The Workbook

1. A body in a fluid is pushed up by a force equal to the weight of the fluid it displaces.

   \(V = 0.5 \times 0.4 \times 0.2 = 0.04\) m\(^3\). \(F_B = V\rho g = 0.04 \times 1000 \times 10 = 400\) N.

2. Vertically upward, through the centre of gravity of the displaced water (the centre of buoyancy). For a uniform block this is its centre.

1. \(F_B = 50 - 30 = 20\) N.

2. \(F_B = V\rho_w g\), so \(V = \dfrac{20}{1000 \times 10} = 2 \times 10^{-3}\) m\(^3\).

3. Mass \(= \dfrac{50}{10} = 5\) kg. Density \(= \dfrac{5}{2 \times 10^{-3}} = 2500\) kg/m\(^3\).

1. A floating body displaces its own weight of fluid. With \(V_s\) submerged: \(V_s\rho_w g = V\rho g\), so \(\dfrac{V_s}{V} = \dfrac{\rho}{\rho_w} = \dfrac{600}{1000} = 0.6\).

2. \(\dfrac{V_s}{V} = \dfrac{600}{800} = 0.75\). In a lighter liquid more of the block must go under to displace the same weight.

Fraction submerged \(= \dfrac{\rho_{\text{ice}}}{\rho_w} = \dfrac{900}{1000} = 0.9\). Nine tenths of the iceberg is hidden.

Sea water is denser than fresh water, so the same weight is displaced by a smaller volume and a little more of the iceberg shows.

1. The displaced water weighs \(200 \times 10 = 2000\) N: \(V = \dfrac{200}{1000} = 0.2\) m\(^3\).

2. Total mass \(300\) kg: \(V = \dfrac{300}{1000} = 0.3\) m\(^3\).

3. Extra volume \(0.1\) m\(^3\) over an area of \(2\) m\(^2\): the boat sinks \(\dfrac{0.1}{2} = 0.05\) m \(= 5\) cm.

Average density \(= \dfrac{2}{4 \times 10^{-3}} = 500\) kg/m\(^3\), less than water. The ball floats with \(\dfrac{500}{1000} = \dfrac{1}{2}\) of its volume submerged.

A solid steel ball has density \(8000\) kg/m\(^3\). Even fully submerged it displaces less than its own weight of water, so it sinks. The hollow ball's air lowers its average density.

1. In air: \(20\) N. Volume of the block \(= \dfrac{2}{4000} = 5 \times 10^{-4}\) m\(^3\), so \(F_B = 5 \times 10^{-4} \times 1000 \times 10 = 5\) N.

   In water the balance reads \(20 - 5 = 15\) N.

2. The water pushes up on the block with \(5\) N, so the block pushes down on the water with \(5\) N (Newton's third law). The scale reading rises by \(5\) N.

1. \(F_B = V\rho_{\text{air}}g = 100 \times 1.2 \times 10 = 1200\) N.

2. Weight of helium \(= 100 \times 0.2 \times 10 = 200\) N. Weight of fabric and basket \(= 600\) N.

   Spare lift \(= 1200 - 200 - 600 = 400\) N, so the balloon can lift an extra \(40\) kg.

1. \(\rho = \dfrac{3}{4} \times 1000 = 750\) kg/m\(^3\).

2. Fraction \(= \dfrac{750}{1500} = 0.5\).

1. Fraction submerged \(= \dfrac{500}{1000} = \dfrac{1}{2}\), so \(5\) cm of the \(10\) cm cube is under water.

2. Fully under, the buoyant force is \(V\rho_w g = 10^{-3} \times 1000 \times 10 = 10\) N. The weight is \(10^{-3} \times 500 \times 10 = 5\) N.

   \(F = F_B - W = 10 - 5 = 5\) N. The force equals the weight of the extra water displaced.

1. The block must displace the extra weight \(0.8 \times 10 = 8\) N of water: extra volume \(= \dfrac{0.8}{1000} = 8 \times 10^{-4}\) m\(^3\).

2. The metal now displaces its own volume, \(\dfrac{0.8}{8000} = 10^{-4}\) m\(^3\), and gets a buoyant force of \(1\) N. The string then pulls the block down with only \(8 - 1 = 7\) N.

   The block sinks by the volume that displaces \(7\) N: \(7 \times 10^{-4}\) m\(^3\).

3. No. In both cases the block and the metal float together. The total displaced volume equals their total weight divided by \(\rho_w g\), which is \(9 \times 10^{-4}\) m\(^3\) either way. The level is the same.

1. No. The floating ice displaces exactly its own weight of water. When it melts it becomes that same weight of water, which fills exactly the volume it displaced. The level does not change.

2. Falls. While frozen in, the stone is floated by the ice and displaces its weight of water. After melting the stone sinks and displaces only its volume, which is less than its weight of water. The level drops.

The hydrometer always displaces its own weight. In petrol, the lighter liquid, it must displace a larger volume, so it sinks deeper. Mark \(B\) is therefore higher on the stem than \(A\).

Buoyant force in water \(= 100 - 80 = 20\) N, so \(V = \dfrac{20}{1000 \times 10} = 2 \times 10^{-3}\) m\(^3\).

Buoyant force in oil \(= 100 - 84 = 16\) N \(= V\rho_{\text{oil}}g\), so \(\rho_{\text{oil}} = \dfrac{16}{2 \times 10^{-3} \times 10} = 800\) kg/m\(^3\).

Loss of mass in water \(= 0.1\) kg, which is the mass of water displaced: \(V = \dfrac{0.1}{1000} = 10^{-4}\) m\(^3\).

Density \(= \dfrac{1}{10^{-4}} = 10^{4}\) kg/m\(^3\). Pure gold would be \(19300\) kg/m\(^3\), so the crown is not pure gold.

In water: \(\rho = 0.6 \times 1000 = 600\) kg/m\(^3\).

In the liquid: \(0.75\rho_L = 600\), so \(\rho_L = 800\) kg/m\(^3\).

1. The centre of buoyancy is the centre of gravity of the displaced water. When the boat tilts, more hull goes under on the low side, so the displaced water's centre shifts towards that side.

2. \(M\) is the point where the vertical line through \(B\) meets the boat's own axis. The weight down through \(G\) and the buoyancy up through \(B\) are equal but not in line. They form a couple that turns the boat back towards upright when \(M\) is above \(G\).

3. Stable when \(M\) lies above \(G\). If \(M\) falls below \(G\) the couple turns the boat further over.

Cargo on the floor lowers \(G\), so \(M\) stays above \(G\) even for large tilts. Cargo on the deck raises \(G\) towards \(M\) and can put it above \(M\).

A wide boat shifts a large volume of hull under water when it tilts, so \(B\) moves far sideways and \(M\) is high. A narrow boat shifts little, so \(M\) is low.

Fully submerged, the displaced volume is the whole body, so \(B\) is fixed at the centre of the body's volume. The body is stable when \(B\) is above \(G\).

Then any tilt puts the weight (down through \(G\)) and the buoyancy (up through \(B\)) out of line. The couple turns the body back. A submarine keeps its heavy machinery low to keep \(G\) below \(B\).

1. Fraction submerged \(= \dfrac{16}{20} = 0.8\), so \(\rho = 0.8 \times 1000 = 800\) kg/m\(^3\).

2. In oil of the same density the cylinder displaces its own weight only when fully submerged. It floats with its top level with the surface, just under.

Net upward force \(= F_B - mg = 1200 - 80 \times 10 = 400\) N.

\(a = \dfrac{400}{80} = 5\) m/s\(^2\) upward. Air resistance soon limits the speed.