Proper usage of trig functions in force problems

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In force problems involving free body diagrams, the choice between sine and cosine depends on the orientation of the angle relative to the axes. The component of a vector along an axis is calculated using cosine, while the perpendicular component uses sine. It is essential to practice solving various problems to develop an intuition for identifying the correct trigonometric function. Revisiting the definitions of sine and cosine can aid in understanding their application in these scenarios. Ultimately, mastering the identification of right triangles in diagrams is crucial for accurate calculations.
Matt Armstrong
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Not a particular problem to wonder about but more of a general question, when one has a free body diagram, when is it best to use sine and when is it best to use cosine?

I am reviewing some of my tests for a final, and having previously re-read my forces chapter, I thought that angles rising from the x-axis used sine and angles 'rising' from the y-axis used cosine. However, I have found multiple exceptions to both of these conceptions.

So when is it best to use one or the other?
 
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As you've discovered, it varies. If a vector F makes an angle θ with some axis, then the component of that vector along the axis (parallel to it) is found using Fcosθ; the component perpendicular to the axis is found using Fsinθ.

But the best thing is to solve a number of problems and practice.
 
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A check I often use is the extreme value test. E.g. if the slope were zero, would my expression for the force give the right result.
 
It's not "best" to use one of the other. One is usually correct and the other incorrect. Revise the definition of sine and cosine, for example...

Cos(X) = adjacent/hypothenuse

And how to rearrange these equations to give expressions like...

Adj = hyp * Cos(X)

In short you have to get good at looking at drawings and identifying the right angled triangle so you can work out if you need to calculate the adjacent side or the opposite side and if Sin or Cos is needed.

I also use the trick Haruspex mentions.
 
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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