To find the component of the force of gravity in the $y$-direction (perpendicular to the plank), we'll decompose the gravitational force into components along the rotated coordinate axes.

Definitions:

1. $m$: Mass of the block = 2.07 kg.
2. $\mu_s$: Coefficient of static friction = 0.59.
3. $\theta$: Angle of the plank with respect to the ground.
4. $g$: Acceleration due to gravity = 9.8 m/s$^2$.

Gravity Force:

The gravitational force acting on the block is $F_g = mg$. This force acts vertically downward.

Rotated Coordinate System:

• $x$-axis: Parallel to the plank.
• $y$-axis: Perpendicular to the plank.

Components of Gravity:

1. Along the $y$-axis (perpendicular to the plank): $F_{gy} = mg \cos \theta$

2. Along the $x$-axis (parallel to the plank): $F_{gx} = mg \sin \theta$

Explanation:

• The component $F_{gy}$ is responsible for the normal force exerted by the plank on the block.
• The component $F_{gx}$ is responsible for trying to move the block down the plank due to gravity.

In terms of given variables:

$F_{gy} = mg \cos \theta$

where:

• $m$ is the mass of the block (2.07 kg).
• $g$ is the acceleration due to gravity (9.8 m/s$^2$).
• $\theta$ is the angle between the plank and the horizontal.

Do you need further details or have any questions?

Here are 5 related questions to explore the concept further:

1. What is the normal force acting on the block if the angle $\theta$ is 30°?
2. How does the frictional force change as the angle $\theta$ increases?
3. At what angle $\theta$ does the block start to slide down the plank?
4. How would the static friction coefficient affect the critical angle at which the block begins to slide?
5. What is the component of the gravitational force along the $x$-axis at $\theta = 45^\circ$?

Tip: Always resolve forces into components along the axes that simplify your problem, especially when dealing with inclined planes or non-standard coordinate systems.