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

Question

Two blocks of masses $m_1$ and $m_2$ are connected by a massless string over a frictionless pulley.  $m_1$ rests on a rough inclined plane with an angle of inclination $	heta$. The coefficient of static friction between the block and the plane is $\mu_s$.  If the system is in equilibrium, which of the following is the correct range for the ratio $\frac{m_2}{m_1}$?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option A
## CONCEPT FRAMEWORK
	**Primary Concept**: Static Equilibrium on an Inclined Plane with Friction
	**Related Topics**: Forces, Free Body Diagrams, Static Friction
	**Prerequisites**: Newton's Laws of Motion, Trigonometry
## SOLUTION PATHWAY
**Given Information**:
	* Two blocks with masses $m_1$ and $m_2$
	* $m_1$ on an inclined plane with angle $	heta$
	* Coefficient of static friction: $\mu_s$
	* System is in equilibrium
**Method & Steps**:
	1. Analyze forces on $m_1$:
		Reasoning: Draw a free body diagram for $m_1$. The forces are gravity ($m_1g$), normal force ($N$), tension ($T$), and static friction ($f_s$).
		Formula/Rule: $N = m_1g\cos	heta$, $f_s \le \mu_s N$,  $T = m_2g$
	2. Resolve forces along the incline:
		Working:  The component of gravity parallel to the incline is $m_1g\sin	heta$. For equilibrium, the forces along the incline must balance.  The tension in the string is equal to the weight of $m_2$, so $T = m_2g$. The friction force can act up or down the incline depending on whether $m_2$ is tending to pull $m_1$ up or down the incline.
		Key Point: Consider the two extreme cases: (1) $m_2$ is just large enough to overcome static friction and (2) $m_2$ is just small enough to allow static friction to prevent motion.
	3. Case 1: $m_2$ is large, friction acts down the incline:
		Working: $m_2g = m_1g\sin	heta + f_s$. Since $f_s \le \mu_s N = \mu_s m_1g\cos	heta$, the maximum value of $f_s$ is  $\mu_s m_1g\cos	heta$. Thus, $m_2g = m_1g\sin	heta + \mu_s m_1g\cos	heta$. Dividing by $m_1g$, we get $\frac{m_2}{m_1} = \sin	heta + \mu_s\cos	heta$.
	4. Case 2: $m_2$ is small, friction acts up the incline:
		Working: $m_2g + f_s = m_1g\sin	heta$.  Since $f_s \le \mu_s N = \mu_s m_1g\cos	heta$, the maximum value of $f_s$ is $\mu_s m_1g\cos	heta$. Thus,  $m_2g + \mu_s m_1g\cos	heta = m_1g\sin	heta$. Rearranging and dividing by $m_1g$, we get $\frac{m_2}{m_1} = \sin	heta - \mu_s\cos	heta$.
	5. Combine the results:
		Result: The range of  $\frac{m_2}{m_1}$ is $ \sin	heta - \mu_s\cos	heta \le \frac{m_2}{m_1} \le \sin	heta + \mu_s\cos	heta$
		Verification: This result incorporates both maximum and minimum values of $m_2$ for which the system is in equilibrium.
## OPTION ANALYSIS
[A] ✓: The correct range for the ratio $\frac{m_2}{m_1}$ is indeed $\sin	heta - \mu_s\cos	heta \le \frac{m_2}{m_1} \le \sin	heta + \mu_s\cos	heta$, as derived in the solution.
[B] :x: This is only true if $\mu_s = 0$ (frictionless surface) and the system is in equilibrium.  It doesn't account for the range due to static friction.
[C] :x: This only represents a specific case and not the range, it is related to friction but doesn't account for gravity.
[D] :x: This is not relevant to the force balance equations for the given problem.
## LEARNING AIDS
	**Common Mistakes**:
		1. **Forgetting the direction of friction**: Friction opposes the tendency of motion. It can act up or down the incline. Always consider both cases.
		2. **Using kinetic friction instead of static friction**: The problem states the system is in equilibrium, therefore static friction applies. Not kinetic.
	**Quick Tips**:
		1. **Free body diagram**: Always draw a free body diagram when dealing with forces. This will help visualize all forces acting on the object.
		2. **Equilibrium condition**: For static equilibrium, the net force in any direction must be zero.

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\sin	heta - \mu_s\cos	heta \le \frac{m_2}{m_1} \le \sin	heta + \mu_s\cos	heta$"

Mr. ANIL KUMAR.G · Answer

$\frac{m_2}{m_1} = \sin	heta$

Mr. ANIL KUMAR.G · Answer

$\frac{m_2}{m_1} = \mu_s\cos	heta$

Mr. ANIL KUMAR.G · Answer

$\frac{m_2}{m_1} = 	an	heta$

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