A √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 E f and F w are the reaction forces of the floor and the wall, then ratio of F_w/F_f will be :
(Use g = 10 m/s².)

Solution

Physics sir · 4 steps

Take torques about the foot of the ladder: the floor forces drop out and F_w falls straight out.

1. Geometry. The ladder, wall and floor form a right triangle. Height of the top is √(34 - 9) = 5 m. The weight mg = 100 N acts at the midpoint, which is 1.5 m from the wall's foot horizontally, so 1.5 m from the ladder's foot too.
2. Forces. The wall is frictionless, so F_w is horizontal. The floor gives a normal force N upward and a friction f toward the wall. Balancing forces gives N = 100 N and f = F_w.

   F_w × 5 = 100 × 1.5 ⇒ F_w = 30 N

3. Total floor reaction. The floor reaction F_f is the resultant of N and f: F_f = √(100² + 30²) = √10900 = 10√109 N.
4. So. F_w/F_f = 30/10√109 = 3/√109, which is option (c).

The idea to keep. For a ladder in equilibrium, take torques about the point with the most unknown forces. And remember the floor's reaction is normal plus friction combined, not the normal force alone.

Answer from the NTA final answer key. Solution by PrepWiser.