Chapter 13.2, Problem 13.83P

(a)

To determine

The work done from D to O by the force P $\left(U_{OD}\right)$.

Answer to Problem 13.83P

The work done from D to O by the force P $\left(U_{OD}\right)$ is $\underset{\_}{\sqrt{3}a}$.

Explanation of Solution

Given information:

The potential function associated with force P is $V\left(x,y,z\right)=-{\left(x^2+y^2+z^2\right)}^{1/2}$.

Calculation:

Write the potential function associated with force P:

$V\left(x,y,z\right)=-{\left(x^2+y^2+z^2\right)}^{1/2}$ (1)

Calculate the force P in ‘x’ component using the relation:

Partial differentiate the equation (1) with respect to ‘x’.

$P_x=\frac{\partial V}{\partial x}$

Substitute $-{\left(x^2+y^2+z^2\right)}^{1/2}$ for V.

$\begin{array}{c}{P}_x=\frac{\partial \left(-{\left(x^2+y^2+z^2\right)}^{1/2}\right)}{\partial x}\\ =\frac{1}{2\sqrt{x^2+y^2+z^2}}\times \left(2x\right)\\ =x{\left(x^2+y^2+z^2\right)}^{-1/2}\end{array}$

Calculate the force P in ‘y’ component using the relation:

Partial differentiate the equation (1) with respect to ‘y’.

$P_y=\frac{\partial V}{\partial y}$

Substitute $-{\left(x^2+y^2+z^2\right)}^{1/2}$ for V.

$\begin{array}{c}{P}_y=\frac{\partial \left(-{\left(x^2+y^2+z^2\right)}^{1/2}\right)}{\partial y}\\ =\frac{1}{2\sqrt{x^2+y^2+z^2}}\times \left(2y\right)\\ =y{\left(x^2+y^2+z^2\right)}^{-1/2}\end{array}$

Calculate the force P in ‘z’ component using the relation:

Partial differentiate the equation (1) with respect to ‘z’.

$P_z=\frac{\partial V}{\partial z}$

Substitute $-{\left(x^2+y^2+z^2\right)}^{1/2}$ for V.

$\begin{array}{c}{P}_z=\frac{\partial \left(-{\left(x^2+y^2+z^2\right)}^{1/2}\right)}{\partial x}\\ =\frac{1}{2\sqrt{x^2+y^2+z^2}}\times \left(2z\right)\\ =z{\left(x^2+y^2+z^2\right)}^{-1/2}\end{array}$

Calculate the force (P) using the relation:

$P=P_x+P_y+P_z$

Substitute $x{\left(x^2+y^2+z^2\right)}^{-1/2}$ for $P_x$, $y{\left(x^2+y^2+z^2\right)}^{-1/2}$ for $P_y$ and $z{\left(x^2+y^2+z^2\right)}^{-1/2}$ for $P_z$.

$\begin{array}{c}P=x{\left(x^2+y^2+z^2\right)}^{-1/2}+y{\left(x^2+y^2+z^2\right)}^{-1/2}+z{\left(x^2+y^2+z^2\right)}^{-1/2}\\ =\frac{xi+yj+zk}{{\left(x^2+y^2+z^2\right)}^{1/2}}\end{array}$

Write the expression for the position vector of the cube (r) as follows:

$r=xi+yj+zk$

Differentiate the above equation.

$dr=dxi+dyj+dzk$

Calculate the work done from D to O by the force P $\left(U_{OD}\right)$ using the relation:

$U_{OD}={\displaystyle \underset{0}{\overset{a}{\int }}P\cdot dr}$

Substitute $\frac{xi+yj+zk}{{\left(x^2+y^2+z^2\right)}^{1/2}}$ for $P$ and $dxi+dyj+dzk$ for $dr$.

$U_{OD}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{xi+yj+zk}{{\left(x^2+y^2+z^2\right)}^{1/2}}\left(dxi+dyj+dzk\right)}$

Over the diagonal, $x=y=z$.

Substitute x for y and x for z.

$\begin{array}{c}{U}_{OD}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{xi+xj+xk}{{\left(x^2+x^2+x^2\right)}^{1/2}}\left(dxi+dxj+dxk\right)}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\frac{3x}{{\left(3x^2\right)}^{1/2}}dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\sqrt{3}dx}\end{array}$

$\begin{array}{c}=\sqrt{3}{\left[x\right]}_0^a\\ =\sqrt{3}\left[a-0\right]\\ =\sqrt{3}a\end{array}$

Therefore, the work done from D to O by the force P $\left(U_{OD}\right)$ is $\underset{\_}{\sqrt{3}a}$.

(b)

To determine

Verify the work done by a conservative force around the closed path OABDO $\left(U_{OABDO}\right)$ is zero.

Explanation of Solution

Given information:

The potential function associated with force P is $V\left(x,y,z\right)=-{\left(x^2+y^2+z^2\right)}^{1/2}$.

Calculation:

Write the coordinates of O-A using the relation:

The line O-A is lies on the z-axis. The term x and y are zero.

Calculate the work done on O-A $\left(U_{O-A}\right)$ using the relation:

$U_{O-A}={\displaystyle \underset{0}{\overset{z}{\int }}{P}_{z}dz}$

Since the force $P_x$ and $P_y$ are perpendicular to the O-A and do not work.

Substitute $z{\left(x^2+y^2+z^2\right)}^{-1/2}$ for $P_z$.

$U_{O-A}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{z}{\sqrt{\left(x^2+y^2+z^2\right)}}dz}$

Substitute 0 for x and 0 for y.

$\begin{array}{c}{U}_{O-A}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{z}{\sqrt{\left({\left(0\right)}^2+{\left(0\right)}^2+z^2\right)}}dz}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\frac{z}{\sqrt{\left(z^2\right)}}dz}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\frac{z}{z}dz}\end{array}$

$\begin{array}{c}={\displaystyle \underset{0}{\overset{a}{\int }}dz}\\ ={\left[z\right]}_0^a\\ =a-0\\ =a\end{array}$

The co-ordinates A-B is lies on the x-axis. Then, the term y is zero and z tends to ‘a’.

Calculate the work done on A-B $\left(U_{A-B}\right)$ using the relation:

$U_{A-B}={\displaystyle \underset{0}{\overset{a}{\int }}{P}_{x}dz}$

Since the force $P_z$ and $P_y$ are perpendicular to the A-B and do not work.

Substitute $x{\left(x^2+y^2+z^2\right)}^{-1/2}$ for $P_x$.

$U_{A-B}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{x}{\sqrt{\left(x^2+y^2+z^2\right)}}dx}$

Substitute a for z and 0 for y.

$\begin{array}{c}{U}_{A-B}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{x}{\sqrt{\left({\left(x\right)}^2+{\left(0\right)}^2+{\left(a\right)}^2\right)}}dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\frac{x}{\sqrt{\left(x^2+a^2\right)}}dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\frac{1}{2\sqrt{x^2+a^2}}dx}\end{array}$

Simplify the Equation:

$\begin{array}{c}{U}_{A-B}=\frac{1}{2}{\left[\frac{{\left(x^2+a^2\right)}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]}_0^a\\ ={\left[\frac{1}{2}{\left(x^2+a^2\right)}^{\frac{1}{2}}\times \frac{2}{1}\right]}_0^a\\ ={\left[\sqrt{\left(x^2+a^2\right)}\right]}_0^a\end{array}$

$\begin{array}{c}=\left(\sqrt{a^2+a^2}\right)-\left(\sqrt{0^2+a^2}\right)\\ =\sqrt{2a^2}-\sqrt{a^2}\\ =\sqrt{2}\sqrt{a^2}-\sqrt{a^2}\\ =\sqrt{2}a-a\\ =a\left(\sqrt{2}-1\right)\end{array}$

The co-ordinates B-D is lies on the y-axis. Then, the term x is zero and z tends to ‘a’.

Calculate the work done on B-D $\left(U_{B-D}\right)$ using the relation:

$U_{B-D}={\displaystyle \underset{0}{\overset{a}{\int }}{P}_{y}dz}$

Since the force $P_z$ and $P_x$ are perpendicular to the B-D and do not work.

Substitute $y{\left(x^2+y^2+z^2\right)}^{-1/2}$ for $P_y$.

$U_{B-D}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{y}{\sqrt{\left(x^2+y^2+z^2\right)}}dy}$

Substitute 2a for z and 0 for x.

$\begin{array}{c}{U}_{B-D}={\displaystyle \underset{0}{\overset{a}{\int }}\frac{y}{\sqrt{\left({\left(0\right)}^2+{\left(y\right)}^2+{\left(2a\right)}^2\right)}}dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\frac{y}{\sqrt{\left(y^2+2a^2\right)}}dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\frac{1}{2\sqrt{y^2+2a^2}}dx}\end{array}$

Simplify the Equation,

$\begin{array}{c}{U}_{B-D}=\frac{1}{2}{\left[\frac{{\left(y^2+2a^2\right)}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]}_0^a\\ ={\left[\frac{1}{2}{\left(y^2+2a^2\right)}^{\frac{1}{2}}\times \frac{2}{1}\right]}_0^a\\ ={\left[\sqrt{\left(y^2+2a^2\right)}\right]}_0^a\end{array}$

$\begin{array}{c}=\left(\sqrt{a^2+2a^2}\right)-\left(\sqrt{0^2+2a^2}\right)\\ =\sqrt{3a^2}-\sqrt{2a^2}\\ =\sqrt{3}\sqrt{a^2}-\sqrt{2}\sqrt{a^2}\\ =\sqrt{3}a-\sqrt{2}a\\ =a\left(\sqrt{3}-\sqrt{2}\right)\end{array}$

Calculate the work done by P $\left(U_{OABD}\right)$ using the relation:

$U_{OABD}=U_{O-A}+U_{A-B}+U_{B-D}$

Substitute a for $U_{O-A}$, $a\left(\sqrt{2}-1\right)$ for $U_{A-B}$, and $a\left(\sqrt{3}-\sqrt{2}\right)$ for $U_{B-D}$.

$\begin{array}{c}{U}_{OABD}=a+\left[a\left(\sqrt{2}-1\right)\right]+\left[a\left(\sqrt{3}-\sqrt{2}\right)\right]\\ =a+\sqrt{2}a-a+\sqrt{3}a-\sqrt{2}a\\ =a\sqrt{3}\end{array}$

The work done from D to O along the diagonal is negative of the work done from O to D.

$\begin{array}{l}{U}_{OD}=\sqrt{3}a\\ U_{OD}=-\sqrt{3}a\end{array}$

Calculate the work done by a conservative force around the closed path OABDO $\left(U_{OABDO}\right)$ using the relation:

$U_{OABDO}=U_{OABD}+U_{OD}$

Substitute $a\sqrt{3}$ for $U_{OABD}$ and $-\sqrt{3}a$ for $U_{OD}$.

$\begin{array}{c}{U}_{OABDO}=\sqrt{3}a-\sqrt{3}a\\ =0\end{array}$

Therefore, the work done by a conservative force around the closed path OABDO $\left(U_{OABDO}\right)$ is zero.