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.
$\sum \tau_clockwise = \sum \tau_anticlockwise$

Step 2: Identify the weights (take $g = 10 \, m/s^2$).
Weight of boy A: $W_A = 40 \times 10 = 400 \, N$
Weight of boy B: $W_B = 50 \times 10 = 500 \, N$
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): $400x \, N \cdot m$
Moment due to B (clockwise): $500(3 - x) \, N \cdot m$

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.
$x = \frac{1500}{900} = \frac{5}{3} \approx 1.67 \, m$
Distance from fulcrum to A: $\frac{5}{3} \, m$
Distance from fulcrum to B: $3 - \frac{5}{3} = \frac{4}{3} \approx 1.33 \, m$

Part (b)

Step 1: Assume boy A (lighter, 40 kg) leaves the see-saw. Only boy B remains.
The fulcrum is already at $\frac{4}{3} \, m$ 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: $\frac{5}{3} \, m$.

Step 2: If boy B leaves instead, unsupported length would be from fulcrum to B's position: $\frac{4}{3} \, m$.
Typically, the longer unsupported arm is considered when the lighter boy leaves.

Final answer for unsupported length (A leaves): $\frac{5}{3} \, m$
\dfrac{5{3} m from A, \dfrac{4}{3} m from B}