Answer:
The answers to the question is
The magnitude and direction of the tension that the right section of the wire sustains is 435.835 N at an angle ∡5.644 ° to the horizontal
Explanation:
The net downward force = 151 N
Tension in left section of wire = 447 N
Angle subtended with the horizontal for the left section = 14 °
To solve the question we use the sum of forces at equilibrium = 0
That is ∑F = 0
Therefore sum of vertical forces = 0 and
sum of horizontal forces = 0
Therefore
ΣFy gives 151 N = 447·sin 14 + x·sin θ ....................(1) and
ΣFx gives 447·cos 14 = x·cos θ..................................(2)
Solving the above two equations should provide the answers to the question
We have x·sin θ = 151 - 447·sin 14 = 42.861 N
and x·cos θ = 433.722 N
Therefore or tanθ = 0.099 and
tan⁻¹0.099 = θ = 5.644 ° to the horizontal
Therefore from x·cos θ = 433.722 we have
x·cos 5.644 ° = 433.722 or x = = 435.835 N
Therefore the magnitude of the force is 435.835 N at an angle ∡5.644 ° to the horizontal