Question

A 150-N bird feeder is supported by three cables. Find the tension in each cable.

Answer 1

see the attached figure with letters to better understand the problem

we know that

in the x-coordinates

$TACx=TBCx\\ TAC*cos60=TBC*cos30\\\\ TAC*\frac{1}{2} =TBC*\frac{\sqrt{3}}{2}$

$TAC=TBC*\sqrt{3}$ -----> equation $1$

in the y-coordinates

$TCD=150 N$

$TACy+TBCy=TCD\\ TAC*sin60+TBC*sin 30=150\\ \\ TAC*\frac{\sqrt{3}}{2} +TBC*\frac{1}{2} =150\\ \\ TAC*\sqrt{3} +TBC=300$

$TAC*\sqrt{3} +TBC=300$ -----> equation $2$

substitute equation $1$ in equation $2$

$[TBC*\sqrt{3} ]*\sqrt{3} +TBC=300\\ 3TBC+TBC=300\\\\ TBC= \frac{300}{4} \\ \\ TBC=75 N$

$TAC=TBC\sqrt{3}\\ TAC=75\sqrt{3} N$

therefore

the answer is

$TAC=75\sqrt{3} N\\ TBC=75N\\ TCD=150N$

Answer 2

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

the tree cables meet are zero.

• The tension in the 30° inclined cable is 75.0 N

• The tension in the 60° inclined cable is approximately 129.9 N

• The tension in the bird feeder tension cable 150 N

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;

$150 = \dfrac{\sqrt{3} }{2} \cdot T_2 + \dfrac{1 }{2} \cdot T_3...(1)$

$\dfrac{1 }{2} \cdot T_2 = \dfrac{\sqrt{3} }{2} \cdot T_3 ...(2)$

T₂ = √3·T₃

Which gives;

$150 = \dfrac{\sqrt{3} }{2} \cdot \sqrt{3} \cdot T_3 + \dfrac{1 }{2} \cdot T_3 = 2 \cdot T_3$

• $T_3 = \dfrac{150}{2} = 75.0$

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

T₂ = √3·T₃

• T₂ = √3 × 75.0 N ≈ 129.9 N

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

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