PhysSandbox

Newton’s Second Law: Net Force, Mass and Acceleration

A 45-minute guided lesson: students combine forces on a free-body diagram, measure the acceleration, discover that doubling the mass halves it, add perpendicular forces as vectors, and find the force that brings a body into equilibrium.

Level: Grades 9–11 · AP Physics 1 · IB Physics SL45 minSimulator used in this lesson: Force Diagram Builder

Learning objectives

  • Draw and read a free-body diagram and identify the net (resultant) force.
  • Use the simulator to show that acceleration is proportional to net force and inversely proportional to mass: vec a = sum vec F / m.
  • Add forces that are not along one line by components and by the parallelogram rule.
  • Find the single extra force that puts a body into equilibrium (sum vec F = 0).

Students should already know

  • Vectors: magnitude, direction, components with sine and cosine.
  • The difference between velocity and acceleration.

Standards

  • NGSS HS-PS2-1
  • AP Physics 1 · Unit 2 (Dynamics)
  • IB Physics · Topic A.2
  • Common Core HSN-VM.B.4
Warm-up

1.What does a force do?

~5 min

A crate sits on frictionless ice. You push it with a steady force. Before touching the simulator, decide which statements are true:

Explore 1

2.One force, one mass

~8 min

The diagram below shows a 2 kg object with a single force of 12 N pointing along +x. The readouts give the net force and the acceleration it produces.

Your task

  1. Read |a| for 12 N and 2 kg. Then drag the Magnitude |F| slider of Force 1 to 24 N. Did the acceleration double?
  2. Put the force back to 12 N and change the Mass to 4 kg, then 1 kg. Fill in a small table: mass, force, acceleration.
  3. Find a mass and force combination (other than 12 N / 2 kg) that gives exactly 6.000 m/s². How many such combinations exist?

You should see 6.000 m/s² at 12 N / 2 kg, 12.000 m/s² when the force is doubled and 3.000 m/s² when the mass is doubled. Acceleration is proportional to force and inversely proportional to mass — the content of Newton’s second law. Any pair with F / m = 6 gives the same acceleration: 6 N / 1 kg, 18 N / 3 kg, 30 N / 5 kg.

Newton’s second law

( sum vec F = m vec a\)

Solved for acceleration

( vec a = dfracsum vec Fm\)

Unit of force

( 1\ textN = 1\ textkg · textm/s^2\)

Explore 2

3.Forces that are not in line

~12 min

Now two forces act on the 2 kg object: 12 N along +x and 8 N along +y (90°). Forces are vectors, so the net force is neither 20 N nor 4 N.

Your task

  1. Read ΣF_x, ΣF_y and |ΣF|. Check with Pythagoras: sqrt12^2 + 8^2.
  2. Read |a| and its components. In what direction does the object accelerate? Estimate the angle from the dashed resultant arrow, then compute tan^-1(8/12).
  3. Change the angle of Force 2 from 90° to 0°, then to 180°. For which angle is |ΣF| largest? Smallest? Write a one-line rule.

The readouts give ΣF = (12, 8) N, |ΣF| = 14.42 N and |a| = 7.211 m/s² at 33.7° above the x-axis. The acceleration always points along the net force, never along any single force. Forces in the same direction add (20 N); opposite forces subtract (4 N); anything in between lies between these limits.

Components

( sum F_x = sum F_i cosθ_i, quad sum F_y = sum F_i sinθ_i\)

Magnitude

( |sum vec F| = sqrt(sum F_x)^2 + (sum F_y)^2\)

Direction

( θ = tan^-1\!( dfracsum F_ysum F_x )\)

Explore 3

4.Equilibrium: making the acceleration zero

~10 min

A 5 kg object is pulled by 10 N at 0° and 10 N at 120°. The third force is switched off. Your job is to find the single force that cancels the first two.

Your task

  1. First read the net force of the two active forces: ΣF_x, ΣF_y, |ΣF| and its direction.
  2. Predict the magnitude and angle of a third force that will make |ΣF| = 0. (Hint: equal magnitude, opposite direction to the current resultant.)
  3. Tick Include in diagram for Force 3 and set your predicted values. Adjust until |a| = 0.000 m/s². Record the values that work.
  4. Now move the third force’s angle 20° away from the balancing value. Is the object still in equilibrium? What is the acceleration now?

Two 10 N forces at 120° to each other have a resultant of exactly 10 N at 60° (the three vectors form an equilateral triangle). The balancing force — the *equilibrant* — is therefore 10 N at −120° (equivalently 240°). With it switched on the diagram closes into a triangle and the acceleration is zero: the object stays at rest, or keeps moving at constant velocity if it was already moving.

Worked example

5.A sled pulled at an angle

~5 min

Two ropes on a sled

A 15 kg sled on smooth ice is pulled by two ropes: 30 N at 20° above the +x axis and 20 N at −40° (i.e. 40° below the +x axis). Find the magnitude and direction of the acceleration.

  1. Components of rope 1: F_1x = 30 cos 20° = 28.2\ textN, F_1y = 30 sin 20° = 10.3\ textN.
  2. Components of rope 2: F_2x = 20 cos(-40°) = 15.3\ textN, F_2y = 20 sin(-40°) = -12.9\ textN.
  3. Net force: sum F_x = 43.5\ textN, sum F_y = -2.6\ textN, so |sum vec F| = sqrt43.5^2 + 2.6^2 = 43.6\ textN.
  4. Acceleration: a = 43.6 / 15 = 2.90\ textm/s^2, direction tan^-1(-2.6/43.5) = -3.4° — almost exactly along +x, slightly downward.

a ≈ 2.90 m/s² at −3.4° to the +x axis. Check in the simulator: m = 15, Force 1 = 30 N at 20°, Force 2 = 20 N at −40° — |a| should read 2.90.

Quiz

6.Check your understanding

Answer each question, then press Check. Numeric answers from the simulator are accepted within the stated tolerance.

1.A constant net force acts on a box sliding on frictionless ice. The box…

2.Forces of 12 N (along +x) and 8 N (along +y) act on a 2 kg object. What is the magnitude of its acceleration?

m/s²±3 % accepted

3.The same 12 N and 8 N forces now act in opposite directions on a 5 kg object. What is the acceleration?

m/s²±3 % accepted

4.Two 10 N forces act on a body at 120° to each other. The force needed to keep the body in equilibrium is…

5.Which objects have zero net force acting on them? (Select all that apply.)

6.A 2 kg object accelerates at 6 m/s². What acceleration will the same net force give a 3 kg object?

m/s²±3 % accepted

7.Homework

  1. Three forces act on a 4 kg object: 15 N at 30°, 15 N at 150° and one unknown force. Find the unknown force that makes the acceleration zero, then check it in the simulator (switch Force 3 on and enter your values). Open simulator with these settings ↗
  2. Set four forces of 10 N at 0°, 90°, 180° and −90°. Explain why the acceleration is zero. Then change only the mass. Does anything on the readouts change? Why not? Open simulator with these settings ↗
  3. A 0.5 kg hockey puck is hit with a 40 N force for 0.1 s. Using a = F/m and Δv = a·t, find the speed it leaves the stick with. Then decide: does the puck keep accelerating after it leaves the stick? Open simulator with these settings ↗

Notes for teachers

  • Pacing: Warm-up 5 min, Explore 1 8 min, Explore 2 12 min, Explore 3 10 min, worked example 5 min, quiz 5 min. Explore 3 (equilibrium) is the one to shorten if needed — ask for the prediction only and reveal 10 N at −120° on the board.
  • The simulator shows the diagram and the readouts only — nothing moves. This is deliberate for a first lesson on the second law: students focus on ΣF and a without being distracted by position and velocity. The Free Fall and Projectile Motion simulators show the resulting motion.
  • Angles are measured from the +x axis, counter-clockwise, in the range −180° … 180°. Students sometimes enter 240° — it is the same direction as −120°; the slider stops at 180°, so they must use the negative value.
  • Misconceptions targeted: (1) “force causes velocity” rather than acceleration — Explore 1 and quiz 1; (2) adding force magnitudes as numbers — Explore 2 and quiz 2; (3) “equilibrium means at rest” — Explore 3 callout and quiz 5.
  • Differentiation: students who finish early can derive the equilateral-triangle result of Explore 3 using components, or find two different equilibrants for a three-force system (there is only one — a good discussion of why).
  • Numeric quiz answers accept ±3 %, so values read from the readouts are accepted alongside calculated ones.