16 worked answers · short steps first, explanations when you need them
Chapter 8 Exercises
Textbook pages 178–179
Question 1
A 4.7 m steel wire (area 3.0×10⁻⁵ m²) and a 3.5 m copper wire (area 4.0×10⁻⁵ m²) stretch by the same amount under the same load. Find Ysteel/Ycopper.
Steps only
For extension ΔL=FL/(AY), equal force and extension mean Y is proportional to L/A.
Answer
Ysteel/Ycopper≈1.8.
Question 2
Figure 8.9 gives a stress–strain curve. Find (a) Young’s modulus and (b) approximate yield strength from the graph.
Steps only
Read a point in the straight-line part: stress≈150×10⁶ Pa at strain≈0.002. Young’s modulus is stress/strain.
Answer
(a) 7.5×10¹⁰ Pa; (b) about 3×10⁸ Pa.
Question 3
Stress–strain graphs for materials A and B are drawn to the same scale. (a) Which has greater Young’s modulus? (b) Which is stronger?
Steps only
Young’s modulus is the slope in the initial straight part of a stress–strain graph. Curve A is steeper there, so A has the greater modulus.
Answer
(a) Material A; (b) material A.
Question 4
Say whether these are true or false, with reasons: (a) Rubber’s Young’s modulus is greater than steel’s. (b) Stretching a coil is determined by its shear modulus.
Steps only
(a) False. Steel deforms much less than rubber under the same stress, so steel has the greater Young’s modulus. (b) True. When a helical coil is pulled, its wire mainly twists, which is a shear deformation.
Answer
(a) False; (b) True.
Question 5
Two wires, each 0.25 cm in diameter, are connected as shown in Fig. 8.11: a 6 kg load hangs at the top of the steel wire and a 4 kg load hangs below it, at the end of the brass wire. The unloaded lengths are 1.5 m (steel) and 1.0 m (brass). Find each extension. Use Ysteel=2.0×10¹¹ Pa, Ybrass=0.91×10¹¹ Pa, and g=9.8 m/s².
Steps only
The steel supports both loads: Fsteel=(6+4)g=98 N. The brass supports only the lower load: Fbrass=4g=39.2 N. Radius of each wire is 0.00125 m, so A=πr²≈4.91×10⁻⁶ m².
Answer
Steel extension≈1.5×10⁻⁴ m; brass extension≈8.8×10⁻⁵ m.
Question 6
An aluminium cube has 10 cm edges. One face is fixed to a wall and a 100 kg mass is attached to the opposite face. Find the face’s vertical deflection if aluminium’s shear modulus is 25 GPa. Use g=9.8 m/s².
Steps only
The sheared face has area A=(0.1)²=0.01 m²; cube height L=0.1 m. The sideways force is F=mg=980 N.
Answer
About 3.9×10⁻⁷ m (0.39 micrometre).
Question 7
Four identical hollow mild-steel columns support a 50,000 kg structure equally. Each has outer radius 0.60 m and inner radius 0.30 m. Find the compressional strain. Use g=9.8 m/s² and Ysteel=2.0×10¹¹ Pa.
Steps only
Each column carries one quarter of the weight: F=50,000×9.8/4=1.225×10⁵ N. Its metal cross-sectional area is π(R²−r²)=π(0.60²−0.30²)≈0.848 m².
Answer
Compressional strain≈7.2×10⁻⁷.
Question 8
A copper piece has cross-sectional dimensions 15.2 mm × 19.1 mm and is pulled with 44,500 N, staying elastic. Find its strain. Use Ycopper=1.1×10¹¹ Pa.
Steps only
Area=(15.2×10⁻³)(19.1×10⁻³)=2.90×10⁻⁴ m². Stress=F/A≈1.53×10⁸ Pa.
Answer
Strain≈1.4×10⁻³ (about 0.14%).
Question 9
A steel cable has radius 1.5 cm. If stress must stay at or below 1.0×10⁸ Pa, what is its maximum load?
Steps only
Cross-sectional area A=πr²=π(0.015)²≈7.07×10⁻⁴ m².
Answer
About 7.1×10⁴ N.
Question 10
A rigid bar is supported symmetrically by two copper wires at its ends and an iron wire at its centre. All three wires are 2.0 m long and are to have equal tension. Find the ratio of the copper-wire diameter to the iron-wire diameter. Use Ycopper=1.1×10¹¹ Pa and Yiron=1.9×10¹¹ Pa.
Steps only
For a horizontal rigid bar, the wires must stretch equally. Extension is FL/(AY). With equal F and L, equal extension requires AcopperYcopper=AironYiron.
Answer
dcopper/diron≈1.31.
Question 11
A 14.5 kg mass is attached to a 1.0 m steel wire (area 0.065 cm²) and whirled vertically at 2 revolutions per second. Find the wire’s extension at the lowest point. Use Ysteel=2.0×10¹¹ Pa and g=9.8 m/s².
Steps only
Convert area: 0.065 cm²=6.5×10⁻⁶ m². Angular speed ω=2×2π=4π≈12.57 rad/s. At the bottom, tension supplies weight and centripetal force: F=m(g+ω²r), taking r≈1.0 m.
Answer
About 1.9×10⁻³ m (1.9 mm), using the stated 2 rev/s.
Question 12
Water’s volume changes from 100.0 L to 100.5 L when pressure rises by 100 atm. Find its bulk modulus and compare it with air’s isothermal bulk modulus. Use 1 atm=1.013×10⁵ Pa and Bair≈1.0×10⁵ Pa.
Steps only
The printed final volume is larger even though pressure rises; water should compress. Using 0.5 L as the intended compression gives strain 0.5/100=0.005 and B=1.013×10⁷/0.005≈2.03×10⁹ Pa.
Answer
Using 0.5 L as the intended compression, Bwater≈2.03×10⁹ Pa, about 20,000 times Bair. The printed final volume (100.5 L) conflicts with compression; it would need to be lower than 100.0 L.
Question 13
Water has density 1.03×10³ kg/m³ at the surface. Find its density where the pressure is 80.0 atm. Use Bwater=2.2×10⁹ Pa and 1 atm=1.013×10⁵ Pa.
Steps only
The pressure rise above the surface is (80−1) atm≈8.00×10⁶ Pa. The fractional volume decrease is ΔV/V≈Δp/B≈8.00×10⁶/(2.2×10⁹)=0.00364.
Answer
About 1.034×10³ kg/m³.
Question 14
Find the fractional volume change of a glass slab under hydraulic pressure 10 atm. Use glass bulk modulus B=37×10⁹ Pa and 1 atm=1.013×10⁵ Pa.
Steps only
Pressure=10×1.013×10⁵=1.013×10⁶ Pa. For bulk compression, ΔV/V=−p/B.
Answer
Fractional volume change≈−2.7×10⁻⁵ (a contraction).
Question 15
A solid copper cube has edge 10 cm. Find its volume contraction under pressure 7.0×10⁶ Pa. Use copper bulk modulus B=140×10⁹ Pa.