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: F_x = 0 (Forces right = Forces left) F_y = 0 (Forces up = Forces down) 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 θ and F_y = F 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 F = 0 and M = 0. Vector Triangle or Components?
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 Tension Scenario: 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.) Solution 1. Resolve Tension: Vertical component: T_y = T 45^ Horizontal component: T_x = T 45^ 2.
Apply F_y = 0: (To just lift, vertical upward force must balance weight) T 45^ = 200 T = 200 / 45^ T 283 N Example 2: Hanging Sign with Bar Scenario: 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.) Solution 1. 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 30^. 4. Moment Equation ( M = 0): Clockwise Moment (Sign) = Anticlockwise Moment (Wire) 98.1 × 0.80 = (T 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 () or adjacent () 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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