Engineering Mechanics Chapter 1

Step-by-Step Solution Step 1 Given that the mass of first particle is $8\;{\rm{kg}}$ and the mass of second particle is $12\;{\rm{kg}}$. Also the spacing between the particles is $800\;{\rm{mm}}$ We are required to determine the gravitational force between particles and we need to compare the force with weights. Step 2 The length $1\;{\rm{mm}}$ is equal […]

Step-by-Step Solution Step 1 Given that the weight of the man on Earth is $155\;{\rm{lb}}$, and the acceleration due to gravity on the moon is ${g_m} = 5.30\;{\rm{ft/}}{{\rm{s}}^2}$. We are required to find out the mass of man on Earth in slugs, kilograms and man’s weight in Newton. Also it is required to find out

Step By Step Solution Step 1 We have the given diameter $350\;{\rm{mm}}$, length $2\;{\rm{m}}$, and the density of the concrete $2.45\;{{{\rm{Mg}}} \mathord{\left/ {\vphantom {{{\rm{Mg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}}$. We are required to determine the weight of column in pounds. Step 2 Convert the unit of diameter $\left( d \right)$ into SI units.\[\begin{array}{c} d = 350\;{\rm{mm}}\\

Step-by-Step Solution Step 1 The given quantities are (a) ${{354\;{\rm{mg}}\left( {45\;{\rm{km}}} \right)} \mathord{\left/ {\vphantom {{354\;{\rm{mg}}\left( {45\;{\rm{km}}} \right)} {\left( {0.0356\;{\rm{kN}}} \right)}}} \right. } {\left( {0.0356\;{\rm{kN}}} \right)}}$, (b) $\left( {0.00453\;{\rm{Mg}}} \right)\left( {201\;{\rm{ms}}} \right)$, and (c) ${{435\;{\rm{MN}}} \mathord{\left/ {\vphantom {{435\;{\rm{MN}}} {23.2\;{\rm{mm}}}}} \right. } {23.2\;{\rm{mm}}}}$ We are asked to determine the value of given quantities in SI units in

Step-by-Step Solution Step 1 We are given that the density of the water is ${\rho _w} = 1.94\;{\rm{slug/f}}{{\rm{t}}^3}$ We are asked to determine the density of water in SI units. Step 2 The mass $1\;{\rm{slug}}$ is equal to $14.59\;{\rm{kg}}$ in SI units, and the length $1\;{\rm{ft}}$ is equal to $0.3048\;{\rm{m}}$. Step 3 Now, convert the

Step-by-Step Solution Step 1 We have the relation $1\;{\rm{Pa}} = 1\;{\rm{N/}}{{\rm{m}}^2}$, and the seal level pressure is $14.7\;{{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{i}}{{\rm{n}}^2}}}} \right. } {{\rm{i}}{{\rm{n}}^2}}}$. We are required to find the value of $1\;{\rm{Pa}}$ in ${{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{f}}{{\rm{t}}^2}}}} \right. } {{\rm{f}}{{\rm{t}}^2}}}$, and the value of sea level pressure in ${{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}}

Step-by-Step Solution Step 1 We have the equation of force which is given by, $F = G\frac{{{m_1}{m_2}}}{{{r^2}}}$ We have the mass of each sphere, that is ${m_1} = {m_2} = 200\;{\rm{kg}}$. The radius of each sphere is: ${r_s} = 300\;{\rm{mm}}$ We are required to show that the given equation is dimensionally homogeneous. Also, we are

Step-by-Step Solution Step 1 The given physical quantities are ${\left( {212\;{\rm{mN}}} \right)^2},\;{\left( {52800\;{\rm{ms}}} \right)^2}$ and ${\left[ {548\left( {{{10}^6}} \right)} \right]^{{1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}\;{\rm{ms}}$ We are required to determine the given physical quantities in SI units in three significant figures. Step 2 The value of $1\;{\rm{mN}}$ is equal to $1 \times {10^{

Step-by-Step Solution Step 1 We are given with the density of Aluminium ${\rho _{Al}} = 5.26\;{\rm{slug/f}}{{\rm{t}}^3}$ We are asked to compute the density of Aluminium in SI units. Step 2 “slug” is the unit of mass, and $1\;{\rm{slug}}$ is equal to $14.59\;{\rm{kg}}$ of SI units. Whereas, “ft” or feet is the unit of length, and

Step-by-Step Solution Step 1 We are given the units $\left( {684\;{\rm{\mu m}}} \right)/\left( {43\;{\rm{ms}}} \right)$, $\left( {{\rm{28}}\;{\rm{ms}}} \right)\left( {{\rm{0}}{\rm{.0458}}\;{\rm{Mm}}} \right){\rm{/}}\left( {348\;{\rm{mg}}} \right)$ and $\left( {2.68\;{\rm{mm}}} \right)\left( {426\;{\rm{Mg}}} \right)$. We are asked to evaluate each of the following significant figure and SI units. Step 2 (a) To find the SI unit we need to use the