In the adjacent figure is a cube having a smooth grove at an inclinati...
Understanding the Problem
The scenario involves a cube with a smooth groove inclined at 30 degrees, allowing a cylinder to slide without friction. The cube accelerates horizontally, and we need to determine the normal reaction force and the cylinder's acceleration in the ground frame.
1. Normal Reaction Force
- The cylinder experiences gravitational force (mg) acting vertically downward.
- The components of the gravitational force along the groove:
- Perpendicular to the groove: mg * cos(30)
- Along the groove: mg * sin(30)
- The normal reaction (N) must balance the perpendicular component of the gravitational force:
- N = mg * cos(30)
Here, m = 2 kg, g = 9.81 m/s², and cos(30) = √3/2. Thus, the normal reaction can be calculated as follows:
- N = 2 kg * 9.81 m/s² * (√3/2) ≈ 16.97 N.
2. Acceleration of the Cylinder in Ground Frame
- Since the cube accelerates horizontally with acceleration a0, the cylinder remains in equilibrium along the horizontal direction due to the absence of horizontal force acting on it.
- The cylinder will have an acceleration component down the groove due to gravity:
- a_cylinder = g * sin(30)
Given that sin(30) = 1/2, we find:
- a_cylinder = 9.81 m/s² * 1/2 = 4.905 m/s².
- This acceleration is along the incline of the groove.
Conclusion
- Normal Reaction (N) = 16.97 N.
- Acceleration of the Cylinder (in ground frame) = 4.905 m/s² down the groove.
This analysis reveals the forces at play and their effects on the cylinder within the accelerating cube.