Question

A ladder of length \(3\,m\) and mass \(20\,kg\) leans on a frictionless wall with its feet at rest on the floor \(1\,m\) away from the wall. The reaction force of the wall on the ladder is nearly:

• A. \(34.6\,N\)
• B. \(98\,N\)
• C. \(196\,N\)
• D. \(28.3\,N\)

Hint:

For ladder equilibrium problems, taking moments about the foot of the ladder eliminates unknown floor reactions immediately.

Answer 1

The Correct Option is A

Concept: For a ladder in equilibrium: \[ \sum F_x=0 \] \[ \sum F_y=0 \] \[ \sum \tau=0 \] The wall is frictionless, therefore it exerts only a horizontal reaction.

Step 1: Find the height of the top end. Using Pythagoras theorem: \[ h=\sqrt{3^2-1^2} \] \[ h=\sqrt8 \] \[ h=2.828\,m \]

Step 2: Locate the centre of gravity. The ladder is uniform. Hence its weight acts at the midpoint. Horizontal distance of midpoint from the foot: \[ \frac12=0.5\,m \] Weight: \[ W=20\times9.8 \] \[ W=196\,N \]

Step 3: Apply torque balance about the foot. Clockwise moment due to weight: \[ 196\times0.5 \] Counterclockwise moment due to wall reaction \(R\): \[ R(2.828) \] Equilibrium: \[ R(2.828)=196(0.5) \] \[ R=\frac{98}{2.828} \] \[ R\approx34.6\,N \]

Step 4: Final answer. \[ \boxed{34.6\,N} \]