What Is the Tension Force in a Clothesline Holding Big Bird?

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The discussion revolves around calculating the tension force in a clothesline supporting Big Bird, who exerts a gravitational force of 720 N. The clothesline forms a symmetrical 'V' with an internal angle of 150 degrees. Participants suggest using trigonometric principles and free body diagrams to resolve the tension forces into components. The key approach involves ensuring equilibrium by balancing the horizontal and vertical forces, leading to the equation Tcos(angle) + Tcos(angle) = Fg. Ultimately, the correct angle for the tension forces needs to be determined to solve for the tension accurately.
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Homework Statement


Big Bird’ has a force of gravity on him of 720 N. He is perched on a clothesline such that the
clothesline forms a symmetrical 'V' with the angle within the 'V' being 150degrees. What is the
tension force in the rope on each side of ‘Big Bird’?


Homework Equations



Trigonometry

The Attempt at a Solution



I tried to draw a scale diagram which failed. I also tried to figure this all out but I'm really stumped.

sin150 = x /720N
sin150 x 720N = x
0 = x
I don't know what to do from there, or even if I'm doing this right?
 
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Draw the V with two tensions acting, they are the same.

Now split the two tensions into components. For equilibrium, sum of horizontal forces=0
and similarly, sum of vertical forces=0
 
So he (big bird) is standing on a rope and has a force straightdown in gravity, there are 2 forces of Tension, so try drawing the free body diagram (hint: the tension forces form a V and also break the tension forces into vectors after)

Then imagine a line going through the center of big bird and think about wut the angle should be for each tension force (its not 150)

In the end u should find tht the forces in the x-direction cancel and the forces in the y-direction are Tcos(angle) + Tcos(angle) = Fg

then solve.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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