After Isolate: inventory the interactions that cross the chosen body's boundary.
START HERE → choose the body
A reasoning model: situation → complete force list → equations
Isolate
Inventory
Choose axes
Solve & check
Cut away everything outside the chosen body/system.
Show every external interaction that remains.
Make components and signs readable.
Use mechanics laws; test the result.
contacts become forces
axes split components
External to what? The chosen system sets the boundary.
Why draw an FBD? It replaces a complicated scene with a traceable model. Include every external force on the chosen system (including support reactions); omit internal forces between parts inside that system. The picture is not decoration—it records the assumptions used to write mechanics equations.
cut frees body
Solve & check · choose equations for the motion
Earth outside
STATIC / equilibrium ΣF = 0; ΣM = 0 (no linear or angular acceleration)
ACCELERATING rigid body (planar) ΣF = m a_G; ΣM_G = I_G α Use force and moment laws, not zero by default.
A force is external when its source lies outside the chosen system. Include every crossing interaction; omit forces between parts both inside it. The partner force acts on the source object, not on this body's FBD (Newton's third law).
The diagram records forces ON the isolated body—not the equal-and-opposite forces it exerts on others.
One body isolated: 5 kg block at rest on a level table
+y
+x
N = 49 N upward
Force inventory · what crosses the boundary?
W = mg = 49 N downward
Worked check · do the force arrows agree with rest?
With +y upward: ΣFy = N − W = 0.
Assume Earth gravity g = 9.8 m/s²; level surface; no other vertical loads; static equilibrium. Weight: W = mg = (5 kg)(9.8 m/s²) = 49 N downward. Vertical balance: ΣFy = N − W = 49 N − 49 N = 0 N; therefore N = 49 N upward.
What the model does—and does not—do An FBD helps formulate mechanics equations; it does not by itself determine unknown forces or prove the chosen body/system and idealizations are valid. Check the boundary, assumptions, signs, and units against the physical situation. Takeaway: isolate first, account for every external interaction, choose useful axes, then solve and check with the right law.
Friction may be static or kinetic. Its actual value depends on motion and contact conditions; an FBD lists candidate interactions, not unknown magnitudes.
10 kg on a 30° incline · force directions & components
Ideal rigid block; g = 9.8 m/s²; contact maintained. Ramp angle is stated as 30°.
center of mass
f points up-slope only if motion or impending motion is down-slope
Takeaway: list real interactions on the body; components only re-express one force.
Force pair: equal and opposite, on different bodies
At the block–ramp contact, the ramp pushes the block; the block pushes the ramp. Only the first arrow belongs on the block's FBD.
Ramp on block: N → include on block FBD
Reading the component arrows as one vector sum
49 N down slope
W∥ + W⊥ = W: the dotted sides complete a vector parallelogram. The two components re-express weight; do not add them as separate forces on the FBD.
Inventory continues: replace contacts with external reactions.
Frictionless roller
R ⟂ surface
Ideal pin / hinge
Fixed support (planar)
Blocks motion into the surface; allows sliding along it and rotation.
Prevents translation in x and y; allows rotation.
Ax → Ay ↑
Prevents x/y translation and rotation; supplies a resisting couple.
gravity → W
contact → N, f
Each arrow is the support's force on the isolated body—not a known applied load. Internal forces between parts of that body are not added. Reaction directions follow ideal constraints; their sizes must be solved.
ramp outside
Worked example · simply supported beam
chosen body (system)
Ay = 200 N
400 N ↓
Ax = 0 N
4 m total
2 m
ΣFx: Ax = 0 | ΣFy: Ay + By − 400 = 0
Pin A: Ax, Ay
About A: By(4 m) − 400(2 m) = 0 → By = 200 N
Then Ay = 200 N; Ax = 0 N. Equal halves share the centered load.
Model: planar rigid beam, ideal pin + frictionless roller, static equilibrium; load acts at midspan.
Weight W — at the center of mass; vertically downward (near Earth).
Real supports deform; friction may alter what reactions a contact can sustain. Support names give reaction types—not their magnitudes. Takeaway: constrain motion to list reactions; solve their sizes from loads and equilibrium.
Normal N — contact force perpendicular to the surface; pushes away from it.
Friction f — tangent to contact; opposes relative motion or its impending tendency.
Tension T — along a taut cable; it pulls away from the body, never pushes.
Applied load F — draw the force as specified (direction, point, and size if given); don't assume its direction.
Direction and point of action come from the interaction—not from a guess.
N ⟂ ramp
Resolve W in slope-aligned axes
W = 98 N ↓
W∥ = 98 sin 30° = 49 N down slope
W⊥ = 98 cos 30° ≈ 84.9 N into plane
W = 98 N
These perpendicular components add vectorially to W; they are not extra forces. Friction is not automatically 49 N: its value depends on whether contact sticks or slips and on contact conditions.
chosen block
ramp
Block on ramp: partner force; not on block FBD
T on crate
At an attachment, T follows the taut cable and pulls away from the chosen body. An applied load follows the problem's stated direction.
84.9 N into plane
From Inventory: keep each real force; choose how to describe it.
From Choose axes: write balances in declared directions.
One 120 N force · components depend on the frame
Planar vector; θ = 35° from +x. Diagram scale: 2 px per N.
At rest: neither translation nor turning
Ax, Ay, MA
+y up
Three scalar conditions (2D)
F = 120 N
35°
Fy = +120 sin 35° ≈ +68.8 N
+x right
Fx = +120 cos 35° ≈ +98.3 N
By = 200 N
Project the arrow onto the declared axes: Fx = F cos θ and Fy = F sin θ here because θ is measured from +x. The components are signed coordinates of the same vector—not extra forces to add to F.
Roller B: By
A rotated frame can simplify projections
Choose horizontal/vertical axes by default; rotate them to follow an incline or member when that makes components or equations simpler.
+v ⟂ u
Fv = 0
+u along F
same origin
F = Fu = +120 N
Application · choose axes along a ramp
+v (normal)
+u along ramp
For a body on this incline, u-parallel and v-normal axes often make the contact directions and force components simpler to read than x-y axes.
φ = 30°
Declare the frame once; project every force onto the same signed axes. The frame changes coordinates, never the physical forces.
Signed projections follow the declared + directions
Example: 100 N at θ = 210° from +x (down-left)
Fy = 100 sin 210° = −50.0 N
Fx = 100 cos 210° = −86.6 N
F = 100 N
Both projections point opposite the declared positive directions, so both carry minus signs. A component pair is another coordinate description of this same 100 N force—not two new forces.
ΣFx = 0
ΣFy = 0
ΣM_O = 0
Force sums in x and y balance translation. The moment sum balances rotation about chosen point O. For a planar rigid body at rest, both linear and angular acceleration are zero.
Continue from equilibrium: two applications, one FBD workflow.
Opposing components cancel → no net force.
A. Two cables share a 300 N sign load
Moment is a force’s turning effect: its size depends on force and perpendicular distance to the line of action. Its sign records clockwise/counterclockwise sense.
Junction FBD: cable pulls + sign weight.
T₃₀ = 220 N · 30°
T₄₅ = 269 N · 45°
+x → right +y ↑ up
T₃₀ pulls up-left; T₄₅ pulls up-right. Cables pull, never push.
Components (N): ΣFx: −T₃₀ cos30° + T₄₅ cos45° = 0 ΣFy: T₃₀ sin30° + T₄₅ sin45° − 300 = 0
W = 300 N
Solve: T₄₅ = 1.225T₃₀; then (0.500 + 0.866)T₃₀ = 300 → T₃₀ ≈ 220 N, T₄₅ ≈ 269 N. Positive answers confirm both assumed forces pull.
Moments: balance turning about a point
Moment M = turning effect
F ↓
Check (unrounded): horizontal 219.6cos30° ≈ 190.2 N left = 269.2cos45° ≈ 190.4 N right; vertical 219.6sin30° + 269.2sin45° ≈ 300 N up. Rounded forces give small residuals.
R at O
|M_O| = F d⊥
Choose O at the line of action of an unknown reaction: its perpendicular arm is zero, so that force contributes no moment about O. This can remove that unknown from the moment equation.
Takeaway: horizontal components cancel; upward components sum to 300 N.
O
B. Method of joints · isolate one truss joint
Sign convention: choose clockwise or counterclockwise positive and keep it consistent. Moment is force × perpendicular distance (N·m), not merely force × any drawn length.
d⊥
Pin-jointed, massless; static. +x right, +y up.
3 left : 4 up : 5 diagonal
Assume both pull away
horizontal bar
600 N down
For 3 unknowns, 3 independent equations may determine one answer.
Let diagonal D pull up-left: components (−3D/5, +4D/5). Assume horizontal H pulls left. ΣFy: (4/5)D − 600 = 0 → D = 750 N, tension. ΣFx: −(3/5)(750) − H = 0 → H = −450 N: opposite the assumed pull, so the bar is in 450 N compression.
UNIQUE 2u = 6 u = 3
DEPENDENT u + v = 4 2u + 2v = 8
INCONSISTENT u + v = 4 u + v = 7
Count scalar equations against unknowns, but check independence and consistency: repeated equations leave freedom; contradictions have no solution. Even a unique algebraic answer can still fail the physical model or friction limit.
Action–reaction lives on neighboring joints
Tension on left joint: pull left
Equal opposite force on right joint
10 kg block · 30° incline
Shown member in tension: each joint FBD includes its own outward member-end pull. The pair acts on different joints; never draw both on one joint FBD.
N outward
Components are projections, not extra forces
f required up-slope
Assume rest; write balances
Fast diagnosis · prevent five FBD traps
Along u: f − mg sin30° = 0
Normal: N − mg cos30° = 0
Missed interaction? Account for every external contact/load once.
Components? They re-express one force; do not add them as new forces.
10 kg
f needed = (10)(9.8)sin30° = 49 N N = (10)(9.8)cos30° ≈ 84.9 N
Friction? Oppose actual or impending relative slip, not a memorized direction.
Cable? Tension must pull along the cable—never push.
49 N down-slope
Negative assumed force? Reverse its direction; a negative member axial force means compression.
+u up-slope; +v outward
W=98 N
Practical close-out check
PHYSICAL CHECK · Coulomb static friction f_s ≤ μ_s N = 0.30(84.9 N) ≈ 25.5 N. Equilibrium would require 49 N, more than the contact can supply. No static solution exists: the block cannot remain at rest under this model. Do not force an algebraic answer past the friction bound.
One isolated body/joint only • every interaction once • axes and signs declared • enough independent equations for the unknowns • units and force balance checked.
Model boundary: ideal planar, static system; these joint balances do not prove whole-truss stability or member strength. Real cables and supports can depart from ideal behavior.
Takeaway: equations propose the forces rest would require; the inequality decides whether that assumed state is physically possible.
Takeaway: isolate a joint, solve its force balance, and pass each member force to the next joint with the opposite direction.
5. Solvable check: 5 kg on a level table
Same 5 kg table case from the workflow: a quick balance check.
N=49 N
W = mg = (5 kg)(9.8 m/s²) = 49 N ΣFy = N − W = 49 − 49 = 0 N No horizontal load: ΣFx = 0. Rest is consistent.
5 kg
W=49 N
Direction diagnosis · read the situation, then the sign
Friction follows relative slip
slip →
Truss result · not a failed FBD
body
If slip tends right, friction points left; reverse the tendency, reverse friction.
H = −450 N ⇒ actual force points right: 450 N compression.
Arrows start as model assumptions; equations and physical meaning decide the final direction and state.
Friction threshold · a second way to check
available μs = 0.30
needed tan30° = 0.577
0
0.30
0.577
0.70
Static friction can hold only if μs ≥ tanθ. Here 0.30 < 0.577, so the rest assumption fails regardless of mass.
Moment-center zoom · reuse the 4 m beam example
Choose A: Ax, Ay pass through A (zero arm).
About A: ΣMA = 0 By(4 m) − 400(2 m) = 0 By = 200 N. Ax and Ay vanish here because their arms about A are zero.
Ax, Ay at A
400 N
A
2 m to load · 4 m to B
By unknown
Takeaway: a smart moment center removes forces through it from that moment equation—not from the FBD.
Read the moment signs · centered 400 N beam load
CCW positive: the load turns clockwise; B reaction, counterclockwise.
2 m arm 4 m arm
ΣMA = (+200 N)(4 m) − (400 N)(2 m) = +800 − 800 = 0 N·m. Equal opposite turning effects cancel; the reaction at A contributes zero moment because its line of action passes through A.
From section 6: same workflow, now your turn to build and check.
cut body free
name forces
set signs
Eight steps: from identifying interactions to a full, checked FBD.
When drawing an FBD for the block alone, which force belongs on it at the block–ramp contact?
A 120 N force is 35° above +x. Which component pair matches the board’s x-y axes?
A force points down-left in axes where right and up are positive. What signs should its x- and y-components have?
For a planar rigid body at rest, which set of conditions expresses equilibrium?
A 10 kg block rests on a 30° incline with μs = 0.30. The required static friction is 49 N, while the maximum available is about 25.5 N. What follows?
For the 4 m beam, why does choosing A as the moment center remove Ax and Ay from the moment equation about A?
At a junction supporting the sign, which statement about a cable’s tension arrow is correct?
At the truss joint, the assumed horizontal member force solves to H = −450 N when assumed to pull left. What does this indicate?