Question

A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.60 N. 1. What is the density of the metal? (Include units) \rho =?
2. What did the cube weigh before you drilled the hole in it? (Include units) \omega =?

Answer 1

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

$V_c = l^3$

$V_c = 6^3 = 216cm^3$

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

$V_d = \pi r^2*l$

$V_d = \pi (\frac{2}{2})^2*6$

$V_d = 6\pi cm^3$

In this way the net volume would be

$\Delta V = V_c-V_d$

$\Delta V = 216cm^3-6\pi cm^3$

$\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3$

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

$F= mg$

$m = \frac{F}{g}$

$m = \frac{6.6}{9.8}$

$m = 0.673kg$

PART A) From the relation of density as a unit of mass and volume we have to

$\rho = \frac{m}{V}$

$\rho = \frac{0.673}{197.15*10^{-6}}$

$\rho = 3413.64kg/m^3$

PART B) To find the weight of the cube then we apply the ratio of

$W = mg$

$W = V\rho g$

$W = (216*10^{-6})(3413.64)(9.8)$

$W = 7.22N$