The Workbook

1. Stress \(= \dfrac{F}{A} = \dfrac{100}{10^{-6}} = 10^{8}\) Pa.

2. Strain \(= \dfrac{\Delta L}{L} = \dfrac{10^{-3}}{2} = 5 \times 10^{-4}\).

3. \(Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{10^{8}}{5 \times 10^{-4}} = 2 \times 10^{11}\) Pa. This is the value for steel.

1. Stress \(= \dfrac{400}{2 \times 10^{-6}} = 2 \times 10^{8}\) Pa.

2. Strain \(= \dfrac{\text{stress}}{Y} = \dfrac{2 \times 10^{8}}{2 \times 10^{11}} = 10^{-3}\).

3. \(\Delta L = \text{strain} \times L = 10^{-3} \times 4 = 4 \times 10^{-3}\) m \(= 4\) mm.

Stress is force per area: N/m\(^2\), that is, pascal (Pa). Young's modulus is stress divided by strain, so it also has the unit Pa.

Strain is a length divided by a length. The units cancel, so strain is a pure number.

\(\Delta L = \dfrac{FL}{AY}\), and \(A \propto d^2\). So \(\Delta L \propto \dfrac{L}{d^2}\) for the same \(F\) and \(Y\).

\(\dfrac{\Delta L_B}{\Delta L_A} = \dfrac{L_B}{L_A} \times \dfrac{d_A^2}{d_B^2} = 2 \times \dfrac{1}{4} = \dfrac{1}{2}\).

1. Strain \(= \dfrac{10^{-3}}{1} = 10^{-3}\), that is, \(0.1\) percent.

2. \(Y = \dfrac{2 \times 10^{8}}{10^{-3}} = 2 \times 10^{11}\) Pa.

\(F = \dfrac{YA\,\Delta L}{L} = \dfrac{1.1 \times 10^{11} \times 10^{-6} \times 10^{-3}}{1} = 110\) N.

1. \(A\): proportional limit, the end of the straight part. \(B\): elastic limit (yield point). \(C\): ultimate tensile strength, the largest stress the wire can carry. \(D\): fracture point.

2. Hooke's law (stress proportional to strain) holds up to \(A\). The wire returns to its original length on unloading up to \(B\).

3. Beyond \(C\) the wire thins at one place (a neck forms). The stress needed falls, the neck grows, and the wire breaks at \(D\).

On the straight part \(OA\), \(Y\) is the slope. At \(A\): stress \(= 2 \times 10^{8}\) Pa, strain \(= 0.001\).

\(Y = \dfrac{2 \times 10^{8}}{10^{-3}} = 2 \times 10^{11}\) Pa.

1. \(P\) is ductile: after its elastic limit it takes a large extra strain before breaking. \(Q\) is brittle: it breaks almost as soon as it leaves the straight line.

2. \(P\). Its straight part is steeper, and the slope of the straight part is \(Y\).

3. \(P\). Its curve reaches a greater stress before the breaking point.

Steel has the larger Young's modulus. A large \(Y\) means a large stress is needed for a small strain; steel is hard to stretch, rubber is easy.

In physics a body is more elastic when it resists deformation more strongly and returns more exactly. Steel is more elastic than rubber. Rubber is only more stretchable.

1. No. The wire stays a little longer than before. The strain that remains at zero stress is the permanent set.

2. The unloading line is straight and parallel to \(OA\), but it reaches the strain axis at the permanent set instead of at \(O\).

1. \(B = -\dfrac{\Delta P}{\Delta V/V}\), so \(\dfrac{\Delta V}{V} = -\dfrac{\Delta P}{B} = -\dfrac{2 \times 10^{7}}{2 \times 10^{9}} = -0.01\).

   The volume falls by \(1\) percent. The minus sign shows a decrease.

2. Compressibility \(= \dfrac{1}{B} = \dfrac{1}{2 \times 10^{9}} = 5 \times 10^{-10}\) Pa\(^{-1}\).

\(\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B} = \dfrac{10^{7}}{2 \times 10^{9}} = 5 \times 10^{-3}\).

\(\Delta V = 5 \times 10^{-3} \times 1000\) mL \(= 5\) mL. Water is nearly incompressible: a huge pressure squeezes a litre by only a teaspoon.

The mass \(m\) is fixed and \(\rho = m/V\). A small change gives \(\dfrac{\Delta\rho}{\rho} = -\dfrac{\Delta V}{V}\).

With \(\dfrac{\Delta V}{V} = -\dfrac{\Delta P}{B}\) this is \(\dfrac{\Delta\rho}{\rho} = \dfrac{\Delta P}{B}\).

From Question 12, \(\dfrac{\Delta P}{B} = 0.01\). The density rises by \(1\) percent to \(1010\) kg/m\(^3\).

\(Y = \dfrac{\text{tensile stress}}{\text{longitudinal strain}}\), \(B = \dfrac{\text{pressure change}}{\text{volume strain}}\), \(\eta = \dfrac{\text{shear stress}}{\text{shear strain}}\).

A liquid resists a change of volume, so it has a bulk modulus. It flows under any tangential force and has no fixed length or shape to restore, so it has no shear modulus and no Young's modulus.

1. Area of the top face \(= (0.1)^2 = 0.01\) m\(^2\). Shear stress \(= \dfrac{F}{A} = \dfrac{200}{0.01} = 2 \times 10^{4}\) Pa.

2. \(\theta = \dfrac{\text{shear stress}}{\eta} = \dfrac{2 \times 10^{4}}{2 \times 10^{6}} = 0.01\) rad.

   \(x = L\theta = 0.1 \times 0.01 = 10^{-3}\) m \(= 1\) mm.

1. Longitudinal strain \(= \dfrac{2 \times 10^{-3}}{2} = 10^{-3}\).

2. Lateral strain \(= \sigma \times\) longitudinal strain \(= 0.25 \times 10^{-3} = 2.5 \times 10^{-4}\).

3. \(\Delta d = 2.5 \times 10^{-4} \times 2\) mm \(= 5 \times 10^{-4}\) mm. The wire gets thinner by half a micrometre.

1. \(V = \pi r^2 L\). For small changes, \(\dfrac{\Delta V}{V} = 2\dfrac{\Delta r}{r} + \dfrac{\Delta L}{L}\).

   \(\dfrac{\Delta L}{L} = \epsilon\) and \(\dfrac{\Delta r}{r} = -\sigma\epsilon\) (the radius shrinks). So \(\dfrac{\Delta V}{V} = -2\sigma\epsilon + \epsilon = (1 - 2\sigma)\epsilon\).

2. The volume is unchanged when \(1 - 2\sigma = 0\), that is, \(\sigma = 0.5\) (rubber is close to this).

   For \(\sigma = 0.25\): \(\dfrac{\Delta V}{V} = (1 - 0.5) \times 10^{-3} = 5 \times 10^{-4}\). The volume grows a little.

1. A change of length under a pull: Young's modulus \(Y\).

2. A change of volume under pressure from all sides: bulk modulus \(B\).

3. Layers sliding across each other: shear modulus \(\eta\).

4. A change of volume of a gas: bulk modulus \(B\).

1. Yes. The wires are in series, so the same \(100\) N acts through both.

   Steel: \(\Delta L = \dfrac{FL}{AY} = \dfrac{100 \times 1}{10^{-6} \times 2 \times 10^{11}} = 5 \times 10^{-4}\) m \(= 0.5\) mm.

   Copper: \(\Delta L = \dfrac{100 \times 1}{10^{-6} \times 10^{11}} = 10^{-3}\) m \(= 1\) mm. The softer wire stretches more.

2. Total extension \(= 0.5 + 1 = 1.5\) mm.

1. The bar stays horizontal, so both wires stretch by the same amount. They have the same length, so the strains are equal.

2. Equal strain means \(\dfrac{T_s}{AY_s} = \dfrac{T_c}{AY_c}\), so \(\dfrac{T_s}{T_c} = \dfrac{Y_s}{Y_c} = 2\).

   \(T_s + T_c = 300\) N gives \(T_c = 100\) N and \(T_s = 200\) N. The stiffer wire carries the larger share.

1. Largest load \(=\) breaking stress \(\times\) area \(= 4 \times 10^{8} \times 2 \times 10^{-6} = 800\) N.

2. Safe load \(= \dfrac{800}{4} = 200\) N, a mass of \(\dfrac{200}{10} = 20\) kg.