Newton’s First Law: Study Guide (Physics C: Mechanics 2.4)

Which course this is for. This guide is for the calculus-based Physics C: Mechanics course. Physics 1 also covers Newton’s first law; here we also connect it to velocity as a derivative. Keep the two separate when you revise.

Net force: add every force as a vector

In Topic 2.3 you learned to find every force on a system and its partner. Now add up the forces on the system only. The net force is their vector sum:

ΣF = F₁ + F₂ + F₃ + …

Forces are vectors, so add them by components. Choose axes first, then:

ΣF_x = F₁ₓ + F₂ₓ + F₃ₓ + … and ΣF_y = F₁ᵧ + F₂ᵧ + F₃ᵧ + …

Two rules keep this honest:

• Include only forces exerted on the system by things outside it. Third-law partners act on other objects, so they never enter this sum. Internal forces cancel in pairs (Topic 2.3).
• Keep the signs. A force of 40 N in −x contributes −40 N to ΣF_x.

Translational equilibrium

A system is in translational equilibrium when the forces on it add to zero:

ΣF = 0, which in components means ΣF_x = 0 and ΣF_y = 0 (and ΣF_z = 0 in three dimensions).

Each component equation is a separate condition. That is useful: in two dimensions, equilibrium gives you two equations, so you can find two unknowns, such as two tensions.

Equilibrium does not require the system to be at rest. A lift moving up at a steady 2 m/s is in equilibrium, exactly like a lift parked at a floor.

Newton’s first law

If the net force on a system is zero, the velocity of the system stays constant. Constant means both speed and direction stay the same. Rest (v = 0) is one special case.

In the language of Unit 1: constant velocity means

a = dv/dt = 0, so the position is a straight-line function of time: r(t) = r₀ + v t.

So you can read the first law in both directions on a graph:

• If an x–t graph is a straight line (constant slope), the forces along x are balanced.
• If the slope of x–t is changing, the net force along x is not zero.

Three consequences are worth stating plainly.

1. Motion does not need a force. A puck sliding on smooth ice keeps going with no forward push. Everyday objects slow down because friction or drag acts on them.
2. A change in velocity needs a net force. Speeding up, slowing down or turning all need ΣF ≠ 0. A car rounding a bend at constant speed is not in equilibrium, because its direction is changing.
3. “At rest for an instant” is not equilibrium. A ball at the top of its flight has v = 0 for an instant, but gravity acts on it, so its velocity keeps changing.

Balanced in one direction, unbalanced in another

The component equations work independently. Forces can be balanced along x while unbalanced along y. Then:

• the velocity component along the balanced axis stays constant;
• the velocity component along the unbalanced axis changes;
• the velocity changes only in the direction of the unbalanced force.

A projectile with air resistance ignored is the familiar case. Gravity acts only vertically, so ΣF_x = 0: the horizontal velocity stays constant while the vertical velocity changes. Worked example 2 shows the same idea for a drone.

Inertial reference frames

Velocity depends on who is measuring it (Topic 1.4). So “constant velocity” needs a reference frame. An inertial reference frame is one in which an observer finds the first law holds: objects with zero net force keep a constant velocity.

Test a frame by watching an object with no net horizontal force. Picture a ball resting on the smooth floor of a train carriage. When the train brakes:

• A platform observer sees the ball keep moving forwards at the same velocity, because no horizontal force acts on it. The carriage slows down underneath it. The first law holds: this frame is inertial (to a very good approximation).
• A passenger sees the ball start to move forwards from rest, even though nothing pushes it forwards. The first law fails in the braking carriage’s frame: that frame is not inertial.

Any frame moving at constant velocity relative to an inertial frame is also inertial. Any frame that accelerates (speeds up, slows down or turns) relative to an inertial frame is not. In this course, the ground is treated as an inertial frame unless a question says otherwise.

Worked example 1: a banner held by two cables

Question. A banner of mass 12 kg hangs at rest from two cables. Cable 1 pulls up and to the left at 35° above the horizontal; cable 2 pulls up and to the right at 50° above the horizontal. Take +x to the right and +y upward. Find the two tensions.

1. Weight: F_g = mg = 12 × 9.8 = 117.6 N, in −y.
2. Components of the tensions: T₁ = (−T₁ cos 35°, T₁ sin 35°) and T₂ = (T₂ cos 50°, T₂ sin 50°).
3. The banner is at rest, so ΣF_x = 0: −T₁ cos 35° + T₂ cos 50° = 0, which gives T₂ = T₁ cos 35° / cos 50° ≈ 1.274 T₁.
4. ΣF_y = 0: T₁ sin 35° + T₂ sin 50° − 117.6 = 0.
5. Substitute: T₁(0.5736 + 1.274 × 0.7660) = 117.6, so T₁ ≈ 76 N and T₂ ≈ 97 N.

Check. Horizontal components: 76 cos 35° ≈ 62 N left and 97 cos 50° ≈ 62 N right, so they balance. Vertical components: 43.5 N + 74.1 N = 117.6 N, matching the weight. The steeper cable carries more tension, which is what you should expect: it does more of the lifting.

Worked example 2: balanced along one axis only

Question. Take +x horizontal, along the flight path and +y upward. A camera drone’s position, measured from its take-off point, is x(t) = (3.0 m/s)t and y(t) = (0.25 m/s²)t² for 0 ≤ t ≤ 4.0 s. After 4.0 s it climbs at a steady rate while x continues as before. Decide in which directions the forces on the drone are balanced, describe the net force, and sketch v_x and v_y against time from 0 to 8.0 s.

1. Differentiate: v_x = dx/dt = 3.0 m/s (constant); v_y = dy/dt = 0.50t m/s.
2. Differentiate again: a_x = 0; a_y = 0.50 m/s².
3. Horizontal: v_x is constant, so by the first law the horizontal forces are balanced: ΣF_x = 0.
4. Vertical (0 to 4.0 s): v_y is changing, so ΣF_y ≠ 0. The net force points in +y (v_y is increasing). The drone is not in equilibrium, even though it moves at a steady 3.0 m/s horizontally.
5. At t = 4.0 s: v_y = 2.0 m/s and y = 4.0 m. After that, v_y stays at 2.0 m/s, so ΣF = 0 in both directions: the drone is in equilibrium, moving at √(3.0² + 2.0²) ≈ 3.6 m/s at about 34° above the horizontal.

We did not need to know the separate forces (rotor thrust, weight, air resistance). The motion alone tells us which components of the net force are zero.

Common misconceptions

• “A moving object needs a force to keep moving.” With zero net force it keeps its velocity. Friction and drag are what slow real objects.
• “Equilibrium means at rest.” It means ΣF = 0. Constant velocity counts.
• “v = 0, so the forces are balanced.” At the top of a throw, v = 0 for an instant but gravity is unbalanced.
• “Constant speed means equilibrium.” Turning at constant speed is a change in velocity, so the net force is not zero.
• “If any force is unbalanced, the whole velocity changes.” Only the component along the unbalanced direction changes (Worked example 2).
• Adding third-law partners into ΣF. Only forces exerted on the chosen system belong in the sum.
• Using the first law in an accelerating frame. In a braking train or a turning car, the first law appears to fail. Analyse from an inertial frame, such as the ground.

Where this leads

Topic 2.5, Newton’s Second Law, tells you what happens when ΣF is not zero: the net force sets the acceleration. Equilibrium problems like Worked example 1 return throughout the course, and in Unit 5 you will add a second condition for objects that can rotate. Try the practice questions now, then use the revision notes and the checklist. Next topic: Newton’s second law. You can also return to the course roadmap.