Inclined force on a beamResolve it before anything else

Why resolve? (30 seconds)

• The beam equations work in two directions: x and y. An inclined force belongs to neither of them — its components do.
• Fy → vertical reactions, S and M. Fx → AX and the N diagram (N — the force that acts along the beam).
• The direction of the force sets the direction of the components: down and to the right → Fx to the right, Fy down.

10 t at 30° — four lines

• Resolving
  Fx = 10·cos30° = 8.66 t to the right · Fy = 10·sin30° = 5 t down
• ΣFx
  Only A (pinned) has a horizontal reaction. Assume AX to the left: ΣFx: 8.66 − AX = 0 → AX = 8.66 t to the left
• ΣMA
  5·2 − BY·6 = 0 → BY = 1.67 t
• ΣMB
  AY·6 − 5·4 = 0 → AY = 3.33 t ✓ 3.33 + 1.67 − 5 = 0

That's it. One resolution — and from there it is an ordinary beam.

Why is Fx not in ΣM? It acts along the beam axis and passes through the support → its lever arm is 0.

4 pitfalls everyone falls into

• A component in the opposite direction. The direction of the original force sets the sign of Fx and of Fy.
• The angle is given from the vertical. Then sin and cos swap — check what α was measured from.
• Not erasing the original force. It gets counted twice — once whole and once as components.
• AX at the roller support. A roller has no horizontal reaction — all of Fx falls on the pinned support.