A 130kg mail bag hangs by a vertical rope 3.9m long. A postal worker then displaces the bag to a position 2.2m sideways from its original position, always keeping the rope taut.
Incomplete.
To determine the tension in the rope when the mail bag is displaced, we can consider the forces acting on the bag.
First, let's visualize the scenario:
- The mail bag is hanging vertically and is initially at rest.
- The vertical rope holding the bag is 3.9m long.
- The postal worker displaces the bag sideways, 2.2m from its original position, while keeping the rope taut.
Now, let's break down the forces acting on the bag:
1. Weight force (Fg):
The weight force acts vertically downward and is determined by the mass of the mail bag (130kg) and the acceleration due to gravity (9.8 m/s²):
Fg = m * g
Fg = 130 kg * 9.8 m/s²
Fg = 1274 N
2. Tension force in the rope (T):
The tension force in the rope holds the mail bag in equilibrium and prevents it from falling or being pulled upward. The tension force has both vertical and horizontal components.
- Vertical component (Tv):
The vertical component of the tension force balances the weight force, as the mail bag remains at rest in the vertical direction:
Tv = Fg
Tv = 1274 N
- Horizontal component (Th):
The horizontal component of the tension force is responsible for keeping the mail bag in its horizontally displaced position. It is related to the horizontal distance (2.2 m) the bag is displaced from its original position and the length of the rope (3.9 m):
Th = (2.2 m / 3.9 m) * Tv
Substituting the value of Tv:
Th = (2.2 m / 3.9 m) * 1274 N
Th ≈ 715.38 N
Therefore, the tension in the rope when the mail bag is displaced 2.2m sideways from its original position while keeping the rope taut is approximately 715.38 Newtons.