Can someone explain this derivation?

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whitejac
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Homework Statement


I've been looking at examples of motion derivations for my class, and it's honestly just very confusing. I heard Dynamics should prep you for this but I must have had a very poor course because we never had to understand geometry and physics to this degree...
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Homework Equations


Equations are listed

The Attempt at a Solution


I understand parts, and parts confuse me greatly. The thing that most confuses me is why they chose to relate side BC as (l-xs). That to me implies that the tripod is almost completely spread out flat because I cannot see how the net displacement is of any noticeable size...
I understand the relarionship between potential energy being .5 (keqx2 = 1.5kxs2 that is a relationship between springs in series, and it is saying that the resulting stiffness times the vertical displacement after mass is added is equal to the sum of individuals displaced... equally?

The real problem is in the third photo where they say the triangle's geometry related side BC to the other 2 sides. This looks like textbook pythagorean's theorem but they use 2LXCos(a) and I don't understand why...
 
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whitejac said:
side BC as (l-xs).
Each leg is shortened by xs under the load, so its new length is l-xs. This is not related to how squat the tripod is, it is true generally.
whitejac said:
that is a relationship between springs in series,
Not specifically, indeed these springs are in parallel. It is merely saying that the total PE is the sum of the PEs in the three springs.
whitejac said:
This looks like textbook pythagorean's theorem
Pythagoras' theorem only applies to right angled triangle. This is using a generalisation of it known as the cosine rule.

Edit: corrected expression in first line.
 
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haruspex said:
Each leg is shortened by xs under the load, so its new length is x-ls. This is not related to how squat the tripod is, it is true generally.
Wouldn't that be a description of point AC then? I thought line BC was designating the horizontal displacement of point B, ie, where the tripod touches te ground with respect to its close position along the vertical axis. That would make sense to be described as it's original length minus the displacement (compression).

haruspex said:
Pythagoras' theorem only applies to right angled triangle. This is using a generalisation of it known as the cosine rule.

Wouldn't this be a case of 3 right triangles each sharing a vertical side? The problem states the legs are fixed equidistant from each other because they're all using same value for α, at least that's how I interpreted it from te image and the question. If that's not the case. Then I can easily see where I made a mistake in not using the law of cosines (which I'd forgotten because I was so set on a right triangle situation)
 
whitejac said:
thought line BC was designating the horizontal displacement of point B, ie, where the tripod touches te ground
You are misreading the perspective. It is a simple side view. The point A is the unloaded position of the top of the tripod, point B is its position under load. It is not a point on the ground. Consequently, CBA is not a right angle.
 
Ah yes. I was just typing that ah-ha! Moment. I was misinterpreting the point B.

Thank you for that! There are so many ways to describe a situation.. This is something I'm struggling with and striving to get some exposure to because it really makes deriving/applying formulas simpler.