The Question is from neet-physics laws-of-motion pulling-or-pushing-a-body, and it was created by the author ABDUL JILANI BASHA

Question

A block of mass $m$ is placed on a rough inclined plane with an angle of inclination $	heta$. A force $F$ is applied parallel to the incline, just sufficient to prevent the block from sliding down. If the coefficient of static friction is $\mu_s$, what is the minimum value of $F$?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option A
## CONCEPT FRAMEWORK
**Primary Concept**: Static Equilibrium and Forces on an Inclined Plane
**Related Topics**: Friction, Normal Force, Free Body Diagrams
**Prerequisites**: Understanding of vector components, Newton's First Law, static friction.
## SOLUTION PATHWAY
**Given Information**:
    * Mass of the block: $m$
    * Angle of inclination: $	heta$
    * Coefficient of static friction: $\mu_s$
    * Force $F$ applied parallel to the incline, preventing sliding down
**Method & Steps**:
    1. **Draw a Free Body Diagram (FBD):** Identify all forces acting on the block. These include:
        * Weight ($mg$) acting vertically downwards.
        * Normal force ($N$) acting perpendicular to the inclined plane.
        * Static frictional force ($f_s$) acting up the incline (since the block is on the verge of sliding down).
        * Applied force ($F$) acting up the incline.
        Reasoning: Identifying all forces is crucial for applying Newton's laws.
    2. **Resolve forces into components parallel and perpendicular to the incline:**
         * The weight component parallel to the incline is $mg\sin	heta$, acting downwards.
         * The weight component perpendicular to the incline is $mg\cos	heta$, acting into the plane.
    3. **Apply Newton's First Law in the perpendicular direction:**
        *  $N - mg\cos	heta = 0$  
        * Thus, $N = mg\cos	heta$
        Reasoning: In equilibrium, the net force in the perpendicular direction is zero.
        Formula/Rule:  $N = mg\cos	heta$
    4. **Apply Newton's First Law in the parallel direction:** Since the block is just about to slide down, the static friction is at its maximum value: $f_s = \mu_s N = \mu_s mg\cos	heta$
        * $F + f_s - mg\sin	heta = 0$
        * $F + \mu_s mg\cos	heta - mg\sin	heta = 0$
        Reasoning: In equilibrium, the net force in the parallel direction is also zero.
         Formula/Rule:  $f_s = \mu_s N$ 
    5. **Solve for F:**
        * $F = mg\sin	heta - \mu_s mg\cos	heta$
        * $F = mg(\sin	heta - \mu_s \cos	heta)$
        Result: The minimum value of $F$ is $mg(\sin	heta - \mu_s \cos	heta)$.
        Verification: The result makes sense physically. The applied force must overcome the component of gravity pulling the block down the incline and the static friction force.
## OPTION ANALYSIS
[A] ✓: The minimum force required to prevent sliding down is $mg(\sin 	heta - \mu_s \cos 	heta)$, which matches our derived result.
[B] :x: This option represents the force required to make the block move *up* the incline, not prevent it from sliding down.
[C] :x: This is only the frictional force and doesn't account for the component of weight down the incline.
[D] :x: This option represents only the component of the weight along the incline. It does not consider the frictional force.
## LEARNING AIDS
**Common Mistakes**:
    1. **Incorrect direction of frictional force:**  The frictional force opposes the impending motion. In this case it is acting *up* the incline.
        How to avoid: Carefully consider the direction the block would slide if there were no applied force.
    2. **Incorrectly resolving weight:**  Ensure to resolve the weight correctly into components parallel and perpendicular to the incline.
        How to avoid: Use trigonometric functions correctly based on the angle of inclination.
**Quick Tips**:
    1. **Draw a clear FBD:**  Always start with a free body diagram to visualize all forces involved.
    2. **Static friction:** Remember that static friction is a force that opposes impending motion, and its maximum value is $\mu_s N$.

ABDUL JILANI BASHA · Answer

The correct Option is "$mg(sin 	heta - \mu_s cos 	heta)$"

ABDUL JILANI BASHA · Answer

$mg(sin 	heta + \mu_s cos 	heta)$

ABDUL JILANI BASHA · Answer

$\mu_s mg cos 	heta$

ABDUL JILANI BASHA · Answer

$mg sin 	heta$

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