Question

If the block is kept on an inclined surface as shown, the condition that the block does not slide up the plane irrespective of the value of F, is

• A. $\tan(\theta+\alpha) \geq \mu$
• B. $\tan(\theta+\alpha) \geq \frac{1}{\mu}$
• C. $\tan(\theta+\alpha) \leq \mu$
• D. $\tan\theta \geq \frac{1}{\mu}$

Answer 1

The problem is ill-posed, the correct answer should be $\tan \theta \leq -1/\mu$. None of the options match this condition.

Answer 2

Let's analyze the forces acting on the block. The block is on an inclined plane with angle of inclination $\alpha$. A force $F$ is applied to the block at an angle $\theta$ with the inclined plane. The coefficient of static friction between the block and the plane is $\mu$.

We are looking for the condition that the block does not slide up the plane irrespective of the value of F. This means that even if the applied force $F$ is very large and directed to push the block up, the block should not move up.

Let $M$ be the mass of the block.

The forces acting on the block are:

1. Weight (Mg): Acts vertically downwards.

   • Component perpendicular to the plane: $Mg \cos \alpha$ (acting into the plane)
   • Component parallel to the plane: $Mg \sin \alpha$ (acting down the plane)

2. Normal force (N): Acts perpendicular to the plane, outwards from the plane.

3. Applied Force (F): Acts at an angle $\theta$ with the inclined plane.

   • Component perpendicular to the plane: $F \sin \theta$ (acting outwards from the plane)
   • Component parallel to the plane: $F \cos \theta$ (acting up the plane)

4. Static Friction force ($f_s$): Acts parallel to the plane, opposing the tendency of motion.

For the block to be on the verge of sliding up the plane, the static friction force $f_s$ acts down the plane. The maximum static friction force is $f_{s,max} = \mu N$.

Equilibrium perpendicular to the plane:

The sum of forces perpendicular to the plane is zero:

$N + F \sin \theta - Mg \cos \alpha = 0$

$N = Mg \cos \alpha - F \sin \theta$

For the block to remain in contact with the surface, $N \geq 0$. This implies $Mg \cos \alpha \geq F \sin \theta$.

Equilibrium parallel to the plane (impending motion up the plane):

The sum of forces parallel to the plane is zero:

$F \cos \theta - Mg \sin \alpha - f_s = 0$

For the block to be on the verge of sliding up, $f_s = \mu N$.

So, $F \cos \theta = Mg \sin \alpha + \mu N$

Substitute the expression for $N$:

$F \cos \theta = Mg \sin \alpha + \mu (Mg \cos \alpha - F \sin \theta)$

$F \cos \theta = Mg \sin \alpha + \mu Mg \cos \alpha - \mu F \sin \theta$

Rearrange the terms to group $F$:

$F \cos \theta + \mu F \sin \theta = Mg \sin \alpha + \mu Mg \cos \alpha$

$F (\cos \theta + \mu \sin \theta) = Mg (\sin \alpha + \mu \cos \alpha)$

This equation gives the value of $F$ required to initiate upward motion. For the block not to slide up, the applied force $F$ must be less than or equal to this value.

So, $F (\cos \theta + \mu \sin \theta) \leq Mg (\sin \alpha + \mu \cos \alpha)$.

The condition given is that the block does not slide up irrespective of the value of F. This means that no matter how large $F$ is, the block should not slide up.

This can only happen if the term multiplying $F$ on the left side, $(\cos \theta + \mu \sin \theta)$, is non-positive.

If $(\cos \theta + \mu \sin \theta) > 0$, then as $F$ increases, the left side $F (\cos \theta + \mu \sin \theta)$ will eventually exceed the right side $Mg (\sin \alpha + \mu \cos \alpha)$, causing the block to slide up.

Therefore, for the block never to slide up, we must have:

$\cos \theta + \mu \sin \theta \leq 0$

Let's use the angle of friction, $\phi$, defined by $\tan \phi = \mu$. Since $\mu \ge 0$, $\phi$ is in the range $[0, \pi/2]$.

Substitute $\mu = \tan \phi = \frac{\sin \phi}{\cos \phi}$:

$\cos \theta + \frac{\sin \phi}{\cos \phi} \sin \theta \leq 0$

Multiply by $\cos \phi$ (since $\phi \in [0, \pi/2]$, $\cos \phi \geq 0$, so the inequality sign does not change):

$\cos \theta \cos \phi + \sin \phi \sin \theta \leq 0$

Using the trigonometric identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$:

$\cos (\theta - \phi) \leq 0$

For $\cos x \leq 0$, $x$ must be in the range $[\pi/2, 3\pi/2]$ (or $90^\circ$ to $270^\circ$).

So, $\theta - \phi \geq \pi/2$ (considering the smallest positive angle).

$\theta \geq \pi/2 + \phi$

This is the condition. Since $\phi$ is the angle of friction ($\tan \phi = \mu$), and $\mu$ is positive, $\phi$ is positive. So $\theta$ must be greater than $90^\circ$.

This leads to $\tan \theta \leq \tan(\frac{\pi}{2} + \phi) = -\cot \phi = -\frac{1}{\mu}$.

Therefore, $\tan \theta \leq -\frac{1}{\mu}$.

None of the options match this condition.