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 151 N on the wire. The left section of the wire makes an angle of 14 degrees relative to the horizontal and sustains a tension of 447 N. Find the magnitude and direction of the tension that the right section of the wire sustains.

Answer 1

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 $\frac{ xsin\theta}{xcos\theta} =\frac{42.861}{433.722} = 0.099$ 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 = $\frac{433.722 }{cos 5.644 }$ = 435.835 N

Therefore the magnitude of the force is 435.835 N at an angle ∡5.644 ° to the horizontal