Newton's First Law: Study Guide (Physics 1 2.4)

This guide is for the algebra-based Physics 1 course. Everything here uses algebra, trigonometry and diagrams. If you are taking the calculus-based Physics C: Mechanics course, it has its own separate guide.

Net force is a vector sum

Topic 2.2 showed how to draw every force on a system as an arrow from a dot. Topic 2.3 showed that each of those forces has a partner acting on something else. Now we add up the arrows on one free-body diagram.

The net force on a system is the vector sum of all the forces exerted on it:

ΣF = F₁ + F₂ + F₃ + …

The Greek letter Σ (sigma) means “sum of”. Forces are vectors, so you add them as vectors.

• In one dimension, choose a positive direction and add signed values. A 50 N push to the right and a 20 N friction force to the left give ΣF_x = +50 − 20 = +30 N.
• In two dimensions, resolve each force into components (Topic 1.5), add the x-components, add the y-components, then rebuild the vector if you need it.

Example: two children pull a sled across level ice. One pulls 30 N east, the other 40 N north. Take +x east and +y north. ΣF_x = +30 N and ΣF_y = +40 N, so the net horizontal force has size √(30² + 40²) = 50 N, at tan⁻¹(40 ÷ 30) ≈ 53° north of east. (The sled’s weight and the normal force from the ice cancel in the vertical direction.)

Only forces on the system go into the sum. Third-law partners act on other objects, so they never appear.

Translational equilibrium

A system is in translational equilibrium when the net force on it is zero:

ΣF = 0, which in two dimensions means ΣF_x = 0 and ΣF_y = 0.

Both equations must hold. This is the starting point for most force problems: draw the free-body diagram, choose axes, and write one equation per axis.

“Equilibrium” describes the forces, not the position. It includes:

• a book resting on a shelf (velocity zero, and staying zero);
• a lift moving up at a steady 2 m/s;
• a parachutist falling at a steady speed.

Newton’s first law

Newton’s first law: if the net force on a system is zero, the velocity of the system stays constant.

“Constant velocity” means both the speed and the direction stay the same. So:

• an object at rest with zero net force stays at rest;
• an object moving with zero net force keeps moving in a straight line at the same speed.

Nothing “runs out”. A hockey puck slows on real ice because friction and air resistance act on it. On perfectly frictionless ice it would glide for ever. This tendency of objects to keep their velocity is called inertia.

You will also use the law the other way round. If you observe constant velocity, you can conclude that the net force is zero, even if you cannot see all the forces. Topic 2.5 shows why: any non-zero net force changes the velocity.

What the first law looks like on a graph

When ΣF = 0 for a whole time interval:

• the velocity–time graph is a horizontal line (zero slope: no acceleration);
• the position–time graph is a straight line, of any slope. A horizontal one means the object is at rest.

When a non-zero net force acts, the velocity–time graph has a non-zero slope. When that force stops, the graph goes flat again at its new value. It does not drift back to its old value.

Balanced in one direction, unbalanced in another

The equations ΣF_x = 0 and ΣF_y = 0 are separate. Forces can balance along one axis while they do not balance along the other. Then:

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

You met this in projectile motion (Topic 1.5). With air resistance ignored, a ball in flight has no horizontal force on it, so v_x is constant. Gravity acts vertically, so only v_y changes.

Figure 1 shows another case. A toy hovercraft glides across a smooth floor at 3.0 m/s in the +x direction. Between t = 1.0 s and t = 2.0 s it crosses a strip where a fan blows steadily across its path, in the +y direction.

Two things to notice. The sideways push does not slow the hovercraft’s forward motion. And once the fan stops, the hovercraft does not swing back to its old direction: with zero net force, the new velocity stays constant.

Inertial reference frames

Topic 1.4 described an inertial frame as one that is not accelerating. The first law gives a test you can actually carry out:

An inertial reference frame is one in which an observer finds the first law to be true. Objects with zero net force on them stay at rest or keep a constant velocity.

Sit in a bus that brakes sharply. A ball resting on the smooth floor suddenly rolls forwards, relative to the bus. No horizontal force acts on it: the floor is smooth and nothing touches it from the side. In the bus’s frame, the ball’s velocity changes with zero net force. The first law fails, so the braking bus is not an inertial frame.

Seen from the pavement (an inertial frame), there is nothing strange. The ball keeps moving forwards at the bus’s old speed, as the first law says. It is the bus that slows down underneath it.

In this course, treat the ground as an inertial frame unless a question says otherwise.

Worked example 1: a swing held to one side

Question. A child sits on a swing. Treat the swing’s two ideal ropes as a single rope with tension F_T (the total pull of both ropes). A parent holds the child and seat (total mass m) at rest with a horizontal pull F, so that the rope makes angle θ with the vertical. Take +x in the direction of the parent’s pull and +y upward.

(a) Derive expressions for F and F_T in terms of m, g and θ. (b) Evaluate them for m = 30 kg and θ = 25°.

Free-body diagram. Three forces on the child-and-seat system: weight F_g = mg (down); the parent’s pull F (horizontal, +x); the rope’s tension F_T (along the rope, up and back towards the support, at θ from the vertical).

1. At rest, and staying at rest, so equilibrium: ΣF_x = 0 and ΣF_y = 0.
2. y-direction: F_T cos θ − mg = 0, so F_T = mg ÷ cos θ.
3. x-direction: F − F_T sin θ = 0, so F = F_T sin θ = (mg ÷ cos θ) × sin θ, which gives F = mg tan θ.
4. (b) mg = 30 × 9.8 = 294 N. F_T = 294 ÷ cos 25° ≈ 324 N. F = 294 × tan 25° ≈ 137 N.

Interpretation. The rope’s vertical component holds up the weight, so the tension must be more than mg once the rope tilts. Its horizontal component pulls the child back towards the middle, and the parent’s pull balances exactly that.

Check with limiting cases.

• θ = 0 (rope vertical): F = 0 and F_T = mg. No pull is needed, and the rope holds the whole weight. Correct.
• θ close to 90° (rope nearly horizontal): tan θ and 1 ÷ cos θ both grow without limit. No horizontal pull, however large, can hold the rope perfectly horizontal while it supports a weight.
• Units: tan θ has no unit, so F is in newtons, like mg.

Worked example 2: a lamp hung from two cords

Question. A lamp of mass 3.0 kg hangs at rest from two light cords. Each cord rises at 20° above the horizontal, one to the left and one to the right, attached to two hooks at the same height. Find the tension in each cord.

1. At rest, so equilibrium. Take +x to the right and +y up. The lamp’s weight is F_g = mg = 3.0 × 9.8 = 29.4 N.
2. x-direction: F_T2 cos 20° − F_T1 cos 20° = 0, so F_T1 = F_T2. Call it F_T. (Symmetry tells you this too.)
3. y-direction: 2F_T sin 20° − mg = 0, so F_T = mg ÷ (2 sin θ), where θ = 20° is each cord’s angle above the horizontal.
4. Numbers: F_T = 29.4 ÷ (2 × 0.342) ≈ 43 N in each cord.

Check. Each cord’s vertical component is 43 × sin 20° ≈ 14.7 N; two of them make 29.4 N, the weight. The horizontal components (about 40 N each) cancel each other.

Limiting cases.

• Cords vertical (θ = 90°): F_T = mg ÷ 2 = 14.7 N. Each cord holds half the weight.
• Cords nearly flat (θ = 5°): F_T ≈ 170 N, almost six times the weight. As θ → 0, the tension needed grows without limit. A cord pulled perfectly straight and horizontal cannot hold up any weight.

Common misconceptions

• “A moving object needs a force to keep it moving.” No. It needs a force to change its velocity. Constant velocity needs zero net force.
• “Equilibrium means at rest.” It means ΣF = 0. Constant velocity counts too.
• “An object at rest has no forces on it.” The book on a shelf has two forces on it. They add to zero.
• “At the top of its flight, a ball is in equilibrium because v = 0.” Gravity still acts, so ΣF ≠ 0. Its velocity is zero only for an instant.
• “A sideways push slows the forward motion.” Only the component of velocity along the unbalanced force changes (Figure 1).
• “When a force stops, the object slows back to how it was.” The velocity stays at its new value.
• “Forces in a third-law pair add to zero, so the object is in equilibrium.” Pairs act on different objects. Only forces on the same diagram go into ΣF.
• “Something pushed the ball forwards when the bus braked.” Nothing did. The bus is a non-inertial frame.

Where this leads

The first law covers the case ΣF = 0. Topic 2.5, Newton’s Second Law, says exactly how velocity changes when ΣF is not zero. Try the practice questions now, then use the revision notes and the checklist to consolidate. To review force pairs, see Topic 2.3, or return to the course roadmap.