friction force just adds to the required driving force; what matters is the net force.
• Elevator / apparent weight: for a free body with tension/normal up and weight down, $N = m(g+a)$ when decelerating downward or accelerating upward (you feel heavier), and $N = m(g-a)$ when accelerating downward (you feel lighter). A bathroom scale reads the normal force $N$, not the true weight $mg$.
• Contact forces (food tray pushing a carton): analyze the whole system to get $a = F_{\text{net}}/(m_1+m_2)$, then isolate the second body to get the contact force it feels, $F_{\text{contact}} = m_2 a$.
• Connected bodies over a pulley: both share the same magnitude of acceleration; a frictionless pulley only redirects the tension. For a glider $m_1$ on a frictionless table pulled by a hanging mass $m_2$: $a = \dfrac{m_2 g}{m_1+m_2}$ and $T = \dfrac{m_1 m_2 g}{m_1+m_2}$.

Friction

• Kinetic friction (object sliding): $f_k = \mu_k N$ — a fixed force opposing motion, set by the surfaces (via the dimensionless coefficient $\mu_k$) and the normal force $N$.
• Static friction (object not yet moving): $f_s \le \mu_s N$. It adjusts to exactly cancel the applied force, up to a maximum of $\mu_s N$, at which point sliding begins.
• Typically $\mu_s > \mu_k$: it takes more force to start sliding than to keep sliding. Friction always opposes the direction of (attempted) motion. Microscopically it arises from rough surfaces interlocking and from molecular bonding between contacting materials.

Fluid resistance (drag)

• A fluid pushes back on a moving object harder the faster it moves.
• Low speed: $f = kv$ (resistance proportional to velocity).
• High speed (e.g. air): $f = Dv^2$ (proportional to velocity squared); $D$ is the drag coefficient.
• Terminal velocity is reached when the resistive force exactly cancels weight ($a=0$):
  • low speed: $v_t = \dfrac{mg}{k}$
  • high speed: $v_t = \sqrt{\dfrac{mg}{D}}$
• For the low-speed case the equation of motion $m\dfrac{dv}{dt} = mg - kv$ is a separable differential equation. Replacing $mg = k v_t$ lets you factor the right side and integrate; starting from rest the solution is the velocity-vs-time curve $v(t) = v_t\left(1 - e^{-(k/m)t}\right)$, which starts at $0$ and rises asymptotically to $v_t$. Differentiating gives the acceleration and integrating gives the position. (The high-speed $v^2$ case leads to an integral requiring techniques not yet covered.)

Forces in uniform circular motion

• Circular motion at constant speed still has acceleration — the centripetal (radially inward) acceleration $a_{\text{rad}} = \dfrac{v^2}{r} = \dfrac{4\pi^2 r}{T^2}$, where $T$ is the period.
• Some real force must supply it (tension, normal force, friction…); there is no outward "centrifugal" force in the inertial free-body diagram. The net inward force is $F = m a_{\text{rad}}$.
• Vertical circle (Ferris wheel): at the top, $N = m\!\left(g - \dfrac{v^2}{r}\right)$ (you feel lighter); at the bottom, $N = m\!\left(g + \dfrac{v^2}{r}\right)$ (you feel heavier). Always insert $a_{\text{rad}}$ with the sign matching its inward direction.
• Conical pendulum (bob on a wire of length $L$, swinging in a horizontal circle at angle $\beta$ from vertical): the vertical component balances gravity and the horizontal component supplies $a_{\text{rad}}$, giving $T_{\text{wire}} = \dfrac{mg}{\cos\beta}$ and period $\;T = 2\pi\sqrt{\dfrac{L\cos\beta}{g}}$.
• Banked curve: to round a curve of radius $r$ at speed $v$ with no friction needed, the road is banked so the horizontal component of the normal force supplies $a_{\text{rad}}$. Mass cancels, leaving $\tan\theta = \dfrac{a_{\text{rad}}}{g} = \dfrac{v^2}{gr}$, i.e. $\theta = \tan^{-1}\!\left(\dfrac{v^2}{gr}\right)$.

Worked Example

Pulling a crate at an angle against kinetic friction. A crate of weight $w = 500\ \text{N}$ sits on a level floor with $\mu_k = 0.40$. You pull on it with a force $F$ at $30^\circ$ above the horizontal. What $F$ keeps it sliding at constant velocity?

The crate is in equilibrium, so resolve into components. The pull's vertical component $F\sin 30^\circ$ helps lift the crate, reducing the normal force:

Substitute $N$ from the $y$-equation into the $x$-equation and solve for $F$:

(The coefficients themselves come from the earlier measurement: it took $230\ \text{N}$ to start the crate moving, so $\mu_s = 230/500 = 0.46$, and $200\ \text{N}$ to keep it moving, so $\mu_k = 200/500 = 0.40$.)

Summary

• Decide first whether a problem is equilibrium ($\sum F = 0$) or dynamics ($\sum F = ma$); then draw a free-body diagram and resolve every force into well-chosen components.
• On ramps, split the weight into $w\sin\theta$ (along the surface) and $w\cos\theta$ (into the surface, balanced by $N$); for connected bodies, all parts share the same acceleration magnitude and an ideal pulley only redirects tension.
• A scale reads the normal force, not true weight — apparent weight is $m(g\pm a)$ depending on the acceleration's direction.
• Friction opposes motion: kinetic friction is the fixed $f_k = \mu_k N$, while static friction self-adjusts up to a maximum $\mu_s N$ (usually $\mu_s>\mu_k$).
• Fluid resistance grows with speed ($kv$ at low speed, $Dv^2$ at high speed); the object speeds up until drag cancels weight at terminal velocity $v_t = mg/k$ or $v_t=\sqrt{mg/D}$, with $v(t)=v_t(1-e^{-(k/m)t})$ in the low-speed case.
• Uniform circular motion needs a real inward force supplying $a_{\text{rad}} = v^2/r = 4\pi^2 r/T^2$; this underlies the Ferris-wheel apparent weight, the conical pendulum, and the frictionless banked curve $\theta = \tan^{-1}(v^2/gr)$.