Limiting Equilibrium
A horizontal force F just prevents a 10 kg block from sliding down a rough 37° incline (μ_s = 0.5). Find minimum F (g = 10 m/s²).
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- limiting equilibrium
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
18.2 N
- Objective: Find F such that static friction (fs) up the plane is at its maximum to prevent sliding.
- Forces along Plane (+ is up):
- Upwards: Fcos37∘+fs
- Downwards: mgsin37∘
- Equilibrium: Fcos37∘+μsN=mgsin37∘
- Normal Force (N):
- N=mgcos37∘+Fsin37∘ (F's vertical component also pushes against the plane).
- Substitution:
- 0.8F+0.5(100×0.8+0.6F)=100×0.6
- 0.8F+40+0.3F=60
- 1.1F=20
- Solve: F=1.120≈18.18 N.
- Final Result: The minimum force required is approximately 18.2 N.
This hard difficulty physics question is from the chapter laws of motion, covering the topic of limiting equilibrium. It appeared in the 2025 exam.
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