Chapter 10, Problem 10.42QA

Interpretation Introduction

To find:

a) The mass of the brick.

b) The downward pressure due to the gravity exerted by the bottom face of the brick.

c) The downward pressure due to the gravity exerted by the brick when it is stood up on its end.

Answer to Problem 10.42QA

Solution:

a) $2.8 \times {10}^{3}g$

b)

$5.7 \times {10}^{3 } \frac{N}{m^2}$

c)

$2.3 \times {10}^{4 } \frac{N}{m^2}$

Explanation of Solution

1) Concept:

To calculate the mass of the gold brick, we use the length of cube and its volume. We calculate the volume of gold brick from the length of the brick. From the density and volume, we can calculate the mass of gold. The density of solid gold is given in appendix 3.  Earth atmosphere is pulled downward by gravity, and it exerts a force that is spread across the entire surface of the planet. The ratio of force F to the surface area A defines the atmospheric pressure. Acceleration due to gravity is a force that attracts objects towards the earth. When the brick is stood up on its end we need to consider the small dimensions of the brick to calculate the area of the brick.

2) Formula:

i) $F=mg$

ii) $P= \frac{F}{A}$

iii) $Density= \frac{Mass}{Volume}$

3) Given:

i) Density of solid gold is

$19.3 \frac{g}{{cm}^3}$

ii) Acceleration due to gravity is

$9.807 \frac{m}{s^2}$

iii) Wide of brick is 4.0cm.

iv) Length of brick is 12.0cm.

v) High of brick is 3.0cm.

4) Calculations:

a) Calculate the volume of the gold brick.

$Volume=4.0 cm \times 12.0 cm \times 3.0 cm=144 {cm}^3$

Mass of the brick of gold is calculated from the density and the volume of the brick as follows:

$144 {cm}^3\times 19.3 \frac{g}{{cm}^3}=2779.2 g gold brick$

Mass of the gold brick is $2.8 \times {10}^{3}g.$

b) The pressure exerted by the bottom face of the brick due to the gravity is calculated as follows:

$4.0 cm \times \frac{1 m}{100 cm}=0.04 m$

$12.0 cm \times \frac{1 m}{100 cm}=0.12 m$

$P= \frac{F}{A}= \frac{mg}{A}$

$P= \frac{2.78 kg \times 9.807 \frac{m}{s^2} }{0.04 m \times 0.12 m} = 5.7 \times {10}^3 \frac{kg m }{m^2 s^2 } \times \frac{1 N}{1 kg m/s^2}= 5.7 \times {10}^3 N/m^2$

$P=5.7 \times {10}^{3 } \frac{N}{m^2}$

c) The pressure exerted by the brick due to the gravity when it is stood up on its end is calculated as follows:

$P= \frac{F}{A}= \frac{mg}{A}$

 

$P= \frac{2.78 kg \times 9.807 \frac{m}{s^2} }{0.04 m \times 0.03 m}=2.27\times {10}^4 \frac{kg m }{m^2 s^2 } \times \frac{1 N}{1 kg m/s^2}= 2.27 \times {10}^4 N/m^2$

$P=2.3 \times {10}^{4 } \frac{N}{m^2}$

Conclusion:

The mass of the gold brick is calculated from the density and the volume. The pressure is calculated from the mass, the acceleration due to gravity, and the area.