A150-n bird feeder is supported by three cables. find the tension in each cable.

At equilibrium, the sum of forces and moments acting at the point where

the tree cables meet are zero.

Reasons:

Required;

The tension in each cable

Solution:

Let T₁ represent the tension in the cable attached to the feeder, let T₂

represent the tension in the cable inclined 60°, and let T₃ represent the

tension in the 30° cable.

At equilibrium, we have;

The tension in cable T₁ is the weight = 150 N

T₁ = T₂·sin(60°) + T₃·sin(30°)

T₂·cos(60°) = T₃·cos(30°)

By evaluating, we have;

T₂ = √3·T₃

Which gives;

The tension in the 30° inclined cable, T₃ = 75.0 N

T₂ = √3·T₃

The tension in the 60° inclined cable, T₂ ≈ 129.9 N

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see the attached figure with letters to better understand the problem

we know that

in the x-coordinates

-----> equation

in the y-coordinates

-----> equation

substitute equation in equation

therefore

the answer is

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

21/58.

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