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.

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

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//