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A mass [tex]$m$[/tex] is placed on an inclined plane. If the mass is in equilibrium, the maximum inclination of the plane with the horizontal would be (where [tex]\mu[/tex] is the coefficient of friction between the mass and surface):

A. [tex]\tan^{-1} \mu[/tex]
B. [tex]\tan^{-1}\left(\frac{\mu}{2}\right)[/tex]
C. [tex]\tan^{-1}\left(\frac{\mu}{m}\right)[/tex]
D. [tex]\cos^{-1} \mu[/tex]

Sagot :

GinnyAnsweridn
To determine the maximum inclination of the plane with the horizontal such that the mass [tex]\( m \)[/tex] remains in equilibrium, we'll need to analyze the forces acting on the mass and apply the principles of equilibrium.

### 1. Forces Acting on the Mass:
- Weight: The force due to gravity, [tex]\( \vec{W} = mg \)[/tex], acts vertically downward.
- Normal Force: The normal force [tex]\( \vec{N} \)[/tex] acts perpendicular to the surface of the inclined plane.
- Frictional Force: The frictional force [tex]\( \vec{f} \)[/tex] acts parallel to the surface of the inclined plane and opposes the motion of the mass. The maximum static frictional force is given by [tex]\( f_{\text{max}} = \mu N \)[/tex], where [tex]\( \mu \)[/tex] is the coefficient of friction.

### 2. Components of the Gravitational Force:
Consider the inclination angle [tex]\( \theta \)[/tex] with the horizontal. The weight [tex]\( mg \)[/tex] can be resolved into two components:
- Parallel to the inclined plane: [tex]\( mg \sin \theta \)[/tex]
- Perpendicular to the inclined plane: [tex]\( mg \cos \theta \)[/tex]

### 3. Conditions for Equilibrium:
For the mass to be in equilibrium on the inclined plane:
- The component of the weight parallel to the plane [tex]\( mg \sin \theta \)[/tex] must be balanced by the maximum static frictional force [tex]\( f_{\text{max}} = \mu N \)[/tex].
- The normal force [tex]\( N \)[/tex] balances the perpendicular component of the weight, [tex]\( N = mg \cos \theta \)[/tex].

### 4. Setting Up the Equations:
From the equilibrium condition parallel to the inclined plane:
[tex]\[ mg \sin \theta = \mu N \][/tex]

Substitute [tex]\( N = mg \cos \theta \)[/tex]:
[tex]\[ mg \sin \theta = \mu (mg \cos \theta) \][/tex]

### 5. Simplifying the Equation:
Divide both sides by [tex]\( mg \)[/tex]:
[tex]\[ \sin \theta = \mu \cos \theta \][/tex]

[tex]\[ \tan \theta = \mu \][/tex]

### 6. Solving for [tex]\( \theta \)[/tex]:
To find the maximum inclination [tex]\( \theta \)[/tex] where the mass is in equilibrium, we solve for [tex]\( \theta \)[/tex]:

[tex]\[ \theta = \tan^{-1} \mu \][/tex]

Thus, the correct answer is:
[tex]\[ \theta = \tan^{-1} \mu \][/tex]

Therefore, the correct choice is:
[tex]\[ \boxed{\tan^{-1} \mu} \][/tex]
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