A rectangular wooden block floats upright in water, partially submerged. The block has a horizontal base area of 0.040 m^2, a height of 0.25 m, and a mass of 6.0 kg. The density of water is 1000 kg m^-3 and g = 9.81 m s^-2. (a) State the condition, in terms of upthrust and weigh…

Answer: ((a)) Upthrust equals the weight of the block (the resultant vertical force is zero). ((b)) W ≈ 58.9 N ((c)) V = 6.0×10^-3 m^3 ((d)) depth = 0.15 m ((e)) g cancels from both sides of mg = g V, leaving V = m/, which does not involve g.. Part (a): Condition for floating equilibrium For the block to float in vertical equilibrium, the resultant vertical force on it must be zero. The only vertical forces are its weight (downward) and the upthrust from the water (upward), so:…

AS Level Physics 9702: Equilibrium of a Floating Wooden Block

A rectangular wooden block floats upright in water, partially submerged. The block has a horizontal base area of $0.040\text{ m}^2$ , a height of $0.25\text{ m}$ , and a mass of $6.0\text{ kg}$ . The density of water is $1000\text{ kg m}^{-3}$ and $g = 9.81\text{ m s}^{-2}$ .

(a) State the condition, in terms of upthrust and weight, that must be satisfied for the block to float in vertical equilibrium. [1]

(b) Calculate the weight of the block. [1]

(d) Calculate the depth to which the base of the block is submerged. [2]

(e) State one reason why the value of $g$ was not actually needed to answer part (c). [1]

Part (a): Condition for floating equilibrium

For the block to float in vertical equilibrium, the resultant vertical force on it must be zero. The only vertical forces are its weight (downward) and the upthrust from the water (upward), so: $\text{upthrust} = \text{weight}$

Part (b): Weight of the block

$W = mg = 6.0 \times 9.81$ $W = 58.86\text{ N} \approx 58.9\text{ N (3 s.f.)}$

Part (c): Volume of water displaced

By Archimedes’ principle, the upthrust equals the weight of the water displaced: $F = \rho_{\text{water}}\, g\, V$ Since the block floats in equilibrium, $F = W = mg$ , so: $mg = \rho_{\text{water}}\, g\, V$ $V = \frac{m}{\rho_{\text{water}}} = \frac{6.0}{1000}$ $V = 6.0\times10^{-3}\text{ m}^3$

Part (d): Depth submerged

The submerged part of the block is a rectangular prism of base area $0.040\text{ m}^2$ and depth $h$ , so its volume is $V = Ah$ : $h = \frac{V}{A} = \frac{6.0\times10^{-3}}{0.040}$ $h = 0.15\text{ m}$

This is less than the block’s total height of $0.25\text{ m}$ , confirming that the block does indeed float with part of it above the surface, consistent with the assumption made.

Part (e): Why g was not needed in part (c)

Setting the upthrust equal to the weight gives $mg = \rho_{\text{water}}\, g\, V$ . Since $g$ appears as a common factor on both sides of the equation, it cancels, leaving $m = \rho_{\text{water}} V$ , so the displaced volume can be found from the mass and the water’s density alone, without needing a numerical value of $g$ .

Final answers

(a) Upthrust $=$ weight for vertical equilibrium.
(b) $W \approx \boxed{58.9}\text{ N}$
(c) $V = \boxed{6.0\times10^{-3}}\text{ m}^3$
(d) Depth submerged $= \boxed{0.15}\text{ m}$
(e) $g$ cancels from $mg = \rho_{\text{water}}\, g\, V$ , leaving $V = m/\rho_{\text{water}}$ .