A $$\sqrt{34}$$ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If $$F_f$$ and $$F_w$$ are the reaction forces of the floor and the wall, then ratio of $$\frac{F_w}{F_f}$$ will be:
(Use $$g = 10$$ m s$$^{-2}$$.)

$$Length=\sqrt{\ 34},base=3m⇒height=5m$$

$$Weight=10g=100Nactingatmidpoint$$

midpoint coordinates:
horizontal = 3/2
vertical = 5/2

Forces:

• at floor: $$F_f$$ (resultant, has horizontal + vertical)
• at wall: $$F_w$$ (horizontal only, since wall is frictionless)

Take moments about the bottom point:

Torque by weight (clockwise):
$$=100\times(horizontaldis\tan ceofCM)$$

$$=100\times(3/2)=150$$

Torque by wall reaction (anticlockwise):
$$=F_w\times height$$
$$=F_w\times5$$

Equilibrium:
$$F_w\times5=150$$
$$F_w=30N$$

Now vertical equilibrium:

Ff(vertical) = 100

Horizontal equilibrium:

$$F_f(horizontal)=F_w=30$$

So resultant floor reaction:

$$Ff=\sqrt{(100^2+30^2)\ }$$
$$=\sqrt{\left(10000+900)\right)}$$
= 10$$\sqrt{109}$$

Now ratio:

$$Fw/Ff=30/(10\sqrt{109}$$
$$=3/\sqrt{109}$$

Final answer:

$$3/\sqrt{109}$$