B. Dynamics · Dynamics: Forces and Motion in a Plane

Newton's Laws in Two Dimensions: Free-Body Diagrams and Components

B2.3 B2.1 A1.12About 75 minutes

Learning goals

I can draw an accurate free-body diagram for an object acted on by forces in two dimensions.
I can explain why the normal force is not always equal to mg.
I can resolve forces into components and apply ΣFx = max and ΣFy = may.
I can solve equilibrium problems in two dimensions, such as a sign or traffic light hanging from two cables.

Words to know

free-body diagram (FBD)
A diagram of one object, shown as a box or dot, with an arrow for every force acting on it, drawn from the object outward.

normal force (FN)
The push of a surface on an object, perpendicular to the surface. Its size adjusts to whatever is needed; it is not always equal to mg.

tension (FT)
The pulling force of a rope, string, cable or chain, directed along it, away from the object.

net force (ΣF)
The vector sum of all the forces acting on one object.

equilibrium
The state of an object with zero net force: it is at rest or moving at a constant velocity (ΣFx = 0 and ΣFy = 0).

applied force (Fa)
A push or pull by a person or a machine.

The toboggan rope

When you pull a toboggan by its rope, your hand is higher than the toboggan, so the rope pulls up and forward. The forward part speeds the toboggan up. But the upward part does something sneaky: it lifts a little of the toboggan's weight off the snow, so the snow pushes up less and the friction gets smaller.

Newton's laws do not change in two dimensions. What changes is the bookkeeping: we split every force into components and apply Newton's second law separately in each direction.

Remember from Grade 11

Newton's first law: with zero net force, an object stays at rest or keeps a constant velocity. Second law: Fnet = ma. Third law: forces come in equal and opposite pairs acting on different objects. You also used Fg = mg (with g = 9.8 N/kg) and kinetic friction Ff = μk FN, and drew free-body diagrams for one-dimensional problems in the SPH3U Forces unit.

The two-dimensional method

Isolate one object. Draw it as a box or a dot.
Draw every force acting on it, starting from the object: gravity (straight down), normal force (perpendicular to the surface), tension (along the rope), friction (along the surface, opposing sliding), applied forces. Label each one. Never draw ma as a force: it is the result, not a cause.
Choose axes. Put one axis along the direction of the acceleration (if you know it). Then the other axis has a = 0.
Resolve every force that is not along an axis into components.
Write two equations: ΣFx = max and ΣFy = may. Solve.

The normal force is found from the equation for the direction perpendicular to the surface. It equals mg only when nothing else pushes or pulls perpendicular to the surface and there is no acceleration in that direction.

Worked example

Example 1: Pulling a toboggan

A 25 kg toboggan (with a small child) is pulled across level snow by a rope at 30° above the horizontal with a tension of 80 N. The coefficient of kinetic friction is 0.10. Find (a) the normal force, (b) the friction force and (c) the acceleration.

Forces: Fg = 25 × 9.8 = 245 N down; FN up; FT = 80 N at 30°; Ff backward. Axes: x forward, y up. a(y) = 0.
FT components: x = 80 cos 30° = 69.3 N; y = 80 sin 30° = 40.0 N.
(a) ΣFy = 0: FN + 40.0 − 245 = 0, so FN = 205 N (not 245 N!).
(b) Ff = μk FN = 0.10 × 205 = 20.5 N.
(c) ΣFx = max: 69.3 − 20.5 = 25 a, so a = 48.8 ÷ 25 = 1.95 m/s².

(a) 205 N (b) 20.5 N (c) 1.95 m/s² forward (about 2.0 m/s²).

Free-body diagram for Example 1. The rope's upward component (40 N) reduces the normal force from 245 N to 205 N, which in turn reduces friction. (The tension and friction arrows are drawn at twice the scale of the others so they are visible.)

Worked example

Example 2: Push or pull?

A 30 kg box must be moved across a floor with μk = 0.30. A person can apply 150 N at 25° to the horizontal. Compare the acceleration when (a) pushing down on the box at 25° below the horizontal and (b) pulling up on it with a rope at 25° above the horizontal.

Horizontal component in both cases: 150 cos 25° = 135.9 N. Vertical component: 150 sin 25° = 63.4 N. mg = 294 N.
(a) Pushing down adds to the weight: FN = 294 + 63.4 = 357.4 N. Ff = 0.30 × 357.4 = 107.2 N. a = (135.9 − 107.2) ÷ 30 = 0.96 m/s².
(b) Pulling up lifts: FN = 294 − 63.4 = 230.6 N. Ff = 0.30 × 230.6 = 69.2 N. a = (135.9 − 69.2) ÷ 30 = 2.2 m/s².

(a) 0.96 m/s² (b) 2.2 m/s². Pulling upward at an angle is more than twice as effective here, because it reduces friction instead of increasing it. This is why a sled is easier to pull than to push.

Equilibrium in two dimensions

When an object is at rest (or at constant velocity), ΣFx = 0 and ΣFy = 0. Engineers use these two equations for every sign, bridge cable and hanging traffic light. Each cable pulls along its own direction, and the components must balance.

Worked example

Example 3: A traffic light on two cables

A 15 kg traffic light hangs from a knot joining two cables. The left cable makes 30° with the horizontal, and the right cable makes 45°. Find the tension in each cable.

Study the knot. Forces: T₁ up and to the left at 30°, T₂ up and to the right at 45°, and the light's weight 15 × 9.8 = 147 N down.
ΣFx = 0: T₂ cos 45° = T₁ cos 30°, so T₂ = T₁ (0.866 ÷ 0.7071) = 1.225 T₁.
ΣFy = 0: T₁ sin 30° + T₂ sin 45° = 147, so 0.500 T₁ + 0.7071(1.225 T₁) = 147.
0.500 T₁ + 0.866 T₁ = 147, so 1.366 T₁ = 147 and T₁ = 107.6 N.
T₂ = 1.225 × 107.6 = 131.8 N.

T₁ ≈ 108 N (left, 30° cable) and T₂ ≈ 132 N (right, 45° cable). The steeper cable carries more of the load.

Left: the traffic light hangs from a knot. Right: the free-body diagram of the knot, with arrows to scale. The two tensions' horizontal parts cancel, and their vertical parts add to 147 N.

Just for fun

Why did the free-body diagram feel so calm?

Because every force in its life was perfectly balanced. (Only in equilibrium problems, though.)

Hands-on activity

Find the best pulling angle

You need: a laundry basket or cardboard box with a few books in it, a strong rope or string tied to it, a luggage scale or spring scale, and a protractor (or a phone level app to measure angles).

Pull the basket across a smooth floor at a slow, constant speed with the rope horizontal. Read the scale. At constant velocity, the pull equals kinetic friction, so μk = F ÷ (mg). (Weigh the basket on a bathroom scale to get m.)
Repeat at rope angles of about 15°, 30°, 45° and 60° above the horizontal. Have a helper estimate the angle. Record the scale reading each time.
Plot pulling force against angle. Is there an angle that needs the least force?
Theory says the smallest force is needed when tan θ = μk (see the Challenge). Compare your best angle with tan⁻¹(μk).
Explain why very steep angles need a large force again, using the components of the pull.

Keep the path clear and pull slowly.

Your turn

A 10 kg crate is pulled across frictionless ice by a rope at 37° above the horizontal with 50 N. Find its acceleration and the normal force.

Two forces act on a 5.0 kg puck on frictionless ice: 30 N [E] and 40 N [N]. Find its acceleration.

A 2.0 kg lamp hangs from two cords, each at 60° above the horizontal (a symmetric V). Find the tension in each cord.

List every force on a sled being pulled up a snowy hill by a rope parallel to the slope, with friction.

A book rests on a table. Its weight is 12 N. What is the third-law partner of the normal force on the book?

Explain why pushing a lawnmower handle downward at a steep angle makes it harder to move.

More practice: warm-up

In which direction does the normal force always point?

In which direction does a tension force point?

What is the net force on an object moving at constant velocity?

Give the y component of a 40 N force at 30° above the horizontal.

A 5.0 kg box sits on a level floor with nothing else pushing on it. What is the normal force?

More practice: on your own

A 40 kg sled is pulled with 120 N at 20° above the horizontal; μk = 0.15. Find the acceleration.

A 4.0 kg picture hangs from a wire; each half of the wire makes 30° with the horizontal. Find the tension.

Forces of 10 N [E] and 10 N [N] act on an object at rest. A third force keeps it at rest. Find the third force.

A person pushes a 20 kg lawnmower at constant velocity with 90 N directed along the handle, 40° below the horizontal. Find the normal force and the total resistive force.

Two hockey sticks hit a 0.17 kg puck at the same time with 4.0 N [E] and 3.0 N [N]. Find the puck's acceleration during contact.

Challenge

A box of mass m is dragged at constant velocity by a rope at angle θ above the horizontal, with kinetic friction μ. Show that the required tension is F = μmg ÷ (cos θ + μ sin θ). Then, for m = 20 kg and μ = 0.40, compare F at 0°, 21.8° and 40°.

A 60 kg person stands at the middle of a slackline. Each side of the line sags to 10° below the horizontal. Find the tension and explain why slacklines need strong anchors.

A 50 kg crate on a frictionless floor is pulled by two ropes: 100 N [N 30° E] and 100 N [N 30° W]. Find its acceleration.

Check your understanding

Quiz

A rope pulls a sled at 30° above the horizontal. Compared with the sled's weight, the normal force is:

Which of these should never appear as an arrow on a free-body diagram?

A 10 kg block is pushed with 50 N at 30° below the horizontal on a level floor. The normal force is:

A sign hangs at rest from two cables. Which statement is true?

A 2.0 kg object has forces 6.0 N [E] and 8.0 N [S] acting on it. Its acceleration is:

The third-law partner of the Earth's gravitational pull on you is:

For parents and teachers

Big idea: Newton's laws apply separately along each axis, so a two-dimensional problem becomes two one-dimensional ones (B2.3). Free-body diagrams (A1.12) are the most important tool in this unit.

How to help: Before any calculation, ask your teen to show you their free-body diagram and name what exerts each force ("the rope", "the floor", "the Earth"). If they cannot name the source, the force probably is not real.

Common difficulties:

Writing FN = mg automatically, even when a rope pulls up or a person pushes down.
Drawing "ma" or "the force of motion" as an extra arrow.
Mixing up sine and cosine when resolving forces. A quick sketch of the right triangle helps.
Pairing the normal force with weight as a third-law pair (they act on the same object, so they cannot be).