Tension problem - bead sliding on a string attached to a pole

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A 100 g bead slides along an 80 cm string attached to a vertical pole, which rotates, causing the string to form a right triangle. The user is struggling with trigonometry, knowing only one angle (90 degrees) and one side length (0.40 m). They consider using the Pythagorean theorem to find the unknown angle and side lengths but are unsure how to apply it without knowing the bead's position on the string. The discussion emphasizes the need to define one of the unknown sides in relation to the hypotenuse. Understanding the relationship between the sides and the string's total length is crucial for solving the problem.
chelsea526
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Homework Statement



A 100 g bead is free to slide along an 80 cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40 cm apart. When the pole is rotated about its axis, AB becomes horizontal.



Homework Equations



F = ma
Trigonometry formuals

The Attempt at a Solution



I am having trigonometry troubles! I only know one angle (90), and one side length (0.40 m). The bead is somewhere along the 0.80 m string, which is where the unknown angle is formed.

I thought I could use cos90 = 0.40/hyp, however, my calculator doesn't like this formula.
How can I find the lengths of the sides, so that I may then find the unknown angle??
 

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chelsea526 said:
I only know one angle (90), and one side length (0.40 m). The bead is somewhere along the 0.80 m string, which is where the unknown angle is formed.
Take advantage of the fact that you know the length of the string. What does that tell you?
 
Im not sure if I should be trying to use this equation

adj^2 + opp^2 = hyp^2 ?

I really am stumped that you don't know where along the string the bead is.
 
chelsea526 said:
Im not sure if I should be trying to use this equation

adj^2 + opp^2 = hyp^2 ?
That's Pythagorean theorem. It'll do fine.

I really am stumped that you don't know where along the string the bead is.
Call one of the unknown sides X. What's the hypotenuse in terms of X?
 
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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