Chapter 4: Review

The chapter problem, solved

The crane lifts a steel beam of mass 2500 kg on a cable 20.0 m long. The cable has a cross-sectional area of $1.5 \times 10^{-4}\,\text{m}^2$ , the steel has a Young modulus of $2.0 \times 10^{11}\,\text{Pa}$ , and the working stress must not exceed $3.0 \times 10^{8}\,\text{Pa}$ .

1. The load in the cable. The cable carries the weight of the beam.

W = mg = 2500 \times 9.81 = 24\,525\,\text{N}

2. The stress in the cable.

\sigma = \frac{F}{A} = \frac{24\,525}{1.5 \times 10^{-4}} = 1.635 \times 10^{8}\,\text{Pa}

The stress is about $1.64 \times 10^{8}\,\text{Pa}$ , or $164\,\text{MPa}$ .

3. The strain in the cable.

\varepsilon = \frac{\sigma}{E} = \frac{1.635 \times 10^{8}}{2.0 \times 10^{11}} = 8.175 \times 10^{-4}

The strain is about $8.18 \times 10^{-4}$ . It has no unit.

4. The extension of the cable. The strain is the extension divided by the original length.

\begin{aligned} \varepsilon &= \frac{\Delta L}{L} \\ \Delta L &= \varepsilon L = 8.175 \times 10^{-4} \times 20.0 \\ &= 1.635 \times 10^{-2}\,\text{m} \end{aligned}

The cable stretches by about $16.4\,\text{mm}$ . That is a little over a centimetre, so the first estimate of a millimetre was too small and an estimate of a metre was far too large.

5. The stiffness of the cable. The gradient of the force-extension graph is

k = \frac{EA}{L} = \frac{2.0 \times 10^{11} \times 1.5 \times 10^{-4}}{20.0} = 1.5 \times 10^{6}\,\text{N}\,\text{m}^{-1}

6. The energy stored in the cable.

\begin{aligned} E &= \frac{1}{2}F\Delta L \\ &= \frac{1}{2} \times 24\,525 \times 1.635 \times 10^{-2} \\ &= 2.00 \times 10^{2}\,\text{J} \end{aligned}

The cable stores about $200\,\text{J}$ of elastic potential energy.

7. Is the cable safe for the beam? The stress is $1.635 \times 10^{8}\,\text{Pa}$ , and the working limit is $3.0 \times 10^{8}\,\text{Pa}$ . The stress is below the limit, so the cable is safe for the beam under the rule.

8. The heaviest beam. The greatest force is the working stress multiplied by the area.

F_{\text{max}} = 3.0 \times 10^{8} \times 1.5 \times 10^{-4} = 4.5 \times 10^{4}\,\text{N}

The greatest mass is

m_{\text{max}} = \frac{4.5 \times 10^{4}}{9.81} = 4.6 \times 10^{3}\,\text{kg}

The cable may lift a beam of about 4600 kg.

9. The judgement. A second beam of 5000 kg gives a load of $5000 \times 9.81 = 49\,050\,\text{N}$ , so the stress would be

\frac{49\,050}{1.5 \times 10^{-4}} = 3.27 \times 10^{8}\,\text{Pa}

That is above the working limit of $3.0 \times 10^{8}\,\text{Pa}$ , so the cable may not lift the 5000 kg beam under the safety rule. A cable of area at least $1.64 \times 10^{-4}\,\text{m}^2$ would be needed. This final judgement is a reasoned answer. It is not the kind of claim a computer can check, so compare your reasoning with the evidence above.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate.

The beam has mass $2500\,\text{kg}$ and the cable has area $1.5 \times 10^{-4}\,\text{m}^2$ . Find the stress in the cable.
The cable stress is $1.635 \times 10^{8}\,\text{Pa}$ and $E = 2.0 \times 10^{11}\,\text{Pa}$ . Find the strain.
The cable is $20.0\,\text{m}$ long and the strain is $8.175 \times 10^{-4}$ . Find the extension.
The load is $24\,525\,\text{N}$ and the extension is $1.635 \times 10^{-2}\,\text{m}$ . Find the energy stored in the cable.
The working stress is $3.0 \times 10^{8}\,\text{Pa}$ and the cable area is $1.5 \times 10^{-4}\,\text{m}^2$ . Find the greatest mass the cable may lift.
A second beam of mass $5000\,\text{kg}$ is proposed. Decide whether the cable may lift it under the safety rule, and give a reason.
A wire has a cross-sectional area of $4.0\,\text{mm}^2$ . Write this area in $\text{m}^2$ .
Show that a stress has the base units $\text{kg}\,\text{m}^{-1}\,\text{s}^{-2}$ .
A wire has a diameter of $2.00 \pm 0.01\,\text{mm}$ . Find the percentage uncertainty in the cross-sectional area.
A cable pulls a post with a tension of $5000\,\text{N}$ at $30^\circ$ to the horizontal. Find the horizontal and vertical components of the tension.
A wire is stretched, and the force and the extension are measured. Say which graph you would plot to find the Young modulus, and how you would use it. State one measurement you need as well as the graph.
A steel block has a volume of $2.5 \times 10^{-4}\,\text{m}^3$ and a mass of $1.96\,\text{kg}$ . Find its density.
A force-extension graph for a wire is a straight line through $(0, 0)$ and $(1.5\,\text{mm},\, 12\,\text{N})$ . Find the gradient and the energy stored at that extension.
Explain the difference between the limit of proportionality and the elastic limit.

Answers and concise solutions

1. The stress in the cable

Answer: The load is $mg = 2500 \times 9.81 = 24\,525\,\text{N}$ . The stress is $\dfrac{24\,525}{1.5 \times 10^{-4}} = 1.64 \times 10^{8}\,\text{Pa}$ .

2. The strain

Answer: The strain is $\dfrac{1.635 \times 10^{8}}{2.0 \times 10^{11}} = 8.18 \times 10^{-4}$ . It has no unit.

3. The extension

Answer: The extension is $8.175 \times 10^{-4} \times 20.0 = 1.64 \times 10^{-2}\,\text{m}$ , which is $16.4\,\text{mm}$ . The cable stretches by about a centimetre, not a metre.

4. The energy stored

Answer: The energy stored is $\tfrac{1}{2}F\Delta L = \tfrac{1}{2} \times 24\,525 \times 1.635 \times 10^{-2} = 2.00 \times 10^{2}\,\text{J}$ .

5. The greatest mass

Answer: The greatest force is $3.0 \times 10^{8} \times 1.5 \times 10^{-4} = 4.5 \times 10^{4}\,\text{N}$ . The greatest mass is $\dfrac{4.5 \times 10^{4}}{9.81} = 4.6 \times 10^{3}\,\text{kg}$ .

6. The judgement

The second beam weighs 5000 times 9.81 = 49 050 N, so the stress is 49 050 divided by 1.5 times 10 to the minus 4, which is 3.27 times 10 to the 8 Pa.
The working stress must not exceed 3.0 times 10 to the 8 Pa, and 3.27 times 10 to the 8 Pa is larger than that.
The cable may not lift the 5000 kg beam under the safety rule. A cable of area at least 1.64 times 10 to the minus 4 square metres, or a lighter beam, is needed.

Answer: The second beam gives a stress of $3.27 \times 10^{8}\,\text{Pa}$ , which is above the working limit of $3.0 \times 10^{8}\,\text{Pa}$ , so the cable is not safe for it under the rule.

7. The area in square metres

Answer: $4.0\,\text{mm}^2 = 4.0 \times 10^{-6}\,\text{m}^2$ . The millimetre is squared, so the factor is $10^{-6}$ .

8. The base units of stress

Answer: A stress is a force over an area. A newton is $\text{kg}\,\text{m}\,\text{s}^{-2}$ , so $\dfrac{\text{kg}\,\text{m}\,\text{s}^{-2}}{\text{m}^2} = \text{kg}\,\text{m}^{-1}\,\text{s}^{-2}$ , which is the pascal.

9. The percentage uncertainty in the area

Answer: The percentage uncertainty in the diameter is $\dfrac{0.01}{2.00} \times 100\% = 0.5\%$ . The area depends on the square of the diameter, so its percentage uncertainty is doubled: $2 \times 0.5\% = 1.0\%$ .

10. The two components of the tension

Answer: The horizontal component is $5000\cos 30^\circ = 4330\,\text{N}$ . The vertical component is $5000\sin 30^\circ = 2500\,\text{N}$ . The horizontal component is the larger one, because the cable is closer to the horizontal.

11. The graph and the method

Plot the force on the vertical axis against the extension on the horizontal axis. Use only the straight part of the graph.
Find the gradient, which is the stiffness k. Then use E = kL divided by A, where L is the original length.
Measure the original length and the cross-sectional area of the wire as well as the graph.

Answer: Plot force against extension and find the gradient, which is the stiffness k. Measure the original length and the cross-sectional area, then use $E = \dfrac{kL}{A}$ .

12. The density

Answer: The density is $\dfrac{m}{V} = \dfrac{1.96}{2.5 \times 10^{-4}} = 7840\,\text{kg}\,\text{m}^{-3}$ .

13. The gradient and the energy stored

Answer: The gradient is $\dfrac{12}{1.5 \times 10^{-3}} = 8.0 \times 10^{3}\,\text{N}\,\text{m}^{-1}$ . The energy stored is the area of the triangle, $\tfrac{1}{2} \times 12 \times 1.5 \times 10^{-3} = 9.0 \times 10^{-3}\,\text{J}$ .

14. The two limits

The limit of proportionality is the point where the force-extension graph stops being a straight line. Past it the extension is no longer proportional to the force, so Hooke's law no longer holds.
The elastic limit is the point past which the material does not return to its original length when the load is removed. The change that remains is plastic deformation. A material can pass its limit of proportionality before it passes its elastic limit, so the graph can bend while the material still springs back.

Answer: The limit of proportionality is where the graph stops being a straight line, so the extension is no longer proportional to the force. The elastic limit is where the material stops returning to its original length. The graph can bend before the elastic limit is reached.

What you can now do

explain why engineers use stress and strain rather than force and extension;
calculate a tensile stress, a tensile strain and a Young modulus;
use the Young modulus to find an extension;
find the spring constant from the gradient of a force-extension graph;
describe the limit of proportionality, the elastic limit and plastic deformation;
find the energy stored from the area under a force-extension graph;
use $E = \tfrac{1}{2}F\Delta L = \tfrac{1}{2}k(\Delta L)^2$ for a straight-line graph;
check whether a cable is safe for a given load, using a working stress.

Your relevance reflection

Return to the prompt from the chapter opener: think of something that changes shape when a load acts on it, and say whether it returns to its original shape. Write one more example now that you have finished the chapter. Has the chapter changed how you think about the cables, ropes or springs you meet?