Explain air resistance and drag with the particle model, find the drag force from a free-body diagram and acceleration, and analyze terminal velocity

Learning goal: Explain drag as the push-back of fluid particles on a moving object, a Newton's third-law pair; represent air resistance on free-body diagrams; determine the magnitude of the drag force when the other forces and the acceleration are known; identify the factors that affect air resistance; and explain terminal velocity as the condition in which drag balances the other forces.

Fluids push back. Air and water are made of particles. When an object moves through a fluid it pushes the particles in front of it out of the way, and by Newton's third law those particles push back on the object. That push-back is drag, or, for air, air resistance. It always acts opposite to the object's velocity relative to the fluid. The same interaction works in reverse: a helicopter's rotor pushes air downward, and the air pushes the rotor upward, producing lift; a swimmer pushes water backward with each stroke, and the water pushes the swimmer forward. Liquids have far more particles in a given volume than gases and so produce more drag: wading through waist-deep water is slow, while walking through air is not.

What drag depends on. Drag grows with speed, and for most everyday objects it grows roughly with the square of the speed, so doubling the speed roughly quadruples the drag. It grows with the object's cross-sectional area facing the motion, with the density of the fluid, and depends on the object's shape: a streamlined shape lets particles flow around it with fewer collisions. Drag does not depend on the object's mass. Two balls of the same size and shape but different mass feel the same drag at the same speed, which is exactly why the heavier one wins a race through air: its weight is larger while the drag is the same. This course quantifies drag from force diagrams rather than from a drag formula, so the factors are used to reason about comparisons, not to compute coefficients.

Finding drag from a free-body diagram. Because drag is not given by a simple law in this course, it is found from the other forces and the acceleration. Write Newton's second law along the direction of motion with drag as the unknown. A skydiver of mass 80 kg is measured to be accelerating downward at 6.0 m/s². Free-body diagram: weight 784 N down, drag D up. Down positive: 784 − D = 80 × 6.0 = 480, so D = 304 N. A ball of mass 0.10 kg falls with an acceleration of 6.8 m/s²: 0.98 − D = 0.10 × 6.8 = 0.68, so D = 0.30 N. The same method works for a cyclist or a car if the driving force and rolling friction are known.

Terminal velocity: a fully worked model. The skydiver leaves the aircraft at rest relative to the air. At that instant the speed is zero, so the drag is zero, and the acceleration is g = 9.8 m/s². As the speed grows, the drag grows, the net force shrinks, and the acceleration falls. At a certain speed the drag equals the weight, 784 N, the net force is zero, and the acceleration is zero: the skydiver continues at constant speed. That speed is the terminal velocity. Suppose it is 50 m/s for this skydiver in a spread position. Then in the first 2 s the speed reaches about 19 m/s, by 6 s about 41 m/s, by 10 s about 48 m/s, and by 16 s it is within 0.2 m/s of 50 m/s; the approach is gradual because the net force shrinks as the speed rises. Opening a parachute increases the area enormously, so the drag at 50 m/s is now far larger than the weight, the net force is upward, and the skydiver slows until the drag once more equals 784 N at a new, much smaller terminal velocity, a few meters per second. The skydiver never moves upward; slowing down while moving down is an upward acceleration, and the velocity stays downward. The terminal velocity depends on the factors that set drag: a heavier skydiver of the same shape needs more drag to balance the larger weight and so has a higher terminal velocity; a larger area gives a lower one; denser air at low altitude gives a lower one.

Drag on the way up. For a ball thrown upward, drag and weight both point downward while the ball rises, so the net force is larger than the weight and the acceleration is greater than 9.8 m/s². A 0.20 kg ball moving upward with 0.50 N of air resistance has a net downward force of 1.96 + 0.50 = 2.46 N and an acceleration of 12.3 m/s² downward. On the way down the drag is upward, the net force is 1.96 − 0.50 = 1.46 N and the acceleration is 7.3 m/s² downward. The ball therefore takes longer to fall than to rise and lands slower than it was thrown.

Investigation: coffee filters and terminal velocity. Materials: six identical basket coffee filters, a balance, a meter stick, a stopwatch or phone video, a tall space (at least 2.0 m). Steps: measure the mass of one filter (for the example, 1.0 g); mark a point 1.00 m above the floor, drop a single filter from 2.5 m and time only its fall through the lower 1.00 m; then nest two filters together so the shape and area are unchanged and only the mass doubles, and repeat, up to four filters; three trials each. A model in which drag grows with v² predicts that a single filter comes within about 1 percent of its terminal velocity in its first half meter but a stack of four needs about 1.2 m, so timing the whole fall from rest would underestimate the heavier stacks; the 1.5 m run-up before the timed section allows for that. Check it for your filters: time the lower 1.00 m in two halves, and if the second half is not about as fast as the first, raise the drop height. Record: number of filters, mass, weight, mean time, terminal velocity v = 1.00 m ÷ t. At terminal velocity the drag equals the weight, so the drag is known: 0.0098 N per filter. Analyze: plot drag against v and against v². For example (model values), times of 0.80, 0.57, 0.46 and 0.40 s over the lower 1.00 m for one to four filters give speeds of 1.25, 1.75, 2.17 and 2.50 m/s, whose squares 1.56, 3.08, 4.73 and 6.25 are in the ratio of about 1:2:3:4, matching the weights. A result that supports the claim: a straight line through the origin for drag vs. v², a curve for drag vs. v. That establishes which factor, v or v², drag follows for this shape, and shows that terminal velocity rises with weight at fixed area. Safety: keep the drop zone clear; release from a stair landing or a stable step stool with a partner steadying it, never from a chair or desk. What the model cannot show: a filter wobbles and drifts, so the speed is an average; the run-up must be long enough for the heaviest stack to reach terminal velocity; and the result applies to this shape in air, not to a streamlined object or to water.

Worked example 1 — drag on a parachutist at constant speed. A 90 kg parachutist descends at a constant 5.0 m/s. Constant velocity means zero net force, so the drag on the open parachute equals the weight: 90 × 9.8 = 882 N upward. The speed is not needed for this calculation; it would only be needed to compare with a different parachute.

Worked example 2 — drag on a cyclist. A cyclist and bicycle of total mass 75 kg accelerate at 0.40 m/s² while the road exerts a 60 N forward driving force on the bicycle (the reaction to the tire pushing back on the road) and rolling friction is 10 N backward. Forward positive: 60 − 10 − D = 75 × 0.40 = 30, so D = 20 N. At a higher speed the drag would be larger and, with the same drive force, the acceleration smaller; at the speed where D = 50 N the cyclist could no longer accelerate.

Worked example 3 — a falling ball with measured acceleration. A 0.045 kg ball is filmed falling and its acceleration is found to be 8.6 m/s² downward. Weight: 0.441 N. Down positive: 0.441 − D = 0.045 × 8.6 = 0.387, so D = 0.054 N. Later in the fall, when its acceleration has dropped to 2.0 m/s², the drag is 0.441 − 0.090 = 0.35 N; the drag has grown because the speed has grown, and the ball is approaching terminal velocity.

Worked example 4 — comparing terminal velocities. Two skydivers have the same shape and area; one has mass 60 kg and the other 90 kg. Each reaches terminal velocity when drag equals weight: 588 N and 882 N. Since drag at a given speed is the same for both, the heavier diver must go faster to generate 1.5 times the drag. If drag scales with v², the speed ratio is √1.5 = 1.22, so a 50 m/s terminal velocity for the lighter diver implies about 61 m/s for the heavier one.

Representation task: three free-body diagrams of one fall. For the 80 kg skydiver, draw free-body diagrams at three moments: just after leaving the aircraft (speed near zero), partway down when the measured acceleration is 6.0 m/s², and at terminal velocity. On each, draw the weight arrow as 784 N and the drag arrow to scale (0 N, 304 N, 784 N), write the net force and the acceleration, and add a short velocity arrow beside the diagram to show that the velocity is downward in all three, including the one where the acceleration is zero.

Independent performance: the coffee-filter analysis. Carry out the coffee-filter investigation and write a report. Success criteria: the data table lists number of filters, weight, three fall times, mean time and terminal velocity; the drag at terminal velocity is stated to equal the weight, with the reasoning from Newton's second law; both plots (drag vs. v and drag vs. v²) are drawn and the one closer to a straight line through the origin is identified; the report states which factors affecting drag were held constant (shape, area, fluid) and which was varied (weight), and explains why nesting filters rather than using a heavier object of a different shape was necessary; and one limitation of the method is named.