A horizontal beam is supported by a pin at point A (left end) and a roller at point B (right end). How many unknown reaction force components appear in the beam's free-body diagram?
A2 — one vertical reaction at each support
B2 — one force from the pin and one from the roller
C3 — two force components from the pin (Ax and Ay) and one perpendicular force from the roller
D4 — two force components from each support
A pin support can push or pull in any direction, so it contributes two unknown components (Ax and Ay in 2D). A roller only pushes perpendicular to the surface it sits on — one unknown normal force. Total: 2 + 1 = 3 unknowns. This matches the three equilibrium equations available in 2D statics (ΣFx = 0, ΣFy = 0, ΣM = 0), making the beam statically determinate. Knowing how each support type translates into unknowns is as important as the drawing itself.
Question 2 Multiple Choice
A student draws a free-body diagram of a wooden block sitting on a table. Their diagram shows the block, the table surface drawn underneath it, and the weight of the table labeled in the diagram. What fundamental error did they make?
AThey forgot to include the friction force between the block and the table surface
BThey included the table and its weight — a body external to the object of interest. The FBD should show only the block and the forces acting on it, not the table itself
CThey should have combined the table's weight with the block's weight into a single downward force
DThe table's weight is a reaction force and should be shown pointing upward
The fundamental act of an FBD is isolation: you mentally cut away the table (and everything else connected to the block) and replace each connection with the force it exerts. The table appears in the FBD only as a normal force arrow — not as a physical object drawn in the diagram. Including the table itself, or any of its properties, violates the isolation principle. The block's FBD has exactly: its weight downward, a normal force from the table upward, and a friction force if applicable.
Question 3 True / False
Internal forces between parts of a body — such as the tension in a bolt holding two plates together — should be included in the free-body diagram to correctly apply Newton's second law to the body.
TTrue
FFalse
Answer: False
Internal forces always appear in equal-and-opposite pairs within a body, so they cancel when you sum forces for the body as a whole. Newton's second law ΣF = ma involves only the net external force — the sum of all forces acting on the body from outside. Internal forces contribute nothing to this sum. Including them would double-count in both directions and produce no net effect. The FBD is specifically designed to show only external forces, which is why isolation is so important.
Question 4 True / False
When analyzing a multi-body system by drawing separate free-body diagrams for each component, forces at the shared contact surfaces appear as equal-and-opposite pairs across the two diagrams.
TTrue
FFalse
Answer: True
This follows directly from Newton's third law. If body A pushes on body B with force F, then body B pushes back on body A with force −F. In the FBD of body A, you show the force from B on A. In the FBD of body B, you show the force from A on B — equal in magnitude, opposite in direction. These paired forces are how the two FBDs stay consistent with each other. Missing one side of this pair in a multi-body problem is a common source of equilibrium equation errors.
Question 5 Short Answer
What does it mean to 'isolate' an object when drawing a free-body diagram, and what specifically must you do with each physical connection when you isolate it?
Think about your answer, then reveal below.
Model answer: Isolating an object means mentally cutting away every physical connection — supports, cables, contact surfaces, hinges — and replacing each one with the force (or moment) it was exerting on the object. The resulting sketch shows the object alone, surrounded only by labeled force vectors. Nothing that was connected to the object appears in the diagram; only the forces those connections were transmitting.
This replacement step is the core of the methodology. A cable becomes a tension arrow in the direction the cable was pulling. A pin becomes two force components (Fx, Fy) whose directions are initially unknown. A wall support becomes two force components plus a moment. The power of this process is that it converts a complex physical situation into a clean force inventory that can be directly inserted into Newton's equations.