Equilibrium in two dimensions
Section: Forces, Density and Pressure | Syllabus: Cambridge AS Level Physics (9702)
Solving 2D Equilibrium Problems
When objects are suspended by ropes, cables, or hinges in two dimensions, we must ensure equilibrium in both horizontal (x) and vertical (y) directions.
Conditions:
- Σ F_x = 0 (Forces right = Forces left)
- Σ F_y = 0 (Forces up = Forces down)
- Σ M = 0 (Clockwise Moments = Anticlockwise Moments)
Figure 4.15: Resolving Forces
(Show a generic object with diagonal force F at angle θ. Show it resolved into components F_x = F cos θ and F_y = F sin θ.)
Step-by-Step Method
- Draw a Free-Body Diagram (FBD): Isolate the object and draw ALL forces acting on it (Weight, Tension, Normal Reaction, Friction).
- Choose Axes: Usually horizontal/vertical, but for slopes, parallel/perpendicular is better.
- Resolve Forces: Split any diagonal forces into x and y components.
- Solve Equations: Apply Σ F = 0 and Σ M = 0.
- Vector Triangle: Best for 3 forces only. Sketch a closed triangle (head-to-tail) and use sine/cosine rules or Pythagoras.
- Resolution (Components): Best for 4 or more forces or when angles are complex.
Worked Examples
Example 1: Sail and TensionScenario: A surfer pulls a sail (weight W=200 N) out of the water. The rope is at 45^ to the horizontal. Find the tension T required to just lift it.
Figure 4.16: Surfer FBD
(Show the sail. Gravity W down. Tension T acting up-right at 45 degrees. Water resistance/pivot force acting at base.)
Solution1. Resolve Tension:
- Vertical component: T_y = T sin 45^
- Horizontal component: T_x = T cos 45^
2. Apply Σ F_y = 0: (To just lift, vertical upward force must balance weight)
T sin 45^ = 200
T = 200 / sin 45^
T ≈ 283 N Example 2: Hanging Sign with BarScenario: A uniform bar (length d=0.80 m) supports a 10 kg sign at the end. It is held by a wire at 30^ to the bar. Calculate the tension in the wire.
Figure 4.17: Bar and Sign
(Horizontal bar pivoted at wall (left). Sign hanging at right end (W_sign = 10g). Wire converting right end to wall above pivot (30^). Show Tension T along wire.)
Solution1. Identify Forces: Weight of sign W = 10 × 9.81 = 98.1 N acting at 0.8 m. (Ignore bar weight for this specific simplified example or assume light bar).
2. Take Moments about Hinge (Pivot): This eliminates the unknown reaction force at the wall.
3. Find Vertical Component of Tension: The component perpendicular to the bar is T_y = T sin 30^.
4. Moment Equation (Σ M = 0):
Clockwise Moment (Sign) = Anticlockwise Moment (Wire)
98.1 × 0.80 = (T sin 30^) × 0.80
98.1 = T × 0.5
T = 98.1 / 0.5 = 196.2 N Exam Tip: Common Mistakes- Forgetting "g": Do not use mass (kg) in moment equations; always convert to Weight (N) using W=mg.
- Wrong Component: Always check if the component is opposite (sin) or adjacent (cos) to the angle given.
- Reaction Forces: Remember there is often a reaction force (R) at the pivot/hinge, which is why taking moments about the pivot is smart (distance d=0, so moment=0).
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