Finding seesaw angle from weight readings at two ends

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john_Stm
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Hi everyone ,

I have a small question ..

I have attached a ppt . I am having two sensors which can measure the weight of my object at two end position so fmy support ..

My question is that can i find the angle beta if am having the value of X kg and Y kg ..

your answers would really help me ..
 

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That should (almost) work. If you can figure out how high above the sensors the CoM of the box is, you should be able to find the angle. Basically just realize that as you increase beta, the CoM shifts to be further over one sensor or another.

Edit: I almost forgot: Hello and welcome to the forums :)
 
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The file just came up as an unreadable text on my MacBook. :frown:
 
You should be able to open it with Open Office or Libre Office. I'm using the latter, albeit on PC. I don't know what word processing mac comes with by default.
 
FireStorm000 said:
You should be able to open it with Open Office or Libre Office. I'm using the latter, albeit on PC. I don't know what word processing mac comes with by default.

I'll check that out. Leopard comes with TextEdit, but I also have NeoOffice (almost exactly the same as Microsoft Office, but free), and a few other open-source doohickeys. One of them might be able to do something with it. My main point of surprise is that something that I expected to be a graphic showed up as a text. Anyhow, I have a lot of other stuff to do first, but I'll check up on this later.
 
hi firestrom , thanks for your response .. i am curious in finding the angle using the two vertical omponents .. x kg and y kg from my sensor outputs..
 
john_Stm said:
hi firestorm , thanks for your response .. i am curious in finding the angle using the two vertical components .. x kg and y kg from my sensor outputs..
I'm not clear if you're asking me to solve it for you, but I'll absolutely get you started:
To begin, I don't think we have enough information to solve the problem, so I'm going to make some assumptions about the geometry, and we'll go from there.
First, let's define P as the location of the pivot. We'll define CM as the center of mass of the object on the two sensors, plus any anything else the sensors are supporting, such as a plate on which you place the object. For the sake of simplicity, we'll assume that the center of mass is directly over the pivot when the tilt angle, θ, is 0. We'll further assume that the sensors are equidistant from the pivot, at distance l. Finally, the distance between CM and P is the length h.
I'll assume that the sensors exclusively measure force parallel to theta, that is, they rotate on the pivot and measure the full downward force only when θ is 0. If that's the case, we then know that the sum of the measures of the two sensors will equal the cosine of theta times weight: F1+F2=cosθ * m * g
We additionally know that at a certain angle of θ, which we will call [itex]\varphi[/itex], the CM is directly over a given sensor, such that either F1=mg and F2 = 0, or visa versa for -[itex]\varphi[/itex].
When theta is 0, F1=F2=mg/2

See if you can't get started with that setup, and let me know if you have trouble, or any of my assumptions are invalid.