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 292 N on the wire. The left section of the wire makes an angle of 10.5° relative to the horizontal and sustains a tension of 412 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. 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 twofence posts. The limb exerts a downward force of 292 N on the wire. The left section of the wire makes an angle of 10.5° relative to thehorizontal and sustains a tension of 412 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tensionthat the right section of the wire sustains.

Name: During a storm, a tree limb breaks off and comes to rest across a barb
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: 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 292 N on the wire. The left section of the wire makes an angle of 10

Given,Downward force exerts by limb, Fd = 292 NTension in the left section of the wire,TL = 412 N at θ = 10.5°Now from free body diagram, by using equilibrium condition of the forces, ∑Fx = 0hence,TLcos θL = T cosθT cosθ = 412 cos 10.5°T cosθ = 405.10 N ...i) Now along vertical direction, TL sinθL + T sinθ = FdT sinθ = Fd -TL sinθLT sinθ = 292 - 412 sin 10.5°T sin θ = 216.9189 N ...ii)a) Now by squaring and adding equation i) and ii) we get, T2sin2 θ + cos2θ = 405.102 + 216.91892T2 = 211155.958T = 459.517 N b) on dividing ii) from i) we get, tanθ = 216.9189405.1θ = tan-10.53547θ = 28.1677°