The Question is from neet-physics laws-of-motion , 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)

NEET Physics Laws of Motion MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company