Question

A uniform ladder of weight W abd length 2a rests in limiting equilibrium with one end on a rough horizontal ground and the other end on a rough vertical wall. The coefficients of friction between the ladder and the ground and between the ladder and the wall are u and t respectively.
(a) Show that 5u + 6tu - 6 = 0
(b) Find the values of u and t, given that ut = 1/2
(c) Show by integration that tge centroid of a uniform semi-circular lamina of radius a from the centre is 4a/(3*pie).

Answer 1

Explanations & Calculations

a)

• For this consider the figure attached

• Considering the vertical equilibrium,

$\qquad\qquad \begin{aligned} \small R+tS&-W=\small0\\ \small R+tS &= \small W\cdots(1) \end{aligned}$

• Fromt he horizi=ontal equilibrium,

$\qquad\qquad \begin{aligned} \small S-uR&= \small 0\\ \small S&= \small uR\cdots(2) \end{aligned}$

• By (1) and (2),

$\qquad\qquad \begin{aligned} \small R+t(uR)&= \small W\\ \small R &=\small \frac{W}{(1+ut)} \end{aligned}$

• To get a relationship in term only of friction coefficients, another equation is needed. Next is to consider of some torque.
• Consider rightward torque about the point: ladder touches the wall,

$\qquad\qquad \begin{aligned} \small R(2a\cos\theta)-uR(2a\sin\theta)-W(a\cos\theta)&= \small 0\\ \small 2R(1-u\tan\theta)&= \small W\\ \small R&= \small \frac{W}{2(1-u\tan\theta)} \end{aligned}$

• Now removing R we get,

$\qquad\qquad \begin{aligned} \small \frac{1}{2(1-u\tan\theta)}&= \small \frac{1}{(1+ut)}\\ \small \tan\theta&= \small \frac{1-ut}{2u} \end{aligned}$

• Now, to get a relationship between u & t, $\small \tan\theta$ should be known as it depends on the system.
• Therefore, some data is missing from what you have provided.
• Anyway when the $\small \theta=\tan^{-1}(\frac{5}{12})$ , we get the needed result

$\qquad\qquad \begin{aligned} \small\frac{5}{12}&= \small \frac{1-ut}{2u}\\ \small 5u+6ut-6&= \small 0 \end{aligned}$

b)

• When ut=1/2,

$\qquad\qquad \begin{aligned} \small5u+6(\frac{1}{2})-6&= \small0\\ \small u&= \small \frac{3}{5}(<1:\checkmark)\\ \\ \small t (\frac{3}{5})&= \small\frac{1}{2}\\ \small t&= \small \frac{5}{6}(<1:\checkmark) \end{aligned}$

c)

• For this consider the figure attached.

• Consider an infinitesimal strip located at distance $\small x$ from the center.
• If the density of the lamina is $\rho$, then we can write the follows.

$\qquad\qquad \begin{aligned} \small \delta m&= \small \delta A\times \rho\\ \small\delta m&= \small 2r\sin\theta\times\delta x\times\rho\\ \end{aligned}$

• Mass of the lamina

$\qquad\qquad \begin{aligned} \small M&= \small \frac{\pi r^2}{2}\times \rho \end{aligned}$

• Since the lamina is homogeneous, the centroid should lye totally on the x axis. And if it lies at some $\small x'$ distance from the center, we can write as follows for the system & each infinitesimal strips regarding the torque

$\qquad\qquad \begin{aligned} \small Mg\times x'&= \small \int\delta mg\times x\\ \small \frac{\pi r^2\rho}{2}g\times x'&= \small\int 2r\sin \theta.\rho .g. x.dx\cdots (1)\\ \end{aligned}$

• To remove x & dx, we can write,

$\qquad\qquad \begin{aligned} \small x&= \small r\cos\theta\\ \small dx&= \small -r\sin\theta d\theta \end{aligned}$

• By (1),

$\qquad\qquad \begin{aligned} \small \end{aligned}$ $\qquad\qquad \begin{aligned} \small \frac{\pi r^2\rho g}{2}x'&= \small -\int 2r^3\sin^2\theta\cos\theta .\rho.g.d\theta\\ \small x'&= \small \frac{-4r}{\pi}\int\sin^2\theta\cos\theta.d\theta\\ \end{aligned}$

• Evaluating over $\large\frac{\pi}{2}$— 0,

$\qquad\qquad \begin{aligned} \small x'&= \small \frac{-4r}{\pi}\int_\frac{\pi}{2}^0\sin^2\theta .d(\sin\theta)\\ &= \small \frac{-4r}{\pi}\bigg[\frac{\sin^3\theta}{3}\big|_{\frac{\pi}{2}}^0\\ \small x'&= \small \frac{4r }{3\pi} \end{aligned}$