A conveyor belt is moving grain into a bin that is 1.90 m below the top of the conveyor belt. The grain does not slip on the conveyor belt that is inclined at 15.0° and they move at a constant speed of 6.00 m/s. In order for the conveyor belt to get the grain into the bin, what must the horizontal distance between the end of the conveyor belt and the bin be?
Tan15 = 1.90/d. d = ?.
To solve this problem, we can break it down into two components: the horizontal distance and the vertical distance.
First, let's consider the vertical distance. The grain is falling vertically from a height of 1.90 m. The acceleration due to gravity, g, is approximately 9.8 m/s². We can use the equation of motion to find the time it takes for the grain to fall:
Δy = 1/2 * g * t²
Rearranging the equation:
t² = 2 * Δy / g
Substituting the values:
t² = 2 * 1.90 m / 9.8 m/s²
t² = 0.38776 s²
t ≈ 0.62 s
Now, let's consider the horizontal distance. The conveyor belt and the grain are moving at a constant speed of 6.00 m/s. The time it takes for the grain to fall vertically should be the same amount of time it takes for the end of the conveyor belt to reach the bin horizontally. So, we can use the equation:
d = v * t
Substituting the values:
d = 6.00 m/s * 0.62 s
d ≈ 3.72 m
Therefore, the horizontal distance between the end of the conveyor belt and the bin should be approximately 3.72 meters.