Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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Textbook Solution:

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Answer:

To determine

Rank the energy quantities from largest to smallest.

Answer

The ranking is a>b=c−−−−−−−−$\underset{\_}{a>b=c}$ the energy quantities from largest to smallest.

Explanation

a$a$ The absolute value of the average potential energy of the sun-earth system.

b$b$ The average kinetic energy of the earth in its orbital motion relative to the sun.

c$c$ The absolute value of the total energy of the sun-earth system.

Write the expression for the potential energy between the sun-earth system.

    U=−GMSMER$U=-\frac{G{M}_S{M}_E}{R}$        I$I$

Here, U$U$ is the average potential energy of the sun-earth system, G$G$ is the gravitational constant, MS$M_S$ is the mass of sun, ME$M_E$ is the mass of earth, R$R$ is the separation between the sun and earth.

Use equation I$I$ to write the absolute value of U$U$.

    |U|=GMSMER$\left|U\right|=\frac{G{M}_S{M}_E}{R}$        II$II$

Write the expression for the kinetic energy.

    KE=12MEv2E$K_E=\frac{1}{2}{M}_E{v}_E^2$        III$III$

Here, KE$K_E$ is the kinetic energy of the earth, vE$v_E$ is the velocity of the earth.

Write the expression for the gravitational force.

    FG=GMEMSR2$F_G=\frac{G{M}_E{M}_S}{R^2}$        IV$IV$

Here, FG$F_G$ is the gravitational force.

Write the expression for the centripetal force of the earth.

    FC=MEv2ER$F_C=\frac{M_E{v}_E^2}{R}$        V$V$

Here, FC$F_C$ is the centripetal force.

Equate equation IV$IV$ and V$V$ to solve for v2E$v_E^2$.

    v2E=GMSR$v_E^2=\frac{G{M}_S}{R}$        VI$VI$

Use equation VI$VI$ in III$III$ to solve for KE$K_E$.

    KE=GMSME2R$K_E=\frac{G{M}_S{M}_E}{2R}$        VII$VII$

Write the expression for the total energy of the sun-earth system.

    E=KE+U$E=K_E+U$        VIII$VIII$

Here, E$E$ is the total energy of the earth-sun system.

Use equation I$I$ and VII$VII$ in VIII$VIII$ to solve for E$E$.

    E=GMSME2R−GMSMER=−GMSME2R$\begin{array}{c}E=\frac{G{M}_S{M}_E}{2R}-\frac{G{M}_S{M}_E}{R}\\ =-\frac{G{M}_S{M}_E}{2R}\end{array}$        IX$IX$

Use equation IX$IX$ and write the absolute value of E$E$.

    |E|=GMSME2R$\left|E\right|=\frac{G{M}_S{M}_E}{2R}$        X$X$

Conclusion:

The ranking of the energy quantities from largest to smallest is |U|>KE=|E|$\left|U\right|>K_E=\left|E\right|$.

Therefore, the ranking is a>b=c−−−−−−−−$\underset{\_}{a>b=c}$ the energy quantities from largest to smallest.