How Do You Find the Weight of a Beam Using Spring Scales?

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SUMMARY

The discussion centers on calculating the weight of a uniform beam supported by two spring scales, reading 71 lb and 100 lb. The user correctly identifies that the beam is in equilibrium, leading to the conclusion that the weight of the beam equals the sum of the vertical forces exerted by the scales. By applying trigonometric principles, the net upward force is calculated as 137 lb, which represents the weight of the beam.

PREREQUISITES
  • Understanding of static equilibrium in physics
  • Basic knowledge of trigonometric functions (sine and cosine)
  • Familiarity with forces and tension in strings
  • Ability to interpret free-body diagrams
NEXT STEPS
  • Study static equilibrium principles in physics
  • Learn about trigonometric functions and their applications in force calculations
  • Explore free-body diagram techniques for analyzing forces
  • Investigate the mechanics of tension in strings and cables
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of static equilibrium and force analysis in mechanical systems.

melaku
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I am asked to do this.

"A uniform beam is attached to two springs scales as shown. The scales read 71 lb and 100 lb respectively. Find the wight of the beam."

http://img144.imageshack.us/my.php?image=phy8wq.png


I have no idea how to go on about solving this. please someone explain it to me like you would explain it to a child.

Thanks in advance.
 
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I'm not being awkward, but I don't understand the diagram one bit.
 
My guess is that the diagram is meant to show a rod held up by two strings,
one at 45 degrees with a tensiion of 71#, and the other at 60 degrees from the horizontal with a tension of 100#. This is consistent with their being no net horizontal force on the rod.
The net horizontal force is 100 cos 60 - 71 sin 45=0 (as it should).
The net upward force on the stick is 100 sin 60 + 71 sin 45 =87 + 50 = 137#,
which equals the weight of the stick.
 
I reccomended him here for help guys, don't make me look bad:)

That diagram IS funky though
 

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