Why do you draw a free body diagram
A free‑body diagram isolates all forces acting on a single object so you can apply Newton’s laws correctly. It shows the magnitude and direction of each force, making the problem solvable with clear equations.
Physics · Forces
A free‑body diagram (FBD) is a sketch that represents only the object of interest and every external force that touches it. By stripping away surrounding objects, the diagram forces you to list weight, normal, tension, friction, and any applied pushes or pulls. This clarity prevents you from accidentally mixing forces from different bodies and ensures that Newton’s second law can be written as a single vector equation.
Steps to draw a free‑body diagram
Follow these actions in order
- 1Identify the single object you will analyze
- 2Draw a simple shape (usually a box) to represent the object
- 3Choose a convenient coordinate system, often aligned with motion or surfaces
- 4Add arrows for every external force, labeling magnitude and direction
- 5Write ΣF = ma using the drawn forces
Choosing the coordinate axes early simplifies the algebra. Align one axis with the incline, the plane, or the direction of motion so that many force components become zero or parallel to an axis. If the problem involves an incline, let the x‑axis run up the slope and the y‑axis be perpendicular to it. This choice turns the weight component Ω mg sinθ into an x‑force and mg cosθ into a y‑force, reducing the number of trigonometric steps later.
Typical forces that appear in most mechanics problems
- Gravitational force (weight)
- Normal reaction from a surface
- Frictional force (static or kinetic)
- Tension in a rope or cable
- Applied push or pull forces
- Spring force
Example: a 5 kg block rests on a 30° incline, a rope pulls up the plane with a tension of 10 N, and kinetic friction is negligible. First draw a box for the block, then add arrows: weight 49 N pointing down, normal N perpendicular to the plane, tension 10 N up the slope. Resolve the weight into components: Ω mg sin30° = 24.5 N down the plane and Ω mg cos30° = 42.4 N into the plane. Apply Ω = ma_x (block is at rest, so a=0) giving 10 N – 24.5 N = 0, confirming equilibrium, and Ω = 0 gives N – 42.4 N = 0, so N = 42.4 N.
Force summary for the block
| Force | Magnitude (N) | Direction |
|---|---|---|
| Weight | 49 | Vertical down |
| Weight component parallel | 24.5 | Down the plane |
| Weight component perpendicular | 42.4 | Into the plane |
| Tension | 10 | Up the plane |
| Normal | 42.4 | Perpendicular outward |
After the diagram is complete, write Newton’s second law for each axis. Because the block does not accelerate, the sum of forces in both directions equals zero, allowing you to solve for unknowns like the normal force or friction. The FBD thus serves as a bridge between a visual picture and the algebraic equations needed for the test. Mastering this skill saves time and reduces errors on physics problems.
Check yourself
In the example of a 5 kg block on a 30° incline with a 10 N tension up the plane, what is the normal force?
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