The Question is from neet-physics laws-of-motion laws-of-friction, and it was created by the author Mr. ANIL KUMAR.G

Question

A block of mass $m$ is placed on a rough inclined plane with an angle of inclination $	heta$. The coefficient of static friction between the block and the plane is $\mu_s$. A horizontal force $F$ is applied to the block.  For what range of values of $F$ will the block remain at rest?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Static equilibrium with friction
 **Related Topics**: Force resolution, friction limits, inclined plane mechanics
 **Prerequisites**: Newton’s laws, static friction, vector decomposition
## SOLUTION PATHWAY
**Given Information**:
 - Mass: $m$, angle: $	heta$, static friction: $\mu_s$
 - Horizontal force $F$ applied to the block
**Method & Steps**:
 1. **Resolve forces into components**:
  - Weight components: $mg\sin	heta$ (along incline), $mg\cos	heta$ (perpendicular)
  - Horizontal force $F$: $F\cos	heta$ (along incline), $F\sin	heta$ (perpendicular)
  - Normal force: $N = mg\cos	heta + F\sin	heta$
 2. **Maximum $F$ (prevents sliding up)**:
  - Friction acts downward: $f_{	ext{max}} = \mu_s N$
  - Equilibrium: $F\cos	heta = mg\sin	heta + \mu_s(mg\cos	heta + F\sin	heta)$
  - Solve: $F_{	ext{max}} = \frac{mg(\sin	heta + \mu_s\cos	heta)}{\cos	heta - \mu_s\sin	heta}$
 3. **Minimum $F$ (prevents sliding down)**:
  - Friction acts upward: $f_{	ext{max}} = \mu_s N$
  - Equilibrium: $F\cos	heta + \mu_s(mg\cos	heta + F\sin	heta) = mg\sin	heta$
  - Solve: $F_{	ext{min}} = \frac{mg(\sin	heta - \mu_s\cos	heta)}{\cos	heta + \mu_s\sin	heta}$
 4. **Range**: $F_{	ext{min}} \leq F \leq F_{	ext{max}}$
 ## OPTION ANALYSIS
[A] ✓: Matches derived range $F_{	ext{min}} \leq F \leq F_{	ext{max}}$ with static friction ($\mu_s$).
[B] :x: Uses kinetic friction ($\mu_k$), which is irrelevant for static equilibrium.
[C] :x: Only provides an upper bound, ignoring the lower limit.
[D] :x: Only provides a lower bound, ignoring the upper limit.
 ## LEARNING AIDS
 **Common Mistakes**:
  1. Using $\mu_k$ instead of $\mu_s$: Static friction applies here, not kinetic.
  2. Ignoring the direction of friction: Friction opposes impending motion, which changes direction based on $F$.
 **Quick Tips**:
  1. For maximum $F$, friction acts opposite to the applied force component.
  2. For minimum $F$, friction assists the applied force to prevent sliding.

Mr. ANIL KUMAR.G · Answer

The correct Option is "mg(sinθ - μs cosθ)/(cosθ + μs sinθ) ≤ F ≤ mg(sinθ + μs cosθ)/(cosθ - μs sinθ)"

Mr. ANIL KUMAR.G · Answer

mg(sinθ - μk cosθ)/(cosθ + μk sinθ) ≤ F ≤ mg(sinθ + μk cosθ)/(cosθ - μk sinθ)

Mr. ANIL KUMAR.G · Answer

F ≤ mg(sinθ + μs cosθ)/(cosθ - μs sinθ)

Mr. ANIL KUMAR.G · Answer

F ≥ mg(sinθ - μs cosθ)/(cosθ + μs sinθ)

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