91影视

Here we have given that, weight W = 480 N

Angle between rope and posy is$伪=36.9掳$

${渭}_s=0.30$

Height of the post is h

The force that can cause an object to rotate along an axis is measured as torque. Which is given by,

$\begin{array}{r}\stackrel{\mathbf{鈫�}}{\mathbf{蟿}}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{r}}\mathbf{脳}\stackrel{\mathbf{鈫�}}{\mathbf{F}}\\ \mathbf{蟿}\mathbf{=}\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{鈦�}\mathbf{胃}\end{array}$ (1)

Where$\stackrel{\mathbf{鈫�}}{\mathbf{r}}$is torque,$\stackrel{\mathbf{鈫�}}{\mathbf{r}}$ is radius vector,$\stackrel{\mathbf{鈫�}}{\mathbf{F}}$ is force applied on object and is angle between and.

(a)

Here, the object is in static equilibrium. So, by second condition of equilibrium,

$\underset{i}{\overset{n}{鈭�}}鈥�{蟿}_i=0,n=1,2,3,鈥�$ (2)

Free body diagram for given problem is,

From equation (2) and free-body diagram we can write,

For upper point,

$\begin{array}{r}\stackrel{鈫�}{{蟿}_f}+\stackrel{鈫�}{{蟿}_F}=0\\ fh\sin 鈦�\frac{蟺}{2}+Fa\sin 鈦�\left(\frac{鈭�蟺}{2}\right)=0\\ fh鈭�Fa=0\\ f=\frac{Fa}{h}\end{array}$ (3)

Where f is friction force.

Now, for lower point,

$\begin{array}{r}\stackrel{鈫�}{{蟿}_T}+\stackrel{鈫�}{{蟿}_F}=0\\ Th\sin 鈦�(蟺鈭�伪)+Fb\sin 鈦�\left(\frac{蟺}{2}\right)=0\\ 鈭�Th\sin 鈦�伪+Fb=0\\ T=\frac{Fb}{h\sin 鈦�伪}\end{array}$

Now, equalizing the forces acting along the vertical axis will give the expression for the normal reaction of the surface,

$W+T\cos 伪=N\left(5\right)$

Now, for friction force we have,

$f鈮�{渭}_{s}N$ (6)

Now, from equation (5) and (6) we have,

$f鈮�{渭}_s\left(W+T\cos 伪\right)$

Now, put the values of equation (3) and (4) in above equation we get,

$\begin{array}{r}\frac{aF}{h}鈮�{渭}_s\left(W+\left(\frac{Fb}{h\sin 鈦�伪}\right)\cos 鈦�伪\right)\\ \frac{aF}{h}鈭�{渭}_s\left(\frac{Fb}{h\sin 鈦�伪}\right)\cos 鈦�伪鈮�{渭}_{s}W\\ F\left(\frac{a}{h}鈭�{渭}_s\left(\frac{b\cos 鈦�伪}{h\sin 鈦�伪}\right)\right)鈮�{渭}_{s}W\\ F\left(\frac{a}{h}鈭�{渭}_s\left(\frac{b}{h\tan 鈦�伪}\right)\right)鈮�{渭}_{s}W\\ F鈮�\frac{{渭}_{s}W}{\left(\frac{a}{h}鈭�{渭}_s\left(\frac{b}{h\tan 鈦�伪}\right)\right)}\end{array}$ (7)

Now, here we have to suppose that force acting on the middle of the body.

So,$\mathrm{a}=\frac{\mathrm{h}}{2}\mathrm{and}\mathrm{b}=\frac{\mathrm{h}}{2}$

So, from equation (7), we have,

$\begin{array}{r}\left.F鈮�\frac{{渭}_{s}W}{\left(\frac{h/2}{h}鈭�{渭}_s\left(\frac{h/2}{h\tan 鈦�伪}\right)\right.}\right)\\ F鈮�\frac{{渭}_{s}W}{\left(\frac{1}{2}鈭�{渭}_s\left(\frac{1}{2\tan 鈦�伪}\right)\right)}\\ F鈮�\frac{\left(0.30\right)\left(400N\right)}{\left(\frac{1}{2}鈭�\left(0.30\right)\left(\frac{1}{2\tan 鈦�{36.9}^{鈭�}}\right)\right)}\\ F鈮�399.7\mathrm{N}\end{array}$

Hence, the largest value it can have without causing the post to slip if the force is applied at the midpoint of the post is 399.7 N .

(b)

Here we have to suppose that, its point of application is$\frac{6}{10}$of the way from the ground to the top of the post.

So, we have,$a=\frac{4h}{10}andb=\frac{6h}{10}$

So, from equation (7), we have,

$\begin{array}{r}F鈮�\frac{{渭}_{s}W}{\left(\frac{4h/10}{h}鈭�{渭}_s\left(\frac{6h/10}{h\tan 鈦�伪}\right)\right)}\\ F鈮�\frac{{渭}_{s}W}{\left(\frac{4}{10}鈭�{渭}_s\left(\frac{6}{10\tan 鈦�伪}\right)\right)}\\ F鈮�\frac{\left(0.30\right)\left(400\mathrm{N}\right)}{\left(\frac{4}{10}鈭�\left(0.30\right)\left(\frac{6}{10\tan 鈦�{36.9}^{鈭�}}\right)\right)}\\ F鈮�748.77\mathrm{N}\end{array}$

Hence, the maximum force it can be without causing the post to slip if its point of application is $\frac{6}{10}$of the way from the ground to the top of the post is 748.77 N.

(c)

Here we have to suppose that the point of application of the force is too high.

So, suppose that the force is acting on top of the body.

So, we can write, a = h - b

So, from equation (7), we have,

$\begin{array}{r}F鈮�\frac{{渭}_{s}W}{\left(\frac{h鈭�b}{h}鈭�{渭}_s\left(\frac{b}{h\tan 鈦�伪}\right)\right)}\\ F鈮�\frac{{渭}_{s}W}{\left(1鈭�\frac{b}{h}鈭�{渭}_s\left(\frac{b}{h\tan 鈦�伪}\right)\right)}\\ F鈮�\frac{{渭}_{s}W}{\left(1鈭�\frac{b}{h}\left(1+{渭}_s\frac{1}{\tan 鈦�伪}\right)\right)}\end{array}$

Here if $1鈭�\frac{b}{h}\left(1+{渭}_s\frac{1}{\tan 鈦�伪}\right)$Then Force is very high.

Now, to determine critical point let,

$\begin{array}{r}1鈭�\frac{b}{h}\left(1+{渭}_s\frac{1}{\tan 鈦�伪}\right)=0\\ 1=\frac{b}{h}\left(1+{渭}_s\frac{1}{\tan 鈦�伪}\right)\\ \frac{b}{h}=\frac{1}{\left(1+{渭}_s\frac{1}{\tan 鈦�伪}\right)}\\ \frac{b}{h}=\frac{1}{\left(1+\left(0.30\right)\frac{1}{\tan 鈦�{36.9}^{鈭�}}\right)}\\ =0.71\end{array}$

Hence, It is shown that if the point of application of the force is too high, the post cannot be made to slip, no matter how great the force The critical height for the point of application is 0.71 .