Ladder equilibrium conditions A heavy ladder rests on a horizontal floor and against a vertical wall. In which case can the ladder not be in equilibrium under its own weight and contact reactions?

Introduction / Context:
Determining whether a ladder can remain in static equilibrium requires understanding the roles of normal and frictional forces at the two contacts (floor and wall). If both contacts are smooth, the only available contact forces are normals, which may be insufficient to balance all components and moments created by the ladder’s weight.

Given Data / Assumptions:

• Weight W acts at the ladder’s center of gravity.
• Smooth contact → only normal reaction, no friction.
• Rough contact → friction available up to limiting value μN.
• Rigid ladder; planar statics (2D).

Concept / Approach:
For equilibrium in 2D, we need ∑F_x = 0, ∑F_y = 0, and ∑M = 0. With both contacts smooth, the wall provides a horizontal normal and the floor provides a vertical normal. There is no horizontal force at the floor and no vertical friction at the wall. These two normals cannot, in general, generate the required counter-moment to balance the turning effect of the weight unless the geometry is singular, so equilibrium cannot be sustained.

Step-by-Step Solution:

Verification / Alternative check:
Introduce friction at either contact and you can select friction directions/magnitudes to satisfy all three equilibrium equations. Remove friction at both and you lose one necessary degree of freedom.

Why Other Options Are Wrong:

• Floor smooth, wall rough: Wall friction can stabilize the system.
• Floor rough, wall smooth: Floor friction can stabilize the system.
• Both rough: Even more likely to achieve equilibrium.
• None of these: Incorrect since a specific non-equilibrium case exists.

Common Pitfalls:
Assuming “more smoothness” always helps; in ladder statics, at least one frictional contact is typically needed for equilibrium.

Final Answer:
Floor smooth and wall smooth