Apply Newton's laws with free-body diagrams to systems of connected objects, elevators and multiple forces

Learning goal: Draw a free-body diagram for each object in a system, apply a = F_net/m to the object or to the whole system, use Newton's third law to relate the forces between connected objects, and solve for accelerations, tensions and contact forces in problems such as an Atwood machine, a cart pulled by a hanging mass, an elevator and an object under several forces in one dimension.

The three laws as tools. Newton's first law says an object's velocity stays constant unless a net force acts. The second law quantifies the change: a = F_net/m, where F_net is the vector sum of every force on that one object. The third law says that whenever object A exerts a force on object B, B exerts a force of equal magnitude and opposite direction on A; the two forces act on different objects and never cancel each other in one object's free-body diagram. In this course the laws are applied to systems of objects that move together, such as two masses linked by a string over a pulley, and to forces that must first be computed from their own laws: gravitational forces from weight or the universal law, elastic forces from Hooke's law, friction from its coefficients. Those force laws come in later lessons; here the forces are given and the method is the point.

Free-body diagrams. A free-body diagram shows one object as a dot or box and every force acting on it as an arrow pointing in the force's direction, labeled with its name and, where known, its size. It does not show forces the object exerts on other things, and it does not show velocity or acceleration as forces. Draw one diagram per object. Choose a positive direction for each object along the direction it will move, write F_net = ma along that direction, and solve. For a system that moves as one, you may also treat the whole system as a single object: internal forces such as the push or pull between two linked blocks on one line then cancel in pairs by the third law, and only external forces appear. For a string over a pulley the two ends of the string pull on different masses in the same direction (both upward in an Atwood machine), so those tensions are not a third-law pair and do not cancel as vectors; they drop out only when you add forces along the path of the string, where the tension holds one mass back exactly as much as it pulls the other forward.

Signs and the choice of positive direction. For a string over a pulley, the natural choice is "positive in the direction the system accelerates": down for the heavier hanging mass, up for the lighter one. With that choice both masses have the same positive acceleration a, and the equations line up. An ideal string is massless and does not stretch, so the tension is the same everywhere along it and both masses have the same speed and acceleration. An ideal pulley is massless and frictionless, so it changes the direction of the tension without changing its size. Real strings and pulleys break these assumptions slightly; that is one source of discrepancy in the investigation.

A fully worked model: the Atwood machine. Masses of 1.2 kg and 1.0 kg hang from a string over a light, frictionless pulley. Find the acceleration and the tension. Free-body diagram of the 1.2 kg mass: weight 1.2 × 9.8 = 11.76 N down, tension T up; positive is down, so 11.76 − T = 1.2a. Free-body diagram of the 1.0 kg mass: weight 9.8 N down, tension T up; positive is up, so T − 9.8 = 1.0a. Adding the two equations eliminates T: 11.76 − 9.8 = 2.2a, so a = 1.96 ÷ 2.2 = 0.89 m/s². Then T = 9.8 + 1.0 × 0.891 = 10.7 N. Check with the other equation: 11.76 − 1.2 × 0.891 = 10.7 N. Checks on sense: the tension lies between the two weights, as it must, since the heavy mass falls (weight beats tension) while the light one rises (tension beats weight). The system view, adding forces along the string's path, gives the same acceleration at once: net force (11.76 − 9.8) N along the string on a total mass 2.2 kg. In general a = (m₁ − m₂)g/(m₁ + m₂), which reduces to zero for equal masses and to g when m₂ is zero, both sensible limits. If the 1.2 kg mass starts from rest, it falls 1.0 m in t = √(2 × 1.0 ÷ 0.891) = 1.5 s, and that is how the acceleration is measured: time the fall and use a = 2d/t².

Cart and hanging mass. A 2.0 kg cart on a level, low-friction track is tied to a 0.50 kg mass hanging over a pulley at the track's end. Cart: T = 2.0a (only the tension acts horizontally). Hanging mass: 0.50 × 9.8 − T = 0.50a, so 4.9 − T = 0.50a. Adding: 4.9 = 2.5a, a = 1.96 m/s², T = 3.92 N. The tension is less than the hanging weight; if it were equal, the hanging mass could not accelerate. A common error is to treat the hanging weight as the force on the cart alone and get a = 4.9 ÷ 2.0 = 2.45 m/s²; that forgets that the 0.50 kg mass must accelerate too.

Elevators and apparent weight. A scale reads the normal force it exerts on whatever stands on it. When a 60 kg person stands on a scale in an elevator accelerating upward at 2.0 m/s², the free-body diagram has the normal force N up and weight 588 N down: N − 588 = 60 × 2.0, so N = 708 N. The scale reads more than the weight; this reading is called the apparent weight. Accelerating downward at 2.0 m/s²: 588 − N = 60 × 2.0, N = 468 N. Moving at constant velocity, up or down, N = 588 N. The person's actual weight, the gravitational force, is 588 N in all three cases; only the normal force changes.

Investigation: the Atwood machine. Materials: a pulley clamped to a stand at least 1.2 m above a mat, thin string, a 200 g and a 220 g mass (or slotted masses), a meter stick, a stopwatch or phone video, and a mat. Steps: hold the system at rest with the 220 g mass at a measured height of 1.00 m above the mat; release and time the fall; repeat five times; then move 10 g from the light side to the heavy side and repeat, keeping the total mass at 420 g. Record the fall times and compute a = 2d/t² for each mass difference. Analyze: predicted acceleration for 220 g and 200 g is (0.22 − 0.20) × 9.8 ÷ 0.42 = 0.47 m/s², giving a predicted fall time of √(2 ÷ 0.467) = 2.1 s. A result that supports the claim: measured accelerations that grow in proportion to the mass difference at constant total mass, and lie close to the prediction. Discrepancies come from pulley friction and the pulley's own mass (both make the measured acceleration smaller), from the string's mass and from timing. Safety: keep hands and feet away from the falling mass, use a mat, and do not exceed masses the clamp can hold. What the model cannot show: the prediction assumes a massless, frictionless pulley and an inextensible string; the size of the gap between prediction and measurement is a measure of how far the apparatus is from that ideal.

Worked example 1 — two blocks pushed together. A 3.0 kg block and a 2.0 kg block sit touching on a frictionless surface, and a 15 N horizontal push is applied to the 3.0 kg block. System view: a = 15 ÷ 5.0 = 3.0 m/s². The 2.0 kg block is accelerated only by the contact force from the 3.0 kg block: F_contact = 2.0 × 3.0 = 6.0 N. Free-body diagram of the 3.0 kg block: 15 N forward, 6.0 N backward from the 2.0 kg block (third law), net 9.0 N, and 9.0 ÷ 3.0 = 3.0 m/s², consistent. If the push were applied to the 2.0 kg block instead, the contact force would be 3.0 × 3.0 = 9.0 N, because it would then have to accelerate the heavier block.

Worked example 2 — a bucket hauled up. A 2.0 kg bucket is pulled upward by a rope with tension 24 N. Free-body diagram: 24 N up, weight 19.6 N down. Net force 4.4 N up, so a = 4.4 ÷ 2.0 = 2.2 m/s² upward. If the tension were 19.6 N the bucket would move at constant velocity (or stay at rest); if the rope were slack the bucket would fall at 9.8 m/s². The tension needed for a given upward acceleration is m(g + a).

Worked example 3 — apparent weight when an elevator starts down. A 75 kg passenger stands on a scale as the elevator accelerates downward at 1.5 m/s². Weight = 735 N. With down positive: 735 − N = 75 × 1.5 = 112.5, so N = 622.5 N, about 620 N. The scale reads less than the weight because part of the gravitational force is going into accelerating the passenger rather than being balanced by the floor. If the cable snapped and the elevator fell freely, N would be 0: the passenger and scale would fall together.

Worked example 4 — Atwood machine with unequal masses. Masses 0.80 kg and 0.60 kg. a = (0.80 − 0.60) × 9.8 ÷ 1.40 = 1.96 ÷ 1.40 = 1.4 m/s². Tension from the light mass: T = 0.60 × (9.8 + 1.4) = 6.72 N; from the heavy mass: T = 0.80 × (9.8 − 1.4) = 6.72 N, matching. Released from rest, the heavy mass falls 0.70 m in √(2 × 0.70 ÷ 1.4) = 1.0 s.

Representation task: two free-body diagrams and a system equation. For the cart-and-hanging-mass system in the lesson (2.0 kg cart, 0.50 kg hanging mass), draw one free-body diagram for the cart and one for the hanging mass. On each, label every force with its size (weight 19.6 N and normal force 19.6 N on the cart, tension 3.92 N, hanging weight 4.9 N), and write the F_net = ma equation beside it. Do not draw a third, combined diagram for the system: write the two equations along the string's path (cart forward, hanging mass downward) and add them to get the system equation; explain in one sentence why the tension terms cancel in that sum even though the two tension forces point in different directions (horizontal on the cart, upward on the hanging mass).

Independent performance: design and defend a demonstration of one law. Choose one of Newton's laws and design a demonstration that provides clear evidence for it and collects enough data to support the claim. For example: for the second law, a cart pulled by hanging masses of 50, 100 and 150 g with the cart's total mass kept constant, timed over 1.0 m; for the third law, two spring scales hooked together and pulled; for the first law, a cart on a level low-friction track coasting past equally spaced marks. Present it to classmates and invite critique. Success criteria: the claim is stated before the data are collected; the data table has at least three settings with repeated trials; the analysis compares the measured accelerations or forces with the values predicted from F_net = ma or the third law; the discrepancies are given as percentages with at least two physical explanations; and one improvement suggested by a classmate is recorded with your response to it.