Equilibrium and free-body diagrams

When forces balance

Narrated video. Transcript:

A particle is in equilibrium when the net external force on it is zero. Then it is either at rest or moving with constant velocity.

Two forces: a lamp hanging from a cord. The tension pulls up and the weight pulls down. For equilibrium they must be equal and opposite.

Three forces: a small ring pulled by three strings. Draw the three force arrows from the ring. Their vector sum must be zero.

Now slide the arrows so they join head to tail. If the ring is in equilibrium, the three arrows close up into a triangle.

We can also use components. In equilibrium, the forces along x add to zero, and the forces along y add to zero.

Example: a 6 kg mass hangs from a light rope. Someone pulls the middle of the rope sideways with 50 newtons. Watch the upper half of the rope tilt.

At the middle point, three forces meet: 58.8 newtons down, 50 newtons sideways, and the tension along the upper rope.

Vertical balance: T cos theta equals 58.8. Sideways balance: T sin theta equals 50. Divide them: tan theta equals 50 over 58.8, so theta is about 40 degrees.

Free-body diagrams step by step

Narrated video. Transcript:

Most mechanics problems have several bodies: blocks, strings, pulleys, a floor. The trick is to look at one body at a time.

Step one: draw a simple diagram of the whole set-up. Step two: choose one body as your system.

Step three: draw that body alone as a dot or a box. Draw only the forces acting on it, from everything it touches, plus its weight. This is a free-body diagram.

Do not include forces that the body exerts on others. Those belong in the free-body diagram of the other body.

Example: two blocks on a smooth floor, 4 kg in front and 2 kg behind, joined by a light string. Pull the front block with 12 newtons.

Treat both blocks together: 12 newtons moves 6 kilograms, so the acceleration is 2 metres per second squared.

Now the free-body diagram of the back block alone. Only the string pulls it forward. So the tension is 2 kg times 2, which is 4 newtons.

Check with the front block: 12 newtons forward, 4 newtons back from the string. Net 8 newtons on 4 kilograms gives 2 metres per second squared. It matches.

The same method works in a lift. For a person on a weighing scale, N minus m g equals m a. Accelerate upwards and the scale reads more than your weight.

Try it: forces on a block at rest

Interactive simulation. Things to try:

• The inset shows the free-body diagram: the block as a point with its three forces.
• While the block is at rest, the three forces join head to tail into a closed triangle.
• Raise the angle until the block slides. The triangle no longer closes: the gap is the net force ma.
• Check the components: N = mg cos θ, and friction = mg sin θ while the block stays put.

Quick check

A 2 kg block rests on a 30° incline. The coefficient of static friction is 0.7. What is the friction force on the block? (g = 9.8 m/s²)

1. 9.8 N
2. 11.9 N
3. 17.0 N
4. 19.6 N

Weighing in a lift

A 60 kg student stands on a weighing scale in a lift. The lift accelerates upwards at 2 m/s². What does the scale read? (g = 9.8 m/s²)

1. 468 N
2. 588 N
3. 708 N
4. 120 N

Summary

• Equilibrium of a particle: F₁ + F₂ + F₃ + … = 0
• In components: ΣFx = 0, ΣFy = 0, ΣFz = 0
• Person in a lift: N = m(g + a)

• A particle is in equilibrium when the net external force on it is zero.
• Three forces in equilibrium, placed head to tail, form a closed triangle.
• A free-body diagram shows one body and only the forces acting on it. Draw one for each body.
• Apply F = ma to each free-body diagram, and check your answer with another body or with the whole system.