Friction on a ladder: rough ground and smooth wall A ladder rests on rough ground and leans against a perfectly smooth vertical wall. At the upper end (at the wall), the force of friction acts in which manner?

Introduction / Context:
Ladder friction problems test understanding of contact forces at rough and smooth interfaces. When one contact is smooth, it cannot transmit tangential force (i.e., no friction), which simplifies equilibrium equations and determines where friction must act.

Given Data / Assumptions:

• Ground is rough (friction possible at the foot).
• Wall is perfectly smooth (no friction at the top contact).
• Rigid ladder; static equilibrium; no impending motion unless specified.

Concept / Approach:

A smooth surface can only exert a normal reaction, perpendicular to itself. Therefore, at the smooth vertical wall, the contact force is purely horizontal (normal to the wall) and has no vertical component and no friction. The rough ground must provide the necessary frictional force to prevent slipping in the vertical direction and maintain equilibrium of the ladder’s weight and components of forces.

Step-by-Step Solution:

Verification / Alternative check:

Draw a free-body diagram. If any nonzero vertical friction existed at the wall, it would be a tangential component; smoothness forbids it. Hence friction at the top is zero.

Why Other Options Are Wrong:

(a) and (b) imply vertical friction at the wall, impossible for a smooth surface. (c) describes the normal reaction, not friction. (e) is not a valid wall-contact friction direction here.

Common Pitfalls:

Assuming some friction at the wall; mixing the roles of ground and wall contacts; forgetting to align reactions perpendicular to their surfaces.

Final Answer:

Zero at its upper end