To determine the maximum angle $\theta_{\text{max}}$ at which the block will start sliding, we need to consider the forces acting on the block and how they relate to the frictional force.

Forces Analysis

1. Gravitational Force ($mg$): Acts vertically downward, where $m = 2.07 \, \text{kg}$ and $g$ is the acceleration due to gravity.
2. Normal Force ($N$): Perpendicular to the surface of the inclined plane.
3. Frictional Force ($f_s$): Acts parallel to the surface, opposing motion, with a maximum value of $f_s = \mu_s N$, where $\mu_s = 0.79$.
4. Component of Gravity Along the Plane ($mg \sin \theta$): Acts down the slope, causing the block to slide.
5. Component of Gravity Perpendicular to the Plane ($mg \cos \theta$): Balances the normal force, $N$.

Conditions for Maximum Angle

The block will start moving when the component of gravity parallel to the slope ($mg \sin \theta$) equals the maximum static frictional force ($f_s$):

$mg \sin \theta = \mu_s N$

Since $N = mg \cos \theta$, we can substitute it in:

$mg \sin \theta = \mu_s mg \cos \theta$

Simplifying by canceling $mg$ from both sides:

$\sin \theta = \mu_s \cos \theta$

Dividing both sides by $\cos \theta$:

$\tan \theta = \mu_s$

Thus, the maximum angle $\theta_{\text{max}}$ is:

$\theta_{\text{max}} = \tan^{-1}(\mu_s)$

Final Expression

$\theta_{\text{max}} = \tan^{-1}(0.79)$

If you need a numerical value for this angle, let me know!

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Would you like any further details or have specific questions on this?

Related Questions

1. How does increasing the coefficient of static friction affect $\theta_{\text{max}}$?
2. What is the expression for the normal force $N$ at an arbitrary angle $\theta$?
3. How would the solution change if the plank were moving, involving kinetic friction?
4. How does changing the mass $m$ of the block influence the outcome?
5. What are the equilibrium conditions for the block before it starts moving?

Tip: Understanding the breakdown of forces into parallel and perpendicular components helps solve many physics problems involving inclines.