Question

A 1000-kg object hangs from the lower end of a steel rod 5.0 m long that is suspendedvertically. The diameter of the rod is 0.80 cm and Young's modulus for the rod is210,000 MN/m2. What is the elongation of the rod due to this object

Answer 1

Answer:

$4.64\times 10^{-4}$ m

Explanation:

$m$ = mass of the object = 1000 kg

$F$ = weight of the object hanged from rod = $mg$ = $(1000) (9.8)$ = $980$ N

$L$ = length of the steel rod = 5 m

$\Delta L$ = elongation of length of the steel rod = ?

$Y$ = young's modulus of steel = 210,000 MN/m² = 2.1 x 10¹¹ N/m²

$d$ = diameter of the rod = 0.80 cm = 0.0080 m

$A$ = Area of cross-section of rod = $(0.25)\pi d^{2}$ = $(0.25) (3.14) (0.008)^{2}$ = 50.24 x 10⁻⁶ m²

Using the equation

$Y = \frac{FL}{A\Delta L}$

$2.1\times 10^{11} = \frac{(980)(5)}{(50.24\times 10^{-6})\Delta L}$

$\Delta L = 4.64\times 10^{-4}$ m