Question

two rods of equal length (0.55 m) and diameter (1.7 cm) are placed end - to - end. one rod is aluminum, the other is brass. if a compressive force of 8400 n is applied to the rods. how much does their combined length decrease?

a. 3.4×10³
b. 1.8×10³
c. 6.3×10¹
d. 7.5×10⁰
e. 8.2×10⁰

Explanation:

Step1: Calculate the cross - sectional area

The diameter $d = 1.7\ cm=0.017\ m$, and the cross - sectional area $A=\pi(\frac{d}{2})^2=\pi(\frac{0.017}{2})^2\ m^2$.

Step2: Determine Young's modulus values

The Young's modulus of aluminum $Y_{Al}=70\times 10^{9}\ Pa$ and of brass $Y_{brass}=90\times 10^{9}\ Pa$.

Step3: Use the formula for elongation $\Delta L=\frac{FL}{AY}$

For the aluminum rod, $\Delta L_{Al}=\frac{F\times L}{A\times Y_{Al}}$, and for the brass rod, $\Delta L_{brass}=\frac{F\times L}{A\times Y_{brass}}$. The combined decrease in length $\Delta L_{total}=\Delta L_{Al}+\Delta L_{brass}=FL\times(\frac{1}{AY_{Al}}+\frac{1}{AY_{brass}})=\frac{FL}{A}(\frac{1}{Y_{Al}}+\frac{1}{Y_{brass}})$.
Substitute $F = 8400\ N$, $L = 0.55\ m$, $A=\pi(\frac{0.017}{2})^2\ m^2$, $Y_{Al}=70\times 10^{9}\ Pa$, and $Y_{brass}=90\times 10^{9}\ Pa$ into the formula.
First, calculate $\frac{1}{Y_{Al}}+\frac{1}{Y_{brass}}=\frac{Y_{brass}+Y_{Al}}{Y_{Al}Y_{brass}}=\frac{90\times 10^{9}+70\times 10^{9}}{70\times 10^{9}\times90\times 10^{9}}=\frac{160\times 10^{9}}{6300\times 10^{18}}$.
$A=\pi(\frac{0.017}{2})^2\approx2.27\times 10^{-4}\ m^2$.
$\Delta L_{total}=\frac{8400\times0.55}{2.27\times 10^{-4}}\times\frac{160\times 10^{9}}{6300\times 10^{18}}\approx 3.4\times 10^{-5}\ m$.

Answer:

a. $3.4\times 10^{-5}$