Question

Suppose that you are going to measure Young’s modulus for three rods by measuring their stretch when they are suspended vertically and weights are hung from them (Figure 4.56)

(Figure 4.56)

Rod 1 is 2.7 m long and cylindrical with radius 4 mm (1 mm is 0.001 m). Rod 2 is 3.2 m long by 12 mm wide by 6 mm deep. Rod 3 is 3 m long by 6 mm wide by 6 mm deep. The definition of Young’s modulus, Y = (F/A)/(∆L/L), includes the quantity A, the cross-sectional area.

(a) What is the cross-sectional area of rod 1?

(b) What is the cross-sectional area of rod 2?

(c) What is the cross-sectional area of rod 3?

Short answer

The results are $5.03\times {10}^{-5}{\mathrm{m}}^2$,$7.2\times {10}^{-5}{\mathrm{m}}^2$ and$3.6\times {10}^{-5}{\mathrm{m}}^2$ .

Step-by-step solution

Given data

Rod 1,

L = 2.7 m

R = 4mm = 0.004 m

Rod 2,

L = 3.2 m

w = 12mm = 0.012 m

d = 6mm = 0.006 m

Rod 3,

L = 3 m

w = 0.006 m

d = 0.006 m

(a)Determine the cross-sectional area of rod 1 

$\begin{array}{c}{\mathrm{A}}_1={\mathrm{\pi r}}^2\\ =3.14\times {\left(0.004\right)}^2\\ =5.0265\times {10}^{-5}{\mathrm{m}}^2\\ {\mathrm{A}}_1\simeq 5.03\times {10}^{-5}{\mathrm{m}}^2\end{array}$

(b) Determine the cross-sectional area of rod 2 

$\begin{array}{c}{\mathrm{A}}_2=\mathrm{dw}\\ =0.012\times 0.006\\ =7.2\times {10}^{-5}{\mathrm{m}}^2\end{array}$

(c)Determine the cross-sectional area of rod 3

$\begin{array}{c}{\mathrm{A}}_3=\mathrm{wd}\\ =0.006\times 0.006\\ =3.6\times {10}^{-5}{\mathrm{m}}^2\end{array}$