A uniform ladder of weight 200 N and length 5.0 m leans against a smooth (frictionless) vertical wall. The foot of the ladder stands on rough ground at a ho...

Marking Scheme

• (a) [1 mark]: \(h = \sqrt{5.0^2 - 3.0^2} = \sqrt{25 - 9} = \sqrt{16} = 4.0\) m [1]
• (b) [2 marks]: Weight of ladder (downward) [1]; normal reaction from wall (horizontal), normal reaction from ground (vertical upward), friction from ground (horizontal) [1]
• (c) [3 marks]: Clockwise moment about base \(= \text{weight} \times \text{horizontal distance to centre} = 200 \times 1.5 = 300\) N m [1]; anticlockwise moment \(= R_{\text{wall}} \times 4.0\) [1]; \(R_{\text{wall}} = 300 / 4.0 = 75\) N [1]
• (d) [1 mark]: Friction \(= 75\) N (horizontal equilibrium: friction at base equals reaction from wall) [1]
• (e) [1 mark]: Without friction at the base, there is no horizontal force to balance the wall's reaction, so the ladder would slide and fall [1]

Explanation

Part (a): The ladder, wall and ground form a right-angled triangle with hypotenuse 5.0 m and base 3.0 m. By Pythagoras:

\(h = \sqrt{5.0^2 - 3.0^2} = \sqrt{16} = 4.0\) m

Part (b): Four forces act: (1) the ladder's weight acting downward at its centre (2.5 m along its length, which is 1.5 m horizontally from the foot), (2) the normal reaction from the smooth wall acting horizontally (perpendicular to the wall), (3) the normal reaction from the ground acting vertically upward, and (4) friction at the ground acting horizontally.

Part (c) - Moments about the base: Taking moments about the foot of the ladder eliminates the ground reaction and friction (both pass through the pivot). The weight acts at the ladder's midpoint. Its horizontal distance from the base is half of 3.0 m = 1.5 m:

Clockwise moment \(= 200 \times 1.5 = 300\) N m

The wall reaction acts at the top, 4.0 m above the base:

Anticlockwise moment \(= R_{\text{wall}} \times 4.0\)

For equilibrium: \(R_{\text{wall}} \times 4.0 = 300\), so \(R_{\text{wall}} = 75\) N.

Part (d): For horizontal equilibrium, the only two horizontal forces must balance: friction at the base = wall reaction = 75 N, directed away from the wall.

Part (e): If the ground were also smooth, there would be no friction. The wall pushes the ladder's top horizontally outward (75 N), but nothing would oppose this horizontally at the base. With an unbalanced horizontal force, the base slides outward and the ladder falls.