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 283 N on the wire. The left section of the wire makes an angle of 11.8° relative to the horizontal and sustains a tension of 499 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.
2nd law f=ma
To find the tension in the right section of the wire, we can use Newton's second law of motion which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the object we are considering is the tree limb.
First, let's break down the forces acting on the tree limb. We have the weight of the limb acting downward, the tension in the left section of the wire pulling leftward, and the tension in the right section of the wire pulling rightward.
The vertical component of the tension in the left section of the wire must be equal to the weight of the limb since there is no vertical acceleration. We can find this vertical component using trigonometry.
The vertical component of the tension in the left section can be found using the equation:
T_left * sin(11.8°) = weight of the limb
Substituting the given values:
499 N * sin(11.8°) = weight of the limb
(weight of the limb is equal to the downward force exerted by the limb on the wire: 283 N)
Now, solving for T_left:
T_left = (283 N) / sin(11.8°)
T_left ≈ 1598.78 N
The horizontal component of the tension in the left section can be found using the equation:
T_left * cos(11.8°) = T_right
Substituting the known value for T_left:
1598.78 N * cos(11.8°) = T_right
Now, solving for T_right:
T_right ≈ 1586.82 N
Therefore, the magnitude of the tension in the right section of the wire is approximately 1586.82 N (a). To find the angle (b) relative to the horizontal, we can use trigonometry to find the angle whose cosine is equal to T_right/T_total (where T_total is the sum of the tension in the left and right sections of the wire).
Angle (b) = arccos(T_right / T_total)
Substituting the known values:
Angle (b) = arccos(1586.82 N / (1586.82 N + 499 N))
Now, solving for Angle (b):
Angle (b) ≈ arccos(0.758)
Using a calculator, we find that Angle (b) ≈ 40.5°
Therefore, the direction of the tension in the right section of the wire is approximately 40.5° relative to the horizontal.