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Study Notes

Unit 2 — Force and Translational Dynamics

Why objects move the way they do: systems, forces, free-body diagrams, Newton's three laws, gravity, friction, springs, and circular motion. Each section ends with a short practice set — click "Show answer" to check yourself.

2.1

Systems and Center of Mass

A system is an object, or a collection of objects, that you choose to analyze. If you ignore its internal structure, a system can be treated as a single object.

  • Internal forces act between objects inside the system. External forces come from outside it.
  • Only external forces can change the motion of the system's center of mass.
  • The center of mass is the point that moves as if all the mass were concentrated there and all external forces acted on it.
x_cm = (m₁x₁ + m₂x₂ + …) / (m₁ + m₂ + …)
a_cm = F_net,external / m_total
  • For a uniform, symmetric object, the center of mass is at its geometric center.
  • The center of mass does not have to be located on the object (e.g. a ring or a boomerang).
Why it matters: the choice of system decides which forces count as internal (ignore) and which as external (use).
Practice

1. A 2 kg mass sits at x = 0 and a 4 kg mass sits at x = 6 m. Where is the center of mass?

Show answer
xcm = (2·0 + 4·6)/(2 + 4) = 24/6 = 4 m. It lies closer to the heavier mass.

2. Two ice skaters at rest push off each other. Why does their center of mass stay at rest?

Show answer
The forces they exert on each other are internal to the two-skater system. With no net external force (ignoring friction), the center of mass keeps its velocity, which is zero.
2.2

Forces and Free-Body Diagrams

A force is a vector quantity describing an interaction between two objects. Its SI unit is the newton: 1 N = 1 kg·m/s². Forces can be contact (push, tension, friction, normal) or long-range (gravity).

ForceSymbolDirection
Gravitational (weight)Fg = mgToward Earth's center (down)
NormalFNPerpendicular to the surface, away from it
TensionFTAlong the rope, pulling on the object
FrictionFfParallel to the surface, opposing (impending) sliding
SpringFsToward the spring's equilibrium length

A free-body diagram (FBD) shows all forces acting on one object:

  • Draw the object as a dot or box; draw each force as an arrow starting at the object.
  • Arrow length reflects magnitude; label every force with its type.
  • Include only forces acting on the object, never forces it exerts on others.
  • Do not draw the net force, acceleration, or "force of motion" on the FBD.
Common mix-up: there is no force "in the direction of motion" unless something is actually pushing or pulling that way.
Practice

1. Draw (describe) the FBD for a book at rest on a table.

Show answer
Two arrows: Fg pointing down and FN pointing up, of equal length because the book is not accelerating.

2. A block slides down a rough incline. Name the forces on it and their directions.

Show answer
Fg straight down; FN perpendicular to the incline surface; friction parallel to the surface, pointing up the slope (opposing the sliding).
2.3

Newton's Third Law

Whenever object A exerts a force on object B, B exerts a force on A that has equal magnitude, opposite direction, and is the same type of force.

F(A on B) = − F(B on A)
  • The two forces act on different objects, so they never cancel each other on a single FBD.
  • Equal forces do not mean equal accelerations. a = F/m, so the smaller mass accelerates more.
  • Third-law pairs exist at the same instant, and both forces are of the same kind (gravity pairs with gravity, contact with contact).
Tip: to find a pair, say "A pushes/pulls B" and then flip it: "B pushes/pulls A."
Practice

1. A heavy truck collides with a small car. Compare the forces each exerts on the other, and compare their accelerations.

Show answer
The forces are equal in magnitude and opposite in direction. The car has the larger acceleration because it has the smaller mass.

2. A book rests on a table. Is the normal force on the book the third-law partner of the book's weight?

Show answer
No. Both act on the book, and they are different types of force. The partner of the book's weight (Earth pulls book) is the book pulling up on Earth. The partner of the normal force is the book pushing down on the table.
2.4

Newton's First Law

An object keeps its velocity (including a velocity of zero) unless a net external force acts on it. This tendency is called inertia, and mass is its measure.

  • If ΣF = 0, the object is in translational equilibrium: it is at rest or moving at constant velocity.
  • Motion at constant velocity requires no net force. A net force is needed only to change velocity.
  • Equilibrium applies along each axis separately: ΣFₓ = 0 and ΣFy = 0.
Common mix-up: an object moving at constant velocity on a rough floor still has friction. The pushing force and friction simply balance.
Practice

1. A crate slides across a floor at constant velocity while you push it with 40 N. What is the friction force?

Show answer
Constant velocity means ΣF = 0, so friction is 40 N opposite to the push.

2. A 2.0 kg block hangs at rest from a rope. What is the tension? (g = 9.8 m/s²)

Show answer
ΣFy = 0 → FT = mg = 2.0 × 9.8 = 19.6 N.
2.5

Newton's Second Law

The acceleration of a system is directly proportional to the net external force and inversely proportional to its mass, and points in the direction of the net force.

a = ΣF / m        ΣFₓ = m·aₓ        ΣF_y = m·a_y

Problem-solving steps:

  • Choose the system and draw its FBD.
  • Pick axes (along the acceleration or along an incline) and resolve forces into components.
  • Write ΣF = ma for each axis, then solve.
  • Apparent weight is the normal force you feel. In an elevator accelerating upward at a: FN = m(g + a); accelerating downward: FN = m(g − a).
  • Mass is a constant property; weight depends on the local gravitational field.
Practice

1. A 4.0 kg block is pulled right with 20 N and left with 8 N. Find its acceleration.

Show answer
ΣF = 20 − 8 = 12 N right. a = 12/4.0 = 3.0 m/s² to the right.

2. A 60 kg person stands on a scale in an elevator accelerating upward at 2.0 m/s². What does the scale read? (g = 9.8 m/s²)

Show answer
FN − mg = ma → FN = 60(9.8 + 2.0) = 708 N, more than the person's true weight of 588 N.
2.6

Gravitational Force

Every pair of masses attracts each other with a force given by Newton's law of universal gravitation:

F_g = G·m₁·m₂ / r²        G = 6.67 × 10⁻¹¹ N·m²/kg²
g = G·M / r²             (field strength at distance r from mass M)
  • Gravity follows an inverse-square law: doubling r cuts the force to one quarter.
  • r is measured between the centers of the masses.
  • Near Earth's surface, Fg = mg with g ≈ 9.8 m/s². Gravitational mass and inertial mass are numerically equal.
  • All objects in free fall have the same acceleration g, regardless of mass, because m cancels in a = F/m.
Third law reminder: Earth pulls on you with the same magnitude force that you pull on Earth. You just accelerate far more.
Practice

1. If the distance between two planets' centers triples, what happens to the gravitational force between them?

Show answer
It becomes 1/9 of the original, since F ∝ 1/r².

2. A 10 kg object is at a distance of 2R from Earth's center (R = Earth's radius, i.e. one radius above the surface). What is its weight there? (g at surface = 9.8 m/s²)

Show answer
Field strength is g/4 = 2.45 m/s², so Fg = 10 × 2.45 = 24.5 N (vs. 98 N at the surface).
2.7

Kinetic and Static Friction

Friction is a contact force parallel to the surface that opposes relative sliding between two surfaces.

Static frictionKinetic friction
WhenSurfaces are not slidingSurfaces are sliding
MagnitudeFs ≤ μsFN (adjusts as needed)Fk = μkFN (fixed)
DirectionOpposes impending slidingOpposes sliding
  • Typically μs ≥ μk: it takes more force to start an object moving than to keep it moving.
  • Friction depends on the normal force and the surfaces, and not on contact area or speed (in this model).
  • The normal force is not always mg. On an incline, FN = mg cos θ; with a vertical pull or push it changes.
  • Static friction can point in the direction of motion, such as a car's tires pushing the car forward.
Practice

1. A 10 kg box slides on a floor with μk = 0.30. What is the kinetic friction force? (g = 9.8 m/s²)

Show answer
FN = mg = 98 N, so Fk = 0.30 × 98 = 29.4 N.

2. A 20 kg crate (μs = 0.50) is pushed horizontally with 60 N. Does it move? What is the friction force?

Show answer
Maximum static friction = 0.50 × (20 × 9.8) = 98 N. Since 60 N < 98 N, the crate does not move, and static friction is exactly 60 N opposite the push.
2.8

Spring Forces

An ideal spring exerts a restoring force proportional to how far it is stretched or compressed from its equilibrium length. This is Hooke's law:

F_s = − k·x
  • x is the displacement from equilibrium length; k is the spring constant in N/m (stiffer spring = larger k).
  • The negative sign means the force always points toward equilibrium.
  • On a graph of force vs. displacement, the slope equals k.
  • For a mass hanging at rest on a spring: kx = mg, so x = mg/k.
Note: x is the change in length from the natural length, not the total length of the spring.
Practice

1. A spring (k = 200 N/m) is stretched 0.15 m. What force does it exert?

Show answer
|Fs| = kx = 200 × 0.15 = 30 N, directed back toward equilibrium.

2. A 5.0 N force stretches a spring by 0.10 m. Find k, then find how far the spring stretches when a 2.0 kg mass hangs at rest from it.

Show answer
k = F/x = 5.0/0.10 = 50 N/m. At equilibrium, x = mg/k = (2.0 × 9.8)/50 ≈ 0.39 m.
2.9

Circular Motion

In uniform circular motion, an object moves in a circle at constant speed. Its velocity is always tangent to the circle, so its direction keeps changing. That means it is accelerating, even though its speed is constant.

a_c = v² / r            (toward the center)
ΣF_c = m·v² / r        (net force toward the center)
v = 2πr / T             (T = period)
  • The centripetal force is not a new kind of force. It is the net force (or component of net force) pointing toward the center, supplied by tension, gravity, friction, a normal force, etc.
  • Never draw "centripetal force" as an extra arrow on an FBD.
  • If the inward force disappears, the object moves off in a straight line tangent to the circle, not outward.
  • Doubling the speed requires four times the centripetal force at the same radius.
SituationProvides the centripetal force
Ball swung on a stringTension
Car turning on a flat roadStatic friction
Satellite in orbitGravity
Practice

1. A 0.50 kg ball on a string moves in a horizontal circle of radius 2.0 m at 4.0 m/s. Find its centripetal acceleration and the tension.

Show answer
ac = v²/r = 16/2.0 = 8.0 m/s². Tension supplies the centripetal force: FT = mac = 0.50 × 8.0 = 4.0 N.

2. A car rounds a flat curve of radius 50 m with μs = 0.80. What force provides the centripetal force, and what is the maximum speed without skidding? (g = 9.8 m/s²)

Show answer
Static friction. At the limit, μsmg = mv²/r, so v = √(μsgr) = √(0.80 × 9.8 × 50) ≈ 19.8 m/s. The mass cancels.