A box with a mass of 3.5 kg is resting on a horizontal frictionless incline plane which is inclined at 28 degrees. Another force is applied at an angle of 13 degrees above the direction parallel to the incline. What does the magnitude of this force need to be, in Newtons, such that the box begins to leave the incline?
F*cos13=mgSin28
To determine the magnitude of the force required for the box to begin to leave the incline, we need to consider the forces acting on the box.
First, we can break down the weight of the box into two perpendicular components:
- The component parallel to the incline (mg * sinθ), where θ is the angle of the incline.
- The component perpendicular to the incline (mg * cosθ).
Since the incline is frictionless, there is no friction force acting on the box. Therefore, the only other force acting along the incline is the force we need to find.
To analyze the forces acting along the incline, we need to resolve the applied force into two perpendicular components:
- The component parallel to the incline (F * cosα), where α is the angle above the direction parallel to the incline.
- The component perpendicular to the incline (F * sinα).
Now, considering the equilibrium condition, where the forces along the incline cancel each other out, we can set up the following equation:
F * cosα - mg * sinθ = 0
Rearranging the equation, we can solve for the force F:
F = (mg * sinθ) / cosα
Given that the mass of the box is 3.5 kg, the angle of the incline is 28 degrees, and the angle above the direction parallel to the incline is 13 degrees, we can substitute these values into the equation:
F = (3.5 kg * 9.8 m/s^2 * sin(28 degrees)) / cos(13 degrees)
Calculating the equation gives us:
F ≈ 21.86 Newtons.
Therefore, the magnitude of the force required for the box to begin to leave the incline is approximately 21.86 Newtons.