Force and Translational Dynamics

What this unit covers

The topics below follow the published Physics C: Mech course framework for Unit 2. This unit is worth 20–25% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• Newton’s Laws & Free-Body Diagrams13 min · 3 objectivesState Newton’s three laws and apply the second law in the form ΣF = ma = m dv/dt · Draw free-body diagrams and resolve forces into perpendicular components · Solve for acceleration, normal force, and friction on a single object
• Drag Forces & Terminal Velocity15 min · 3 objectivesWrite Newton’s second law as a differential equation when a resistive force depends on speed · Determine terminal velocity from the condition that net force and dv/dt vanish · Describe qualitatively how velocity approaches its terminal value over time
• Systems & Connected Objects14 min · 3 objectivesApply Newton’s second law to systems of objects linked by strings and pulleys · Solve Atwood machines and blocks connected over an ideal pulley · Recognize that objects joined by an inextensible string share the same magnitude of acceleration
• Newton’s Law of Gravitation13 min · 3 objectivesState the law of universal gravitation and compute the force between two masses · Relate a planet’s surface gravity to its mass and radius through g = GM/R² · Predict how gravitational force changes with separation using the inverse-square law
• Friction: Static, Kinetic & the Inclined Plane14 min · 3 objectivesDistinguish static friction as an inequality from kinetic friction as an equality · Resolve forces on an incline into components along and perpendicular to the surface · Determine the angle at which sliding begins from the coefficient of static friction
• Circular Motion Dynamics: Banked Curves, Loops & Conical Pendulums14 min · 3 objectivesApply Newton's second law in the radial direction to circular-motion problems · Analyze banked curves, vertical loops and conical pendulums with free-body diagrams · Determine minimum speeds and critical conditions for circular motion
• Springs, Hooke's Law & Combining Stiffnesses14 min · 3 objectivesApply Hooke's law as a force law with correct sign conventions · Derive the effective stiffness of springs in series and in parallel · Analyze equilibrium positions for systems involving springs and gravity
• Accelerating Frames & Apparent Weight14 min · 3 objectivesCompute the normal force on an object in a vertically accelerating system · Explain apparent weightlessness in terms of the normal force rather than gravity · Analyze systems where the observer is accelerating

Formulas in Unit 2

Every term in Unit 2

All 59 terms we publish for Force and Translational Dynamics, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Newton's second law in general form

ΣF = dp/dt. Reduces to ma only when mass is constant, which is why rocket problems need the general form.

Velocity-dependent drag

With F_drag = −bv, the equation of motion is a separable differential equation whose solution approaches terminal velocity exponentially.

Terminal velocity from a drag law

Set net force to zero: mg = bv gives v_t = mg/b; for quadratic drag mg = cv² gives v_t = √(mg/c).

Solving a separable equation of motion

Write m dv/dt = F(v), separate to dv/F(v) = dt/m, and integrate both sides with initial conditions as limits.

Static and kinetic friction

Static friction adjusts up to μsN; kinetic friction is μkN and constant during sliding. Static friction is generally the larger maximum.

Inclined plane analysis

Resolve weight into mg sin θ along and mg cos θ perpendicular. The angle at which sliding begins gives tan θ = μs.

Circular motion dynamics

The net inward force equals mv²/r = mω²r. Identify which real force supplies it before writing the equation.

Newton's law of gravitation

F = Gm₁m₂/r², measured center to center. Inside a uniform sphere, only the mass at smaller radius contributes.

Gravitational field inside a sphere

g rises linearly with r inside a uniform sphere and falls as 1/r² outside, peaking at the surface.

Orbits and Kepler's third law

Setting gravity equal to the centripetal requirement gives T² ∝ r³. Orbital speed v = √(GM/r) is independent of the orbiting mass.

Free-body diagram discipline

One diagram per object, showing only real interactions. Never draw ma or centripetal force as separate arrows — they are results, not causes.

Systems with multiple bodies

Treat the whole system to find acceleration (internal forces cancel), then isolate one body to find an internal force.

Pulley constraints

An inextensible string ties the accelerations of connected masses together in magnitude, which supplies the extra equation needed.

Non-inertial frames

In an accelerating frame, Newton's laws require a fictitious force. Physics C solves in an inertial frame instead to avoid it.

Solving the terminal velocity differential equation

From m dv/dt = mg − bv, separating and integrating gives v(t) = v_t(1 − e^(−bt/m)), approaching v_t asymptotically and never reaching it.

Impulse from a variable force

J = ∫F dt, the area under the force-time curve. This is how collision problems are handled when the force is not constant.

Vertical circular motion condition

At the top, the minimum speed for contact is where the normal force reaches zero, so mg alone supplies mv²/r and v_min = √(gr).

Banked curve with friction

Friction acts down the bank above the design speed and up the bank below it, which is why a range of safe speeds exists rather than one.

Gravitational potential energy sign

U = −GMm/r is negative because zero is taken at infinity and gravity is attractive. A more negative U means a more tightly bound orbit.

Total energy of a circular orbit

E = −GMm/2r, exactly half the potential energy. Negative total energy is what makes an orbit bound.

Kepler's second law

A line from the sun to a planet sweeps equal areas in equal times — a direct consequence of angular momentum conservation.

Newton's first law and inertial frames

An object with no net force keeps constant velocity. The law also defines what an inertial frame is: one in which it holds. Newton's laws are valid only in such frames, which is why accelerating frames need fictitious forces.

Newton's third law pairs

Forces come in pairs equal in magnitude, opposite in direction, of the same type, and acting on DIFFERENT objects. Because they act on different objects they never cancel each other in a single free-body diagram.

Identifying a third-law partner

Swap the two nouns: "Earth pulls on ball" pairs with "ball pulls on Earth." The normal force on a book from a table pairs with the book pushing down on the table, NOT with the book's weight — that would be the same object twice.

The normal force is not always mg

N is whatever the perpendicular equilibrium condition requires: mg cos θ on an incline, m(g ± a) in an accelerating elevator, and modified by any applied force with a vertical component. Derive it every time.

Tension in an ideal string

A massless string transmits force undiminished, so the tension is the same throughout and at both ends. A string with mass has different tensions along its length.

Massless string approximation

Setting the string mass to zero is what makes the tension uniform: with ΣF = ma and m = 0, the net force on any segment must vanish, so the pulls at its two ends are equal.

Ideal pulley

Massless and frictionless, so it changes the direction of the tension without changing its magnitude. A pulley with mass has a different tension on each side, because a net torque is needed to give it angular acceleration.

Atwood machine

Two masses over a pulley: a = (m₁ − m₂)g/(m₁ + m₂) and T = 2m₁m₂g/(m₁ + m₂). The acceleration is smaller than g by the ratio of the mass difference to the total, which is why it was used to measure g.

Modified Atwood machine

One mass hanging, one on a table: a = m_hanging g/(m₁ + m₂) with a frictionless table. The full system mass resists, but only the hanging weight drives, which is the key setup step.

Two blocks in contact

Treat the pair as one system to get the acceleration, then isolate one block to find the contact force between them. The contact forces are a third-law pair and cancel in the system equation.

Constraint equations

Connected objects have related accelerations — equal magnitudes over a simple pulley, or a factor of two in a movable-pulley arrangement. Writing the constraint explicitly is what makes the simultaneous equations solvable.

Apparent weight in an elevator

N = m(g + a) with a signed by the acceleration, not the velocity. Moving up while slowing gives a downward acceleration and a reading below mg.

Free fall and weightlessness

Apparent weightlessness means every contact force is zero, N = 0, not that gravity is absent. Orbiting astronauts are in continuous free fall where g is still about 90% of its surface value.

Static friction as an inequality

f_s ≤ μ_s N. Static friction takes exactly the value equilibrium demands, up to a ceiling. Writing f_s = μ_s N for an object that is not on the verge of slipping is wrong.

Coefficients of friction are dimensionless

μ is a ratio of two forces, so it has no units, and it is typically between 0 and about 1.5. A computed μ far outside that range signals an arithmetic error.

Angle of repose

tan θ_max = μ_s. The mass cancels, so the angle at which sliding begins is a property of the surfaces alone — and tilting until it slips is a direct way to measure μ_s.

Friction on an incline, up versus down

Friction always opposes the relative motion, so it acts down the incline while the block slides up and up the incline while it slides down. The deceleration going up therefore exceeds the acceleration coming down.

Stacked blocks with friction

Friction from the lower block is what accelerates the upper one. The maximum system acceleration before the top block slips is a = μ_s g, independent of either mass.

Minimum force to prevent slipping

A block held against a vertical wall requires a horizontal push large enough that μ_s N ≥ mg, giving F ≥ mg/μ_s. Note that the normal force here comes from the applied force, not from gravity.

Linear drag

F = −bv, appropriate at low speed. It gives a separable equation of motion whose solution is an exponential, and a terminal velocity reached asymptotically.

Quadratic drag at higher speed

F = −cv², appropriate at higher speed for objects moving through air. The equation is still separable but integrates to a hyperbolic tangent rather than an exponential.

Terminal velocity for linear drag

Set the net force to zero: mg = bv_t, so v_t = mg/b. Heavier objects fall faster in a drag-limited regime, which is why the mass dependence returns once air resistance matters.

Terminal velocity for quadratic drag

mg = cv_t² gives v_t = √(mg/c). The dependence on mass is a square root rather than linear, which is why doubling the mass increases terminal speed by only about 40%.

Approach to terminal velocity

Acceleration starts at g and decreases toward zero as speed rises and drag grows. The speed approaches v_t asymptotically and never exactly reaches it, so graphs of v(t) flatten rather than meeting the line.

Time constant of linear drag

τ = m/b sets the timescale: after one time constant the speed has closed about 63% of the gap to terminal velocity, and after about five it is effectively there.

Banked curve without friction

tan θ = v²/(rg). The mass cancels, so one posted design speed serves every vehicle, and the inward force comes from the horizontal component of the normal force.

How friction widens the safe speeds on a bank

Friction extends the range of safe speeds above and below the ideal. Below the design speed friction acts up the bank to prevent sliding down; above it, friction acts down the bank.

Conical pendulum

A mass swung in a horizontal circle on a string at angle θ to the vertical satisfies tan θ = v²/(rg) — the same relation as the banked curve, because a tilted force is again resolved into inward and vertical components.

Vertical circle at the top

N + mg = mv²/r, so N falls as speed falls. The minimum speed for contact is v = √(gr), where N = 0 and gravity alone supplies the centripetal force.

Vertical circle at the bottom

N − mg = mv²/r, so N = m(g + v²/r), which exceeds the weight. This is why a rider feels heaviest at the bottom of a loop and why track structures are strongest there.

Car cresting a hill

At the top, mg − N = mv²/r, so N < mg and the car feels light. Contact is lost when v = √(gr), the same critical speed as the inside of a loop.

Centripetal force is a role, not a force

No force is "the centripetal force." Tension, gravity, friction or the normal force plays that role. Never draw mv²/r on a free-body diagram — it belongs on the right-hand side of ΣF = ma.

Gravitational field strength

g = GM/r² is the field a mass M produces, so the surface value depends on the planet's mass and radius. It is the acceleration a free object would have, which is why all objects fall alike.

Variation of g with altitude

g falls as 1/r² measured from the center, not from the surface. At an altitude equal to one Earth radius, g is one quarter of its surface value, not zero.

Shell theorem

A uniform spherical shell attracts an external object as though all its mass were at the center, and exerts zero net force anywhere inside it. This is what lets planets be treated as point masses.

Gravity inside a uniform sphere

Only the mass at smaller radius contributes, so g ∝ r inside and the field falls linearly to zero at the center. A tunnel through the Earth would give simple harmonic motion.

Weight versus mass

Mass is the invariant amount of matter and inertia, measured in kilograms; weight is the gravitational force mg, measured in newtons and depending on location. An astronaut's mass is unchanged in orbit.

The spring force as a restoring force

F = −kx is linear and always directed back toward equilibrium. Linearity plus restoration is exactly the condition for simple harmonic motion, which is why springs and small oscillations appear together throughout the course.

What examiners penalize here

• On the AP exam, draw the free-body diagram first and tilt your axes to match the motion. Write ΣF = ma along each axis separately. Getting the components of weight right — mg sin θ along a slope, mg cos θ into it — is where most points are won or lost.
• Keep the two formulas straight: F = Gm₁m₂/r² is the force between two masses, while g = GM/R² is the field (acceleration) one mass produces at distance R. Both are inverse-square — whenever a distance changes, square the ratio before scaling.
• N = mg only on a horizontal surface with no vertical component of applied force. On an incline N = mg cos θ; with a push at an angle, N changes; in an accelerating elevator, N changes. Deriving N from the perpendicular equilibrium condition every time costs ten seconds and prevents a whole family of errors.
• On any apparent-weight question, write ΣF = ma with a stated sign convention before substituting numbers. Deciding "heavier or lighter" by intuition works about as often as it fails; the equation N = m(g + a) with a signed correctly never does.

Practice Physics C: Mech

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 7 units of AP Physics C: Mechanics

1. Unit 1 · Kinematics
2. Unit 2 · Force and Translational Dynamics
3. Unit 3 · Work, Energy, and Power
4. Unit 4 · Linear Momentum
5. Unit 5 · Torque and Rotational Dynamics
6. Unit 6 · Energy and Momentum of Rotating Systems
7. Unit 7 · Oscillations