To determine the tension required to keep the crate moving at a constant velocity, we need to consider the forces acting on the crate. In this case, we have the weight of the crate and the force of friction opposing its motion. Let's break down the forces and calculate the tension.
Weight (W): The weight of the crate is given as 500N. The weight acts vertically downward, so we can represent it as follows: W = 500N (acting vertically downward)
Force of Friction (Ff): The force of friction opposes the motion of the crate and depends on the coefficient of dynamic friction and the normal force. The normal force is the force exerted by the surface perpendicular to the crate's weight.
The normal force (N) can be calculated as the vertical component of the crate's weight: N = W * cos(30°)
The force of friction (Ff) can then be calculated as: Ff = coefficient of dynamic friction * N
Given the coefficient of dynamic friction is 0.40, we can substitute the values and calculate Ff: Ff = 0.40 * (500N * cos(30°))
- Tension (T): The tension in the rope provides the horizontal force required to overcome friction and keep the crate moving at a constant velocity. It can be calculated as: T = Ff + W * sin(30°)
Substituting the known values, we have: T = (0.40 * (500N * cos(30°))) + (500N * sin(30°))
Now, let's calculate the values and find the tension required to keep the crate moving:
N = 500N * cos(30°) ≈ 433.01N Ff = 0.40 * 433.01N ≈ 173.20N T = 173.20N + (500N * sin(30°)) ≈ 432.60N
Therefore, to keep the crate moving at a constant velocity, a tension of approximately 432.60N is required when the rope is pulled at an angle of 30° above the horizontal.