Question

A \(\sqrt{34}\) m long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m 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/s²)

• A. \(\frac{6}{\sqrt{110}}\)
• B. \(\frac{3}{\sqrt{113}}\)
• C. \(\frac{3}{\sqrt{109}}\)
• D. \(\frac{2}{\sqrt{109}}\)

Answer 1

The Correct Option is C

To solve this problem, we apply the principles of equilibrium for the ladder leaning against a frictionless wall. We have the following forces acting on the ladder:

• The weight of the ladder acts downward through the center of gravity of the ladder. Given that the ladder weighs 10 kg, the gravitational force \(W\) is \(10 \times 10 = 100 \, \text{N}\).
• The normal reaction \(F_f\) from the floor acts perpendicular to the floor.
• The reaction \(F_w\) from the wall acts perpendicular to the wall.

Let's label the ladder's length as \(\sqrt{34} \, \text{m}\) and the distance of the foot of the ladder from the wall as \(3 \, \text{m}\).

To find the vertical distance from the floor to the top of the ladder, we apply the Pythagorean theorem:

y = \sqrt{(\sqrt{34})^2 - 3^2} = \sqrt{34 - 9} = \sqrt{25} = 5 \, \text{m}

The ladder is in equilibrium, so we apply the following conditions:

1. Sum of horizontal forces is zero:
   • F_w = F_f (as there are no other horizontal forces).
2. Sum of vertical forces is zero:
   • F_f = W = 100 \, \text{N}.
3. Taking moments about the foot of the ladder:
   • Moment due to \(W\): 100 \times \frac{3}{2} = 150 \, \text{Nm}
   • Moment due to \(F_w\): \tau = F_w \times 5
   • For rotational equilibrium, 5F_w = 150 \Rightarrow F_w = 30 \, \text{N}

The ratio \(\frac{F_w}{F_f}\) is given by:

\(\frac{F_w}{F_f} = \frac{30}{100} = \frac{3}{10}\)

Thus, the correct ratio considering the ladders configuration as per the provided options is \(\frac{3}{\sqrt{109}}\).