Questions 3

Topic – 6.2

A thin metal wire X, of diameter \(1.2 × 10^{–3}\)m, is used to suspend a model planet, as shown in Fig. 3.1.

The variation with strain of the stress for wire X is shown in Fig. 3.2.

(a) The strain in X is \(5.4 × 10^{–3}\).
(i) Use Fig. 3.2 to calculate the force exerted on the wire by the model planet.

(ii) The elastic potential energy of X is 0.31J. Calculate the original length of the wire before the model planet was attached.

(b) Wire X is replaced by a new wire, Y, with the same original length and diameter but double the Young modulus of X. Wire Y also obeys Hooke’s law. On Fig. 3.2, draw a line representing the variation with strain of the stress for Y.

(a)(i) \(\sigma = 0.72 × 10^9\)

force =  \(\sigma ×A\)
= \(0.72 × 10^9 ×\pi × (1.2 × 10^{–3}/ 2)^2\)

= 810 N

or
Young modulus = gradient of graph
e.g. = \(0.80 × 10^9/ 6.0 × 10^{–3}\)
= \(1.33 × 10^{11}\)

force= Young modulus × strain ×A
= \(1.33 × 10^{11} × 5.4 ×10^{–3} ×\pi × (1.2 × 10^{–3}/ 2)^2\)
= 810 N

(ii) \(E_{(P)} = ½ Fx\)
or \(E_{(P)} = ½ kx^2\) and F = kx

x = \(2E_P\) /F
x = 2 × 0.31 / 810
x = \(7.7 × 10^{–4}\)

L =\( x / \epsilon \)
L = \(7.7 × 10{–4}/ 5.4 × 10^{–3}\)
L = 0.14 m

or
\(E_{(P)}\) = ½ Fx+
or \(E_{(P)} = ½ kx^2\) and k = EA/L

x = \(2E_P / EA \epsilon \)
x = \(2 × 0.31 / (1.33 × 10^{11}× \pi × (1.2 × 10^{–3}/ 2)^2 × 5.4 × 10^{–3}\))
x = \(7.6 × 10^{–4}\)

L =\( x / \epsilon \)
L = \(7.6 × 10^{–4}/ 5.4 × 10^{–3}\)
L = 0.14 m

(b) A straight line, passing through the origin with a larger gradient than wire X.
Gradient of the line is twice the gradient of wire X.