Free-Body Diagrams & System Boundaries

Every force problem starts with a decision the question rarely states: what is the system? Draw a closed loop round it, on paper or in your head. Everything inside is the system; everything outside is the environment, which matters only through the forces it exerts across the loop. A worker pushing crate A, which pushes crate B, offers three sensible systems: A alone, B alone, or both together. None is wrong. They differ in which forces are external, and so in what they can tell you. The contact force between the crates is invisible to the boundary round both and is the main unknown for the boundary round B. Choose the system whose external forces include the quantity you want and as few other unknowns as possible.

To list the external forces, trace the boundary and stop wherever something outside touches the system: a floor, a hand, a rope, another crate, the air if drag matters. Each contact supplies one force, often drawn as a normal part and a friction part. Then add the long-range forces, which here means the weight, exerted by the Earth. That is the complete list, and it is short. Label each arrow with its type and its agent, such as N (floor on crate) or W (Earth on crate). The agent test removes the usual intruders. No object exerts a force of motion; the ma on the right of the second law is not a force; and the push the crate exerts on the hand acts on the hand. If you cannot name what exerts a force, or it acts on something else, it comes off the diagram.

With the list made, the system becomes a dot with every arrow's tail on it, drawn roughly to scale so that the diagram predicts the direction of the acceleration before any algebra. If the arrows visibly balance along one line, the acceleration has no component there. This is a particle model. It is exact for the translation of any system whose parts share one acceleration: a rigid crate, two crates moving together, a car with its passengers. What it discards is where each force acts. Whether the crate tips or the car's wheels spin up depends on lines of action and torques, which a dot cannot show; those need the extended diagrams of rotational dynamics. If parts of your system accelerate differently, such as a block sliding on a moving cart, give each part its own dot.

Axes are yours to choose, and a good choice removes algebra. A 2.0 kg block slides down a frictionless 25° slope. Put x down the slope and y out of the surface. The normal force already lies along y and the acceleration along x, so only the weight, 19.62 N, needs resolving: 19.62 sin 25° = 8.292 N along the slope and 19.62 cos 25° = 17.78 N into it. The y equation, N − 17.78 N = 0, gives N at once, and the x equation gives a = 8.292/2.0 = 4.15 m s⁻². Horizontal and vertical axes are just as valid, but then N and the acceleration each have two components, and you must solve a pair of simultaneous equations to reach the same numbers. The physics is unchanged; only the bookkeeping grows. Mark the positive directions on the diagram, because every sign that follows depends on them.

With diagram and axes fixed, the equations are a transcription. For each axis write one term per arrow: the full magnitude if the arrow lies along the axis, a sine or cosine component if it is tilted, a plus sign if it points the positive way and a minus sign if not. The sum goes on the left; M times that component of the acceleration stands alone on the right. If an arrow's direction is unknown, guess, and a negative result means it points the other way. Then count. One system in a plane supplies two independent equations, so two unknowns at most. A third unknown signals a missing relation, not harder algebra: a constraint such as a_y = 0, a force law from later in this unit, or a second diagram for another system. Finish with a limiting case: flatten the slope to θ = 0 and the block's equations must return a = 0 and N = mg.

Carts A (2.0 kg) and B (3.0 kg) touch on a frictionless track. One hand pushes A forward with P = 30 N while another pushes back on B with R = 15 N. Boundary round both carts: the external horizontal forces are P and R, the contact force C is an internal pair, and 30 − 15 = 5.0a gives a = 3.0 m s⁻². Boundary round B alone: now C is external, and C − 15 = 3.0 × 3.0 gives C = 24 N. The boundary round A was not needed, which makes it a genuine check: 30 − 24 = 6.0 N, and 2.0 × 3.0 = 6.0 N. Notice that C equals neither push. A keeps 6.0 N of the 30 N for its own acceleration and passes on the rest. Only when a = 0, or when A's mass is negligible, does a push travel through a body unchanged. The interactive figure uses these two carts.