Minimum effort to just move a block up a rough inclined plane (force applied parallel to plane) A block of weight W rests on a plane inclined at angle θ to the horizontal. A force P is applied parallel to the plane and just sufficient to move the block up the plane. If the angle of friction is φ (with μ = tan φ), what is the minimum P?

Introduction / Context:
For impending upward motion on a rough incline, friction acts down the plane and adds to the component of weight along the plane. This classic statics question tests correct identification of directions and use of friction-limited equilibrium at the verge of motion.

Given Data / Assumptions:

• Weight W on a plane with inclination θ.
• Coefficient of friction μ, or friction angle φ with μ = tanφ.
• Force P applied parallel to the plane to just start upward motion (impending slip up the plane).
• Rigid block, Coulomb friction model.

Concept / Approach:

Resolve forces along and normal to the plane. At impending upward motion, friction direction is down the plane with magnitude F = μN. Use equilibrium of forces along the plane to find the minimum P that balances the downhill components (weight component plus friction).

Step-by-Step Solution:

Verification / Alternative check:

Check limiting cases: μ = 0 ⇒ P = W sinθ (smooth plane). For steep θ approaching 90°, expression approaches W, consistent with vertical lift.

Why Other Options Are Wrong:

(b) uses the wrong sign for friction; (c) and (d) mix sine/cosine placements; (e) does not match the correct parallel-to-plane loading form for impending upward motion.

Common Pitfalls:

Reversing friction direction; using N = W instead of N = W cosθ; forgetting friction acts at its limiting value only at impending motion.

Final Answer:

P = W(sinθ + μ cosθ)