Choose the number of pulleys at the top: ① one pulley with two masses (Atwood machine) · ② two pulleys and a knot, with a third
mass hanging from the knot · ③ three pulleys, all three masses meeting at one knot.
Pick an example (lab set-ups and problems from the cards, the problem sets and the guides) or set the masses yourself. Set a value with ▼ ▲ or type it in
the box; it is used as soon as you stop typing.
Set each value with ▼ and ▲ (hold to go faster; the ↑ ↓ keys or the mouse wheel in the box also work, Shift = ×10), or type a value in the box. ↺ puts it back to its start value. Outside the classroom range the box turns amber ⚠: fine to explore, but the lab can't do this.
Move the pointer over any graph (or tap it): a dashed line shows the values where it cuts each curve, or the values of the nearest data point.
1 pulley: press ▶ Release. The heavier mass falls the height h to the floor. ◂ / ▸ (or the ← → keys) step
through the motion. Choose light gates, ±2 % or a stopwatch for the timing, and add friction in the pulley to see a real pulley.
2 or 3 pulleys:drag a pulley to move it on the board, or change a mass. The knot swings and settles where the forces balance
(Balanced ✔). Read the angles on the protractor around the knot. ↺ puts the knot back to its start position.
The middle column shows the free-body diagrams. With a knot it also shows the vector triangle (it closes when the knot is
balanced) and a table of the horizontal and vertical components.
● Record saves a trial. The Data tab plots a against (m₂ − m₁)/(m₁ + m₂) (gradient = g), or
the components of the forces on the knot (gradient = 1 when they balance).
Work through the Tasks tab. Write explanations on your worksheet.
The physics
A light string over a light, smooth pulley has the same tension all along it. A fixed pulley only changes the direction of the force.
Atwood: a = (m₂ − m₁)g / (m₁ + m₂) · T = 2m₁m₂g / (m₁ + m₂)
Whole system: the resultant force along the string is m₂g − m₁g; it accelerates the total mass m₁ + m₂.
T is an internal force, so it does not appear. The tension lies between the two weights.
Measure g: a = 2h/t² from the drop. Keep m₁ + m₂ constant (move small masses from one side to the other);
then a = g × (m₂ − m₁)/(m₁ + m₂) − (friction / total mass), so the gradient of the a–x line is g even with friction.
Knot at rest: the resultant force is zero, ΣFx = 0 and ΣFy = 0, and the three tension
vectors drawn tip to tail make a closed triangle. Each tension equals the weight of its hanging mass, T = mg.
Masses in the ratio 3 : 4 : 5 make a right angle between the two strings; three equal masses make 120° between every pair of strings.
Balance is only possible when each tension is smaller than the sum of the other two.
Model: g = 9.81 N kg⁻¹; light, inextensible strings. Atwood: pulley friction is a constant force against the motion. Knot: smooth
pulleys of radius 3 cm; the motion of the knot is damped (air, pulley bearings) so that it settles. The free-body diagram of the knot uses
T = mg; while the knot moves, the gap in the vector triangle is the unbalanced force that makes it move.
Newton's laws
The badge above the scene says which law applies at this moment: 1st law — at rest or moving at a constant velocity, the resultant force is zero (ΣF = 0); 2nd law — accelerating, ΣF = ma. The table under the free-body diagram adds up the forces (ΣF) and compares them with m a (▾ folds it). Tick 3rd-law pairs to see every force with its partner: the same size and type, in the opposite direction, acting on the other body (the Earth, the floor, the string, the fluid…).