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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीReferring to the ladder setup shown in Figure, a ladder of length 3 m leans against a frictionless wall with foot at \( A \) and top at \( B \). The normal reaction of the wall is \( F_1 \) and the floor reaction components are normal force \( N \) and friction \( F \). For rotational equilibrium about point \( A \), the equation relating \( F_1 \) and the weight \( W \) of the ladder is:
Correct answer · Option B
\( 2\sqrt{2} F_1 - (1/2) W = 0 \)
Explanation
Taking moments about \( A \): the height \( BC = \sqrt{3^2 - 1^2} = 2\sqrt{2}\text{ m} \) gives torque \( F_1 \times 2\sqrt{2} \). The weight acts at the midpoint with lever arm \( 0.5\text{ m} \), giving torque \( - (1/2) W \). Thus \( 2\sqrt{2}F_1 - (1/2)W = 0 \).
