Explain and predict changes in motion from net force and mass, distinguish equilibrium from acceleration, and identify interacting force pairs

Learning goal: Use the relationship a = F_net ÷ m to explain and predict the motion of a single object acted on by several one-dimensional forces; recognize that an object at rest or moving with constant velocity has zero net force; solve force-diagram problems in newtons, kilograms and meters per second squared; and identify interacting force pairs using "A acts on B, so B acts on A".

Forces change motion; they do not sustain it. The core idea of dynamics is a relationship you can reason with rather than a sentence to recite. When the vector sum of the forces on an object, the net force F_net, is zero, the object does not accelerate. If it is moving, it keeps moving at the same speed in the same direction; if it is at rest, it stays at rest. A hockey puck sliding across nearly frictionless ice needs no push to keep going; it needs a force to stop. The everyday belief that "things stop when you stop pushing" comes from friction, a real force that is easy to forget because nobody applies it on purpose. When the net force is not zero, the object accelerates: it speeds up, slows down or changes direction. The acceleration is proportional to the net force and inversely proportional to the mass: a = F_net ÷ m. Double the net force on the same object and its acceleration doubles; double the mass under the same net force and the acceleration halves. Rearranged, F_net = ma, which is where the newton comes from: 1 N gives 1 kg an acceleration of 1 m/s².

Equilibrium: rest and constant velocity look the same to the forces. An object in equilibrium has zero net force. That covers two cases which students often treat differently: a chair resting on the floor (gravity down, normal force up, equal) and a car cruising at a steady 25 m/s on a straight level road (the forward push of the road on the tires balancing drag and friction, and gravity balancing the normal force). In both, every force is matched, the net force is zero and the velocity does not change. Constant velocity is not evidence of a forward net force; it is evidence of none.

Accelerating objects. Suppose a sled of mass 20 kg is pulled along level snow by a rope with 60 N of tension while 20 N of kinetic friction opposes the motion. From the force diagram, taking the pull as positive, F_net = +60 N + (−20 N) = +40 N. Then a = F_net ÷ m = 40 N ÷ 20 kg = 2.0 m/s² in the direction of the pull. The vertical forces, weight 196 N and normal force 196 N, cancel, as one dimension always must in this course. If the rope is released, the only horizontal force is −20 N of friction, so a = −20 N ÷ 20 kg = −1.0 m/s²: the sled slows at 1.0 m/s each second until it stops, and once stopped, friction stops acting and the sled remains at rest.

A worked model: a vertical case. A 50 kg person stands on a scale in an elevator. Gravity pulls down with 50 kg × 9.8 N/kg = 490 N. The scale pushes up with a normal force equal to its reading. If the reading is 550 N, then with up positive F_net = +550 N + (−490 N) = +60 N, so a = 60 N ÷ 50 kg = +1.2 m/s²: the elevator is accelerating upward, which happens when it starts rising from rest or slows while descending. If the reading is 490 N the net force is zero and the elevator is at rest or moving at constant velocity, up or down; the scale cannot tell which. If the reading is 430 N the net force is −60 N and the acceleration is −1.2 m/s², downward. The person's mass is 50 kg in every case; only the normal force changes.

Interacting force pairs. A force is an interaction between two objects, and both objects feel it: each experiences an equal force in the opposite direction. A truck hits a sign, so the sign hits the truck with an equal force the other way; the outcomes differ, but not because the forces differ. The truck barely slows because its mass is large, so that force gives it only a small acceleration; the sign bends because it is far weaker than the truck's frame and is held at its base. How an object is damaged depends on its strength and how it is supported, not on its mass alone. Earth pulls a book down, so the book pulls Earth up with an equal force; Earth's enormous mass means its resulting acceleration is unmeasurably small. Use the format "A acts on B, so B acts on A" to find the pair. Naming a force (gravity, friction, tension) does not identify the pair; naming the two objects does.

Pairs are not balanced forces. This is the misconception that matters most. Balanced forces act on the same object and can cancel to give zero net force: the table's normal force and gravity, both on the book. An interacting force pair acts on two different objects and can never cancel, because the two forces never appear on the same force diagram. A horse pulling a cart is pulled backward by the cart with an equal force, yet the cart still accelerates, because the cart's motion depends only on the forces on the cart: the horse's forward pull against friction on the cart's wheels. In this course the laws are applied to one object at a time with several forces on it; systems of many objects are reserved for Physics.

Why seat belts and packaging work. Suppose a car moving at 15 m/s is stopped by a collision in 0.5 s, its front crumpling as it stops. A belted 60 kg passenger is slowed with the car: a = (0 − 15) m/s ÷ 0.5 s = −30 m/s², requiring a net force of 60 kg × 30 m/s² = 1800 N from the belt, spread across the chest and hips. An unbelted passenger has no such force acting, so by the first idea above she continues at 15 m/s until the dashboard stops her, perhaps in 0.05 s: a = −15 ÷ 0.05 = −300 m/s², a force of 60 × 300 = 18,000 N applied to a small area. The belt does not remove the need to stop; it lengthens the stopping time, which reduces the acceleration and therefore the force. The same reasoning governs protective packaging: an egg landing on foam stops over a longer time than one landing on concrete, so the force on it is smaller. Both numbers here are illustrative "suppose" values chosen to show the reasoning, not crash measurements.

Investigation: protective packaging by force reduction. Materials: raw eggs in sealed bags, a meter stick, packaging materials (paper, cotton balls, cardboard, drinking straws, tape), a bathroom scale for optional force measurement, safety glasses, a tray. Steps: design a package that must fit in a 15 cm cube; drop it from 1.0 m onto the tray; inspect; redesign and retest from 1.5 m and 2.0 m. Record for each trial the drop height, the package design, whether the egg survived and, if a video is available, the time from first contact to full stop. A design that survives higher drops than a bare egg supports the claim that lengthening the stopping time reduces the force on the egg. What the test cannot show: the actual force in newtons, unless a force sensor is used; the assumption that the egg's stopping time is the same as the package's should be stated. Part 2, the force needed to break an egg: stand a raw egg upright in a bottle cap or ring of modeling clay inside a bag, rest a small board across its top, and add masses to the board in 0.5 kg steps until the shell cracks; the breaking force is about (board mass + added mass) × 9.8 N/kg. Repeat with three eggs and report the range, not one number. State the assumptions: the load is slow and centered, and a slow squeeze is not the same as a sudden impact, so the result is a guide for the package design rather than a prediction of the drop force. Never drop packages near people, wear safety glasses, and clean any breakage immediately.

Worked example 1 — acceleration from a force diagram. A 10 kg crate on a level floor is pulled by a rope with 50 N of tension to the right while 20 N of kinetic friction acts to the left. Gravity is 10 kg × 9.8 N/kg = 98 N down; the normal force is 98 N up. Horizontal: F_net = +50 + (−20) = +30 N. Vertical: +98 + (−98) = 0. Acceleration: a = F_net ÷ m = 30 N ÷ 10 kg = 3.0 m/s² to the right. Units: N ÷ kg = (kg·m/s²) ÷ kg = m/s². If the crate starts from rest, after 2 s it is moving at 6.0 m/s.

Worked example 2 — force from acceleration. A 1200 kg car speeds up, accelerating forward at 3.0 m/s², on a level road. The net force needed is F_net = ma = 1200 kg × 3.0 m/s² = 3600 N in the direction of the acceleration, which here is forward. This is the net force: if drag and rolling friction together are 600 N backward, the forward force of the road on the tires must be 3600 + 600 = 4200 N. The car's weight, 1200 × 9.8 = 11,760 N, is balanced by the normal force and contributes nothing to the horizontal motion.

Worked example 3 — a hanging mass that accelerates. A 2.0 kg mass hangs from a cord whose tension is 24 N. Gravity is 2.0 kg × 9.8 N/kg = 19.6 N down. With up positive, F_net = +24 + (−19.6) = +4.4 N, so a = 4.4 N ÷ 2.0 kg = 2.2 m/s² upward. If instead the tension were 19.6 N, the net force would be zero and the mass would be at rest or moving at constant velocity, which is the equilibrium case; if the tension were 15 N the net force would be −4.6 N and the mass would accelerate downward at 2.3 m/s² while the cord stays taut.

Worked example 4 — identifying the interacting pair. A swimmer pushes backward on the pool wall with her feet. "Swimmer acts on wall, so wall acts on swimmer": the wall pushes forward on the swimmer with an equal force, and that forward push on the swimmer is what accelerates her. The two forces are equal in size and opposite in direction, act on different objects (wall and swimmer) and therefore do not cancel. The swimmer moves and the wall does not because the wall is attached to the building, whose mass is enormous, not because the wall pushes less hard.

Representation task: three-state force diagrams. For a single shopping cart of mass 15 kg, draw three force diagrams with proportional arrows and compute the net force and acceleration for each: (1) at rest on level ground; (2) pushed at a constant velocity of 1.5 m/s against 12 N of friction; (3) pushed with 21 N against 12 N of friction. Include gravity (147 N) and the normal force (147 N) on each. Under each diagram state whether the cart is in equilibrium and, if not, the acceleration (state 3: 9 N ÷ 15 kg = 0.6 m/s²). Then write "A acts on B, so B acts on A" sentences for two interacting pairs in state 3, and explain why neither pair appears together on one diagram.

Independent performance: a seat-belt argument with evidence. Produce a one-page public message about seat belts in which every claim is supported by force reasoning. Success criteria: the message states what happens to an unbelted passenger when a car stops suddenly and explains it with the zero-net-force idea (until the seat belt, seat or dashboard pushes on her, almost no horizontal force acts on the passenger, so she keeps moving forward at nearly the car's original velocity); it uses a = F_net ÷ m to compare the force on a passenger stopped over a long time by a belt with one stopped over a short time by the dashboard, using clearly labeled "suppose" values with units; it draws a force diagram of the belted passenger during the stop; it identifies one interacting force pair (belt on passenger, passenger on belt) and explains why that pair does not cancel; and it names one assumption made in the calculation. Optionally include data from a cart-and-figure crash test with a stopping-time measurement.