Vector component of the weight of an object

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Graduated College over thirty years ago. In physics, I recall a a question where one is required to weigh an object beyond the capacity of the scale. The answer involved resting one edge of the object on scale and the other on a separate support higher than the scale such that the object created and angle from the perpendicular . I believe that the cosine of the angle from the perpendicular played in the solution. Guidance sought
 

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NascentOxygen
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Hi Philpense. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

It sounds do-able. Consider something uniform like a length of sawn timber. Providing you can ensure the upper support prevents the beam slipping, then it appears the force on the lower support (i.e., the scales) will be cos θ x half the weight, where θ is the beam's inclination to the horizontal.
 
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