A wire is fixed at one end so that it hangs vertically. The wire is given an extension x by suspending a load F from its free end. The variation of F with x is shown in Fig. 3.1.

Question 5

(a) For the deformation of a wire under tension, define

(i) stress, [1]

(ii) strain. [1]

(b) A wire is fixed at one end so that it hangs vertically. The wire is given an extension x by suspending a load F from its free end. The variation of F with x is shown in Fig. 3.1.

Fig. 3.1

The wire has cross-sectional area 9.4 × 10^-8 m² and original length 2.5 m.

(i) Describe how measurements can be taken to determine accurately the cross-sectional area of the wire. [3]

(ii) Determine the Young modulus E of the material of the wire. [2]

(iii) Use Fig. 3.1 to calculate the increase in the energy stored in the wire when the load is increased from 2.0 N to 4.0 N. [2]

(c) The wire in (b) is replaced by a new wire of the same material. The new wire has twice the length and twice the diameter of the old wire. The new wire also obeys Hooke’s law.

On Fig. 3.1, sketch the variation with extension x of the load F for the new wire from x = 0 to x = 0.80 mm. [2]

[Total: 11]

Reference: Past Exam Paper – March 2018 Paper 22 Q3

Solution:

(a)

(i) Stress is defined as the force acting per unit (cross-sectional) area.

(ii) Strain is the ratio of the extension to the original length of the wire.

(b)

(i) A micrometer is used to measure the diameter of the wire. Several measurements are taken around the wire and an average value is calculated.

(ii)

Young modulus E = stress / strain = FL / Ax

The gradient of the force—extension graph gives a value for the quantity (ΔF/Δx). So,

E = (L / A) × gradient

Consider the points (0, 0) and (0.8×10^-3, 4)

Gradient = (4 – 0) / (0.8 × 10^-3 – 0)

E = (4 × 2.5) / [(0.8 × 10^-3) × (9.4 × 10^-8)]

E = 1.3 × 10¹¹ Pa

(iii)

Energy stored in wire = Area under force-extension graph from 2.0 N to 4.0 N

Area of parallelogram = ½ × sum of parallel sides × height

E = ½ × (2+4) × 0.4 × 10^-3

E = 1.2 × 10^-3 J

(c)

straight line from the origin and above the original line

straight line passes through (0.80, 8.0)

{The new wire is made of the same material, so they have the same Young modulus E.

E = (L / A) × gradient

Let the gradient of the first wire be m₁.

E = (L/A) × m₁

The second wire has twice the length (2L) and twice the diameter. This means that the area is 4 times greater (4A) since area = πd² / 4.

Let the gradient of wire 2 be m₂.

E = (2L / 4A) × m2

E = (L / 2A) × m2

The wires have the same Young modulus.

(L / 2A) × m2 = (L / A) × m1

m2 / 2 = m1

m2 = 2 × m1

The gradient of the second wire is twice that of the first wire.}