Is the Triangular Lamina Stable When Suspended from Point C?

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The discussion focuses on the stability of a uniform triangular lamina suspended from point C, with dimensions AB = 1m and BC = 2m. The center of gravity is determined using a coordinate system, leading to the calculation of the angle between BC and the downward vertical. By applying the Pythagorean theorem, the length of AC is found to be √5m, and the angle is calculated using the sine function, resulting in approximately 63.4 degrees. This angle suggests that while the lamina is not perfectly stable, it is not at a highly unstable angle either. The overall stability of the lamina depends on the specific forces acting upon it.
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A uniform triangular lamina ABC where AB = 1m and BC = 2m and it is right angled at B. It is suspended from C and hangs freely under gravity. Calculate the angle between BC and the downwards vertical.

Need help fast pls cheers:smile:
 
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Find the center of gravity of the lamina.
The center of gravity will hang directly below C.

Easy way to find the center of gravity: Set up a coordinate system so that B is at (0,0), A is at (0,1), C is at (2, 0). Take the "average" of the three points. Find the slope of the line through (2,0) and the center of gravity and remember that slope = tan of angle the line makes with BC (x-axis).
(I get a remarkably simple result!)

(Off the subject: I once had a professor who put (on the FINAL exam!) the question "What is "lamina" spelled backwards?"! One student got really angry with him because the question "didn't make sense".)
 


The stability of equilibrium in this scenario can be determined by observing the angle between BC and the downwards vertical. In this case, since the lamina is suspended from point C, the angle between BC and the downwards vertical can be calculated using trigonometric principles.

First, we can use the Pythagorean theorem to find the length of AC, which is the hypotenuse of the right triangle. AC = √(AB² + BC²) = √(1² + 2²) = √5m.

Next, we can use the sine function to find the angle between BC and the downwards vertical. sinθ = opposite/hypotenuse = BC/AC = 2/√5.

Using a calculator, we can find the value of this angle to be approximately 63.4 degrees. This means that the lamina is hanging at an angle of 63.4 degrees from the downwards vertical.

In terms of stability, this angle indicates that the lamina is not perfectly stable as it is not hanging straight down. However, it also does not have a very large angle, which could indicate instability. Overall, the stability of the equilibrium in this scenario would depend on the specific conditions and forces acting on the lamina.
 
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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