PHYSICS · LESSON 04 OF 9
Newton laws and force diagrams
Forces change motion. Isolate one object, draw every force acting on it, and apply ΣF = ma along each axis.
What this lesson explains
Newton’s laws explain and predict motion from its causes. Every structure, machine and vehicle is designed by balancing or using forces: cables in tension, supports pushing up, engines pushing forward, friction resisting. The free-body diagram is the single most important tool in mechanics; nearly every mistake in force problems is a mistake in the diagram.
Before you begin
Vector components
A force F at angle θ above the +x axis has Fx = F cos θ and Fy = F sin θ (see Vectors and motion).
Weight
Weight is the gravitational force: W = mg, with g = 9.8 m/s² near the Earth’s surface. A 5 kg mass weighs 49 N. Mass (kg) is not weight (N).
The idea, made visible
First law (inertia): if the net force on an object is zero, its velocity stays constant — it stays at rest or keeps moving in a straight line at constant speed. A force is not needed to keep something moving; it is needed to change its motion.
Second law: the net force equals mass times acceleration, ΣF = ma. It is a vector equation, so it holds separately along each axis: ΣFx = max and ΣFy = may. Mass measures how hard it is to change an object’s velocity. The unit of force is the newton: 1 N = 1 kg·m/s².
Third law: forces come in pairs. If A pushes on B, then B pushes on A with an equal and opposite force. The two forces act on different objects, so they never cancel each other on a single free-body diagram. The Earth pulls you down (your weight); you pull the Earth up equally.
A free-body diagram (FBD) shows one chosen object and only the forces acting on it: weight (mg, down), contact forces (normal force perpendicular to a surface, friction along it), tensions (along ropes, pulling away from the object), and applied pushes or pulls. Forces the object exerts on other things do not belong on its diagram, and neither does “ma” — ma is the result, not a force.
The normal force is whatever the surface must supply to stop the object sinking in. It equals the weight only in the simplest case (horizontal surface, no other vertical forces, no vertical acceleration). If you pull up on a rope at an angle, the floor pushes less; in an accelerating lift it changes too.
| motion of the lift | acceleration a | N = m(g + a) |
|---|---|---|
| at rest or constant velocity | 0 | 588 N |
| starting upward | +2 m/s² | 708 N |
| slowing while going up | −2 m/s² | 468 N |
| cable broken (free fall) | −9.8 m/s² | 0 N (“weightless”) |
The person’s weight mg = 588 N never changes; the normal force changes with the acceleration.
Key terms
- Force
- A push or pull on an object due to an interaction with another object; a vector measured in newtons (N).
- Net force ΣF
- The vector sum of all forces acting on the object.
- Equilibrium
- ΣF = 0, so a = 0: the object is at rest or moving at constant velocity.
- Normal force N
- The contact force perpendicular to a surface, preventing objects from passing through it.
- Tension T
- The pulling force exerted by a rope or cable, directed along it away from the object. For an ideal (massless) rope over an ideal pulley, the tension is the same throughout.
- Free-body diagram
- A sketch of one isolated object with arrows for every external force acting on it, plus chosen axes.
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| ΣF | net (vector sum of) force | N |
| m | mass | kg |
| a | acceleration | m/s² |
| W = mg | weight | N |
| N, T | normal force, tension | N |
Newton’s second law
ΣFx = max, ΣFy = may
Conditions and limits: In an inertial (non-accelerating) reference frame, for a body of constant mass. Sum only forces acting on the chosen body.
Equilibrium
ΣFx = 0, ΣFy = 0
Conditions and limits: Object at rest or moving with constant velocity.
Weight
W = mg
Conditions and limits: Near the Earth’s surface, g ≈ 9.8 m/s².
Third law
FA on B = −FB on A
Conditions and limits: Always; the two forces act on different bodies and are of the same type (both gravitational, both contact, …).
Apparent weight in a lift
N = m(g + a)
Conditions and limits: Vertical acceleration a, up positive. N is what a bathroom scale reads.
Why it works: Why the normal force is not always mg
Use the block in the figure: m = 5 kg, pulled by T = 20 N at 30° above horizontal, on a frictionless floor.
Vertical forces: N up, T sin 30° = 10 N up, mg = 49 N down.
Resolve every force along y.
The block does not leave the floor, so ay = 0: N + 10 − 49 = 0.
ΣFy = may with ay = 0.
N = 39 N, which is less than mg = 49 N.
The rope carries part of the weight.
Horizontally: T cos 30° = 17.3 N = max ⇒ ax = 3.46 m/s².
ΣFx = max.
N is found from the equations, never assumed. On an incline, or with an upward pull, or in a lift, it differs from mg.
A first worked example
Newton’s-law problem procedure
- Choose the object (or objects, separately). Draw a free-body diagram for each: weight, normal forces, friction, tensions, applied forces — only forces acting on it.
- Choose axes; for motion along a surface, put x along the surface. Mark the direction of the acceleration.
- Resolve every force into components along the axes.
- Write ΣFx = max and ΣFy = may for each object.
- For connected objects, use the constraint (same magnitude of acceleration for a taut rope) and the same tension on both ends of an ideal rope.
- Solve, then check units, signs and limiting cases (e.g. does the answer reduce correctly when an angle is 0?).
Acceleration from the net force
Problem. A 4 kg object has horizontal forces of 20 N to the right and 8 N to the left. Find its acceleration.
Take right as positive: ΣFx = 20 − 8 = 12 N.
Add forces with signs.
a = ΣF / m = 12 / 4 = 3 m/s2.
Newton’s second law.
Result: a = 3 m/s2 to the right.
What it means: Only the net force matters; the object could be moving either way at this instant.
A different case
Pulling at an angle
Problem. A 5 kg block on a frictionless floor is pulled by a rope with tension 20 N at 30° above horizontal. Find its acceleration and the normal force.
x: T cos 30° = 5ax ⇒ 17.32 = 5ax ⇒ ax = 3.46 m/s2.
Horizontal component causes the acceleration.
y: N + T sin 30° − mg = 0 ⇒ N = 49 − 10 = 39 N.
No vertical acceleration.
Result: a = 3.46 m/s2; N = 39 N.
What it means: Setting N = mg would be wrong here.
More worked cases
Each case below uses a different skill. Every step and result is shown.
Two connected blocks (Atwood machine)
Problem. Masses of 3 kg and 5 kg hang on either side of an ideal pulley, joined by a light string. Find the acceleration and the tension (g = 9.8 m/s2).
The heavier 5 kg mass goes down, the 3 kg mass up, both with the same magnitude a.
A taut, inextensible string forces equal accelerations.
3 kg (up positive): T − 3g = 3a.
FBD of the lighter mass.
5 kg (down positive): 5g − T = 5a.
FBD of the heavier mass; its positive direction is its direction of motion.
Add: 2g = 8a ⇒ a = 2 × 9.8 / 8 = 2.45 m/s2. Then T = 3(9.8 + 2.45) = 36.75 N.
Adding eliminates T.
Result: a = 2.45 m/s2; T ≈ 36.8 N.
What it means: T lies between the two weights (29.4 N and 49 N), as it must: it holds back the heavy mass and lifts the light one.
Equilibrium with two cables
Problem. A 10 kg sign hangs from two cables, each making 30° with the horizontal, symmetric about the sign. Find the tension in each cable.
Horizontal components cancel by symmetry.
Equal angles and equal tensions.
Vertical: 2T sin 30° − mg = 0 ⇒ 2T(0.5) = 98.
Equilibrium, ΣF_y = 0.
T = 98 N.
Solve.
Result: Each cable has tension 98 N — as large as the whole weight.
What it means: Shallow cables need large tensions: at 10° each tension would be 98/(2 sin 10°) ≈ 282 N.
Apparent weight
Problem. A 60 kg person stands on a scale in a lift accelerating upward at 2 m/s2. What does the scale read?
Forces on the person: N up, mg = 588 N down; a = +2 m/s2 (up positive).
FBD of the person.
N − mg = ma ⇒ N = 60(9.8 + 2) = 708 N.
Second law, vertical.
Result: The scale reads 708 N (about 72 kg on a scale calibrated in kg).
What it means: The scale measures the normal force, not the weight.
Common misunderstandings
Misunderstanding: “A moving object needs a net force to keep moving.”
Correct idea: Constant velocity needs zero net force. Net force causes acceleration, not motion.
Misunderstanding: Cancelling third-law pairs on one FBD.
Correct idea: The pair acts on two different objects. On one object’s diagram only one of them appears.
Misunderstanding: Assuming N = mg always.
Correct idea: Find N from ΣF_y = ma_y. Angled pulls, inclines and accelerating lifts change it.
Misunderstanding: Drawing “ma” as a force on the FBD.
Correct idea: ma is the result of the forces; it is not an extra force.
Misunderstanding: Using mass instead of weight (kg instead of N).
Correct idea: Weight is mg in newtons.
Going further (optional): Torque and rotational equilibrium (check whether your outline includes it)
A force can make an object turn. Its turning effect about a pivot is the torque τ = rF sin φ, where r is the distance from the pivot to where the force acts and φ is the angle between r and F. Only the perpendicular part of the force turns the object: a 10 N force at right angles to a 0.30 m wrench gives τ = 3.0 N·m, but the same force pushed along the wrench gives zero torque.
An object is in static equilibrium only when both the net force and the net torque are zero. Example: a light beam balances on a pivot with a 20 N load 2.0 m to the right and a 10 N load 4.0 m to the left. The clockwise torque 20 × 2.0 = 40 N·m equals the anticlockwise torque 10 × 4.0 = 40 N·m, so the beam does not turn; the pivot pushes up with 30 N so that the forces also balance.
Keep in mind
- ΣFx = max, ΣFy = mayNewton’s second law
- ΣFx = 0, ΣFy = 0Equilibrium
- W = mgWeight
- FA on B = −FB on AThird law
- N = m(g + a)Apparent weight in a lift
Scope of this lesson
- Friction and inclined planes are in the next topic; circular motion (net inward force) in its own topic.
- Ropes and pulleys are ideal (massless, frictionless) unless stated. Only static torque balance appears (optional section); rotational dynamics is not treated.
Next: Friction and inclined surfaces. Friction is the contact force along a surface. Static friction adjusts up to a limit; kinetic friction has a fixed size, μₖN, and opposes sliding.
Original study text. Sources and credits.