State four factors that determine the number and length of underpinning bays. ii. Sketch and label a section through an underpinning bay for a strip foundation.

Part (a) Step 1: State the condition for rotational equilibrium. The sum of clockwise moments about the fulcrum equals the sum of anticlockwise moments about the fulcrum. _clockwise = _anticlockwise Step 2: Identify the weights (take g = 10 \, m/s^2 ). Weight of boy A: W_A = 40 × 10 = 400 \, N Weight of boy B: W_B = 50 × 10 = 500 \, N Distance between A and B: 3 \, m Step 3: Let the distance from fulcrum to A be x \, m . Then distance from fulcrum to B is 3 - x \, m . Moment due to A (anticlockwise): 400x \, N · m Moment due to B (clockwise): 500(3 - x) \, N · m Step 4: Set up the equilibrium equation. 400x = 500(3 - x) Step 5: Expand the right side. 400x = 1500 - 500x Step 6: Add 500x to both sides. 900x = 1500 Step 7: Divide both sides by 900. x = (1500)/(900) = (5)/(3) ≈ 1.67 \, m Distance from fulcrum to A: (5)/(3) \, m Distance from fulcrum to B: 3 - (5)/(3) = (4)/(3) ≈ 1.33 \, m Part (b) Step 1: Assume boy A (lighter, 40 kg) leaves the see-saw. Only boy B remains. The fulcrum is already at (4)/(3) \, m from B (from part (a)). The "unsupported" length is the length on the side without the boy, i.e., from fulcrum to A's position: (5)/(3) \, m . Step 2: If boy B leaves instead, unsupported length would be from fulcrum to B's position: (4)/(3) \, m . Typically, the longer unsupported arm is considered when the lighter boy leaves. Final answer for unsupported length (A leaves): (5)/(3) \, m (5)/(3) m from A, (4)/(3) m from B