Vector component of the weight of an object

AI Thread Summary
The discussion focuses on weighing an object that exceeds a scale's capacity by using a support system to create an angle. The method involves placing one edge of the object on the scale and the other on a higher support, forming an angle with the perpendicular. The force exerted on the scale is calculated using the cosine of the angle, specifically as cos θ multiplied by half the object's weight. This approach is deemed feasible, particularly for uniform objects like timber, provided the upper support is stable. Overall, the technique effectively reduces the weight measured by the scale through the use of angles and supports.
Philpense
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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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Hi Philpense. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

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[/color], where θ is the beam's inclination to the horizontal.
 
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