A wire of length 1.70 m hangs vertically from a fixed point, as shown in Fig. 4.1.

Question 11

(a) Define, for a wire,

(i) stress, [1]

(ii) strain. [1]

(b) A wire of length 1.70 m hangs vertically from a fixed point, as shown in Fig. 4.1.

Fig. 4.1

The wire has cross-sectional area 5.74 × 10^-8 m² and is made of a material that has a

Young modulus of 1.60 × 10¹¹ Pa. A load of 25.0 N is hung from the wire.

(i) Calculate the extension of the wire. [3]

(ii) The same load is hung from a second wire of the same material. This wire is

twice the length but the same volume as the first wire. State and explain how the

extension of the second wire compares with that of the first wire. [3]

Reference: Past Exam Paper – June 2011 Paper 21 Q4

Solution:

(a)

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

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

(b)

(i)

Young modulus E = Stress / Strain

Young modulus E = (F/A) / (e/L)       (= FL / Ae)

Extension e (= FL / AE) = (25 × 1.70) / (5.74×10^-8 × 1.6×10¹¹)

Extension e = 4.6×10^-3 m

(ii)

{Since the wire is of the same material the Young modulus is the same as the previous wire.

The force F is also constant as the same load is used.}

The area A becomes A/2       OR the stress is doubled.

{Volume = Area × length = AL

Since the length is now twice (= 2L), the area must be halved (= A/2) for the volume to remain the same (2L × A/2 = AL = V).

Stress = Force / Area = F / (A/2) = 2 × F/A

The stress is doubled.}

The extension e is proportional to (L / A)   OR Substitute into the full formula

{Extension e = FL / AE

As force F and the Young modulus E are constants, the extension is proportional to L/A

e ∝ L / A

The length is doubled while the area is halved. So,

New extension = 2L / (A/2) = 4 × L/A = 4 × extension of first wire}.

So, the total extension increase (for the second wire) is 4e.