1. Forces Involved
* Gravity (Weight): The force of gravity acts straight down on the crate. We can break this force into two components:
* Force parallel to the incline (F_parallel): This component tries to pull the crate down the plane.
* Force perpendicular to the incline (F_perpendicular): This component presses the crate against the plane.
* Normal Force (F_normal): This is the force exerted by the plane on the crate, perpendicular to the surface. It balances the perpendicular component of gravity.
* Frictional Force (F_friction): This force opposes the motion of the crate and acts parallel to the plane, opposing F_parallel.
2. Calculations
* Force parallel to the incline (F_parallel):
F_parallel = mg * sin(theta)
where:
* m = mass of the crate
* g = acceleration due to gravity (approximately 9.8 m/s²)
* theta = angle of the incline (35 degrees)
* Normal force (F_normal):
F_normal = mg * cos(theta)
* Maximum Static Frictional Force (F_friction_max):
F_friction_max = μ_s * F_normal
where:
* μ_s = coefficient of static friction (0.65)
3. Comparison
* If F_parallel > F_friction_max: The crate will slide down the plane because the force pulling it down is greater than the maximum force that friction can provide to hold it in place.
* If F_parallel ≤ F_friction_max: The crate will remain at rest because the force of static friction is strong enough to counteract the component of gravity pulling it down the incline.
4. Putting it Together
Since we don't know the mass of the crate (m), we can't calculate the exact forces. However, we can determine the condition for the crate to slide:
* The crate will slide down the plane if: mg * sin(35°) > μ_s * mg * cos(35°)
* Simplifying the inequality: sin(35°) > μ_s * cos(35°)
* Substituting the given values: sin(35°) > 0.65 * cos(35°)
Calculation:
* sin(35°) ≈ 0.574
* 0.65 * cos(35°) ≈ 0.532
Conclusion:
Since sin(35°) is greater than 0.65 * cos(35°), the crate will slide down the plane.
