During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 246 N on the wire. The left section of the wire makes an angle of 14.2° relative to the horizontal and sustains a tension of 432 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

Question

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 246 N on the wire. The left section of the wire makes an angle of 14.2° relative to the horizontal and sustains a tension of 432 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

Answer 1

To solve this problem, we can break it down into smaller components and analyze the different forces acting on the wire.

Let's denote the tension in the right section of the wire as T. We need to find the magnitude (T) and direction (angle relative to horizontal) of this tension.

First, let's draw a diagram to understand the situation:

```
|
|
T |
-------- |
| x (tree limb)
----------------------------- (barbed wire fence)
|
|
```

Now, let's analyze the forces acting on the wire:

1. The tension in the left section of the wire (T_left), which is given as 432 N, makes an angle of 14.2° relative to the horizontal.
2. The weight of the tree limb, which exerts a downward force of 246 N on the wire.
3. The tension in the right section of the wire (T), which we need to find.

To determine the magnitude and direction of T, we can resolve the forces horizontally and vertically.

Initially, let's resolve the forces vertically:

ΣF_vertical = 0
T_left * sin(14.2°) + T * sin(θ) = weight of tree limb
T * sin(θ) = 246 N - T_left * sin(14.2°) ... (Equation 1)

Next, let's resolve the forces horizontally:

ΣF_horizontal = 0
T_left * cos(14.2°) + T * cos(θ) = 0
T * cos(θ) = -T_left * cos(14.2°) ... (Equation 2)

Now, we have two equations (Equation 1 and Equation 2) with two unknowns (T and θ), which we can solve simultaneously.

1. Solve Equation 2 for T in terms of T_left:
T = (-T_left * cos(14.2°)) / cos(θ)

2. Substitute this value of T into Equation 1:
((-T_left * cos(14.2°)) / cos(θ)) * sin(θ) = 246 N - T_left * sin(14.2°)

3. Simplify the equation by canceling out common terms:
- T_left * sin(14.2°) * cos(θ) + 246 N * cos(θ) = -T_left * cos(14.2°) * sin(θ)

4. Rearrange the equation to isolate T_left terms on the left side and θ terms on the right side:
T_left * (sin(14.2°) * cos(θ) + cos(14.2°) * sin(θ)) = 246 N * cos(θ) - T_left * sin(14.2°) * cos(θ)

5. Divide both sides by (sin(14.2°) * cos(θ) + cos(14.2°) * sin(θ)):
T_left = (246 N * cos(θ) - T_left * sin(14.2°) * cos(θ)) / (sin(14.2°) * cos(θ) + cos(14.2°) * sin(θ))

Now, you can solve this equation numerically by substituting different values of θ until you find a solution where T_left matches the given tension of 432 N.