91影视

Which says that Net external torque around any location on the body must be zero for it to stay nonrotating.

$\underset{\mathbf{}}{\mathbf{鈭�}}\stackrel{\mathbf{鈫�}}{\mathbf{蟿}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Where$\stackrel{\mathbf{鈫�}}{\mathbf{蟿}}$is net external torque.

Here we have given that, the length of the meter stick L = 1 m

The coefficient of static friction between the end of the meter stick and the wall is ${\mathrm{渭}}_{\mathrm{s}}=0.40$

Let N is the normal force from the wall on the meter stick

${\mathrm{F}}_{\mathrm{s}}$is static friction from the wall on the meter stick.

W is the weight of the meter stick at its center of gravity

T is tension from the cord on the meter stick.

(a)

Let the system is in equilibrium.

W = 0

So, from equation (1),

$\begin{array}{r}\underset{\stackrel{鈫�}{c}}{鈭�}鈥�=\left(0\right)\mathrm{T}+\left(0\right)\mathrm{N}+\left(\frac{\mathrm{L}}{2}\right)\mathrm{W}鈭�{\mathrm{LF}}_s\\ =0\\ {\mathrm{F}}_{\mathrm{s}}=\frac{\mathrm{W}}{2}\end{array}$

Also,

$\begin{array}{r}鈭�\stackrel{鈫�}{{蟿}_B}=\left(0\right)T+\left(0\right)F_s鈭�\left(\frac{L}{2}\right)W+(\tan 鈦�胃)N\\ =0\end{array}$

From equations (2) and (3),

${\mathrm{F}}_{\mathrm{s}}=\mathrm{狈迟补苍胃}$

Now, From the figure,

$\begin{array}{r}N\tan 鈦�胃鈮�{渭}_{s}N\\ \tan 鈦�胃鈮�{渭}_s\\ 胃鈮�{\tan }^{鈭�1}鈦�\left({渭}_s\right)\\ 胃鈮�{\tan }^{鈭�1}鈦�\left(0.40\right)\\ 胃鈮�{21.80}^{鈭�}\end{array}$

So, the maximum value of $胃$ can have if the stick is to remain in equilibrium is$21.80掳$

Suppose

So, from equation (1),

$\begin{array}{r}鈭�\stackrel{鈫�}{{蟿}_c}=\left(0\right)\mathrm{T}+\left(0\right)\mathrm{N}+\left(\frac{\mathrm{L}}{2}\right)\mathrm{W}鈭�{\mathrm{LF}}_{\mathrm{s}}+(\mathrm{L}鈭�\mathrm{x})\mathrm{W}\\ =0\\ \frac{3}{2}\mathrm{LW}鈭�\mathrm{xW}={\mathrm{LF}}_{\mathrm{s}}\\ \frac{3}{2}W鈭�\frac{x}{L}W=F_s\\ W=\frac{F_s}{\left(\frac{3}{2}鈭�\frac{x}{L}\right)}\end{array}$

Also,

$\begin{array}{r}鈭�\stackrel{鈫�}{{蟿}_{\mathrm{B}}}=\left(0\right)\mathrm{T}+\left(0\right){\mathrm{F}}_{\mathrm{s}}鈭�\left(\frac{\mathrm{L}}{2}\right)\mathrm{W}+(\mathrm{Ltan}鈦�胃)\mathrm{N}鈭�\mathrm{xW}\\ =0\\ \left(\frac{L}{2}+x\right)W=(L\tan 鈦�胃)N\\ \mathrm{W}=\frac{\mathrm{NLtan}鈦�胃}{\left(\frac{L}{2}+\mathrm{x}\right)}\end{array}$

Now, from equations (4) and (5),

$\begin{array}{r}\frac{F_s}{\left(\frac{3}{2}鈭�\frac{x}{L}\right)}=\frac{NL\tan 鈦�胃}{\left(\frac{L}{2}+x\right)}\\ F_s=\frac{NL\tan 鈦�胃\left(\frac{3}{2}鈭�\frac{x}{L}\right)}{\left(\frac{L}{2}+x\right)}\end{array}$

We know that ${\mathrm{F}}_{\mathrm{s}}鈮�{\mathrm{渭}}_{\mathrm{s}}\mathrm{N}$.

So, the above equation becomes

$\begin{array}{r}\frac{NL\tan 鈦�胃\left(\frac{3}{2}鈭�\frac{x}{L}\right)}{\left(\frac{L}{2}+x\right)}鈮�{渭}_{s}N\\ \frac{3}{2}L\tan 鈦�胃鈭�x\tan 鈦�胃鈮�\frac{1}{2}{渭}_{s}L+{渭}_{s}x\\ \frac{3}{2}L\tan 鈦�胃鈭�\frac{1}{2}{渭}_{s}L鈮�x\tan 鈦�胃+{渭}_{s}x\\ \frac{L}{2}\left(3\tan 鈦�胃鈭�{渭}_s\right)鈮�x\left(\tan 鈦�胃+{渭}_s\right)\\ x鈮�\frac{1.00m}{2}\left(\frac{3\tan 鈦�胃鈭�{渭}_s}{\tan 鈦�胃+{渭}_s}\right)\end{array}$

Now, we have $胃=15掳,{渭}_s=0.40$,

Substitute these values in the above equation,

$\begin{array}{r}x鈮�\frac{1.00m}{2}\left(\frac{3\tan 鈦�{15}^{鈭�}鈭�0.40}{\tan 鈦�{15}^{鈭�}+0.40}\right)\\ x鈮�0.3023m\end{array}$

Hence, the minimum value of x for which the stick will remain in equilibrium is 0.3023 m

Here we have,$\mathrm{胃}=15掳,\mathrm{x}=0.10\mathrm{m}\mathrm{and}\mathrm{L}=1.00\mathrm{m}$, and

Now, from step (4), we have

$\begin{array}{r}\frac{NI\tan 鈦�胃\left(\frac{3}{2}鈭�\frac{x}{L}\right)}{\left(\frac{L}{2}+x\right)}鈮�{渭}_{s}N\\ {渭}_s鈮�\frac{L\tan 鈦�胃\left(\frac{3}{2}鈭�\frac{x}{L}\right)}{\left(\frac{L}{2}+x\right)}\\ {渭}_s鈮�\frac{\left(1.00m\right)\tan 鈦�{15}^{鈭�}\left(\frac{3}{2}鈭�\frac{0.10m}{1.00m}\right)}{\left(\frac{1.00m}{2}+0.10m\right)}\\ {渭}_s鈮�0.6252\end{array}$

Hence, the coefficient of static friction be ${渭}_s鈮�0.6252$.