A 485-kg crate hangs from the end of a 11.7 m long rope. You pull horizontally with a varying force to move it a distance d = 5.8 m to the right. What is the magnitude of the applied force F when the crate is at rest in its final position?
To find the magnitude of the applied force F when the crate is at rest in its final position, we can use the principles of equilibrium. When the crate is at rest, the net force acting on it is zero.
First, let's consider the forces acting on the crate:
1. Weight: The weight of the crate is given by the formula:
weight = mass * gravity
Since the mass is given as 485 kg and the acceleration due to gravity is approximately 9.8 m/s^2, the weight of the crate is:
weight = 485 kg * 9.8 m/s^2
2. Tension in the rope: The tension in the rope is the force that balances the weight of the crate. It acts in the vertical upward direction.
3. Applied force: This is the force you apply horizontally to move the crate.
Since the crate is at rest in its final position, the vertical forces must be balanced, which means the tension in the rope must equal the weight of the crate. Mathematically, this can be expressed as:
Tension = weight = mass * gravity
Now, let's calculate the magnitude of the tension:
Tension = 485 kg * 9.8 m/s^2
To find the magnitude of the applied force F, we need to consider horizontal forces. Since the crate is at rest, the horizontal forces must also be balanced. Mathematically, this can be expressed as:
Applied force = Friction force
Since the crate is moving horizontally, the friction force opposes the applied force. In this case, the friction force is given as zero, meaning there is no resistance to the crate's motion. Therefore, the magnitude of the applied force F is zero.
So, when the crate is at rest in its final position, the magnitude of the applied force F is zero.