Aim for FivePhysics C: Mechanics

Unit 2 · Topic 2.4

2.4 Newton’s First Law

Newton's first law says that when the net force on a system is zero, its velocity doesn't change: it stays at rest or keeps moving in a straight line at constant speed. That state is translational equilibrium, and solving equilibrium problems means making the forces add to zero along each axis.

Key terms

  • net force
  • Newton's first law
  • translational equilibrium
  • inertia
  • inertial reference frame

Net force

The net force is the vector sum of every force on a system: F⃗net=∑F⃗\vec{F}_{net} = \sum\vec{F}. Add forces by components, as with any vectors.

Forces can be balanced along one axis and unbalanced along another. A block sliding on a frictionless floor has balanced vertical forces (normal force and weight) but may have an unbalanced horizontal one.

The first law

If the net force on a system is zero, its center of mass moves with constant velocity. "Constant velocity" includes being at rest. So zero net force doesn't mean "not moving"; it means "not accelerating".

Inertia is the tendency of an object to keep its velocity. Mass measures inertia: a more massive object is harder to speed up, slow down or turn.

No force is needed to keep something moving. A hockey puck on ice slows down only because of friction and air resistance, not because it "runs out of force". If it seems like a push is needed to keep something moving, that push is balancing friction.

The first law also defines an inertial reference frame: a frame where an object with no net force really does move at constant velocity. Newton's laws in their usual form only work in inertial frames.

Solving equilibrium problems

Translational equilibrium means ∑F⃗=0\sum\vec{F} = 0, which breaks into one equation per axis:

∑Fx=0,∑Fy=0\sum F_x = 0, \qquad \sum F_y = 0

The steps: draw the FBD, choose axes, split any angled forces into components, set each sum to zero, and solve. With two unknowns you need two equations, so use both axes.

Typical setups include a sign hanging from two cables, a box pushed at constant velocity against friction, and an elevator rising at constant speed. In each one, "at rest" or "constant velocity" in the problem is your signal that the net force is zero.

Equilibrium in one direction only

Many problems mix the two cases. A sled pulled across snow at constant velocity by a rope angled above the horizontal is in equilibrium in both directions: the rope's horizontal component equals kinetic friction, and the normal force plus the rope's vertical component equals the weight.

A block sliding down a frictionless incline is in equilibrium only perpendicular to the slope. The normal force balances mgcos⁡θmg\cos\theta, while mgsin⁡θmg\sin\theta is unbalanced and speeds the block up. Decide axis by axis whether the acceleration is zero.

Equilibrium is about the net force, not about any single force being zero. A book at rest on a table has two forces on it; they just add to zero.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    A traffic light on two cables

    A 100 N traffic light hangs at rest from two cables. Cable 1 makes 30° with the horizontal and cable 2 makes 60° with the horizontal, on opposite sides. Find the tension in each cable.

    Show the solution
    1. Step 1: FBD of the knot: weight 100 N down, T1T_1 up and to the left at 30°, T2T_2 up and to the right at 60°.
    2. Step 2: Horizontal: T1cos⁡30∘=T2cos⁡60∘T_1\cos 30^\circ = T_2\cos 60^\circ, so T2=3 T1T_2 = \sqrt{3}\,T_1.
    3. Step 3: Vertical: T1sin⁡30∘+T2sin⁡60∘=100T_1\sin 30^\circ + T_2\sin 60^\circ = 100. Substituting: 0.5T1+1.5T1=1000.5T_1 + 1.5T_1 = 100, so T1=50T_1 = 50 N.
    4. Step 4: Then T2=3(50)≈86.6T_2 = \sqrt{3}(50) \approx 86.6 N.
    5. Step 5: Check: the steeper cable carries more of the weight, which makes sense, since it points more nearly straight up.

    Answer: T1=50T_1 = 50 N (30° cable) and T2≈87T_2 \approx 87 N (60° cable).

  2. Example 2Calculator allowed

    An elevator moving up at constant speed (classic trap)

    An 800 kg elevator car rises at a constant 2.0 m/s. Find the tension in its cable, ignoring friction. Use g = 9.8 m/s².

    Show the solution
    1. Step 1: "Constant speed in a straight line" means zero acceleration, so the net force is zero.
    2. Step 2: Forces: tension up, weight down. T−mg=0T - mg = 0, so T=mg=(800)(9.8)=7840T = mg = (800)(9.8) = 7840 N.
    3. Step 3: The trap is thinking the tension must be bigger than the weight because the elevator is moving up. A bigger tension is needed only while the elevator speeds up.

    Answer: T = 7840 N, equal to the elevator's weight.

Common mistakes

  • Thinking a moving object must have a net force on it. Constant velocity means zero net force.
  • Using only one axis when there are two unknowns. Write ∑Fx=0\sum F_x = 0 and ∑Fy=0\sum F_y = 0 and solve them together.
  • Assuming each cable holding up a weight carries half of it. The angles decide how the load is shared.

On the exam

  • Look for the words "at rest", "constant velocity" or "constant speed in a straight line". They tell you to set the net force to zero.
  • Conceptual questions often ask whether a force is needed to keep something moving. Answer with the first law: only to balance other forces, such as friction.

Connected topics

Videos

Check yourself

4 questions on 2.4 Newton’s First Law. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A 500 kg elevator car rises at a constant 3.0 m/s, held by a single cable. What is the tension in the cable? Use g = 10 m/s².

Question 2 of 4Calculator allowed

A 20 kg traffic light hangs at rest from two cables. Each cable makes a 30° angle with the horizontal, and the arrangement is symmetric. What is the tension in each cable? Use g = 10 m/s².

Question 3 of 4Calculator allowed

A hockey puck slides across ice that is so smooth that friction is negligible. Which statement about the puck is correct?

Question 4 of 4Calculator allowed

A crate slides down a rough ramp and speeds up. Which statement about the forces on the crate is correct?

0 of 4 answered