3

A ladder AB of length 2a and weight w is inclined to a smooth horizontal ground(A) at an angle $\theta$ and rests against a smooth vertical wall(B). The centre of gravity G of the ladder is $\frac{3}{8}$ of the way up . The middle of the ladder, O,is tied to a point in the vertical wall by a horizontal rope of length l. The reactions at A and B are R and S respectively.
Show that tension in the rope is $\frac{3wl}{4\sqrt{a^2-l^2}}$.

I have obtained $\cos(\theta)$=$\frac{l}{a}$ ; $\sin(\theta)$= $\frac{\sqrt{a^2-l^2}}{a}$ ; $\tan(\theta)$ = $\frac{\sqrt{a^2-l^2}}{l}$

JOEF
  • 269

1 Answers1

1

HINT:

You do not need any Lami's theorem, nor any m-n theorem of trigonometry. It's just $$F_{net} = 0$$

and

$$\tau_{net} = 0$$

Since the rod is in equilibrium, the forces and torques cancel out. The horizontal forces are $T$ and $R$ (anti parallel) and the vertical ones are $W$ and $S$. Now, using the first equation you get: $$ R = T$$ and $$ W = S $$

The second equation over here isn't of any consequence, so forget it.

Now, balance the torques. Since we are only interested in $T$, we'd be making our jobs easy by eliminating $S$ - simply consider the torque about point $B$. We are justified in doing so as our rod is in equilibrium.

Balancing the torques gives you the answer.

MORE HINT: (I am basically throwing off the answer with this).

The anti-clockwise torques are:

  1. $\displaystyle (W\cos\theta)\cdot\Big(\frac3 4a\Big)$
  2. $\displaystyle (T\cos(\pi/2 -\theta))\cdot (a)$

The only clock-wise torque is:

  1. $\displaystyle (R\cos(\pi/2 - \theta))\cdot(2a)$
Parth Thakkar
  • 4,452
  • 3
  • 33
  • 50