Circular Motion 12th grade physics

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To solve for the reaction at point A and the tension in the cable at point C, it is essential to break down the forces into their x and y components. A free body diagram of the system, including the block, cable, and support at A, will help visualize the forces involved. The horizontal and vertical components of the cable force are related through trigonometric functions, particularly given the 30-degree angle with the wall. Applying the three equations of equilibrium will allow for the determination of the cable tension and support reactions. This approach will provide a structured method to tackle the problem effectively.
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From the diagram above, solve for the reaction at point A. and the tension in the cable that is attatched at point C. Assume the bar is mass-less. Block B has a mass of 124 kg. The distance from point B to point A is 3.2m, from point A to point D is 1.3 m, and point D to point C 2.1 m. The cable attaches to the wall forming a 30 degree angle with the wall.


This is an equation with an attached pic from my high school physics class. It was a "just for fun" problem that the teacher does not expect us to be able to figure out, but I wanted to know if someone could give me some suggestions, I really have no idea. This is not for a grade, other wise I would have attempted it in a more vigourous level.
 

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It is usually best to break up forces into their x and y components. I don't know if you have studied free body diagrams, but if yes, you should draw one of the 'block/ cable/ support at A' system, isolating it from the wall. Note that the horizontal and vertical components of the cable force are trig related. Then use the 3 equations of equilibrium to solve for the cable tension and support reactions.
 
Thanks so much! I will be working on this later on today, I will let you know if I have anymore questions
 
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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