Question

A 55 g soapstone cube--a whisky stone--is used to chill a glass of whisky. Soapstone has a density of 3000 kg/m3, whisky a density of 940 kg/m3. What is the approximate normal force of the bottom of the glass on a single stone?

Answer 1

Answer:

$N=0.37N$

Explanation:

Mass $m=55g=>0.055kg$

Soapstone Density $\rho_s=3000kg/m^2$

Whisky Density $\rho_w=940kg/m^2$

Generally the equation for Force is mathematically given by

 $F=U+N$

Therefore

 $N=m*g-(\frac{m}{\rho_s})*\rho_w*g$

 $N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}$

 $N=0.37N$