Weight supported by two supports

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SUMMARY

The discussion focuses on calculating the forces exerted on two supports by a horizontal beam weighing 1100 lbs (4893 N) and measuring 1714.5 mm in length, with supports located at 200 mm and 1150 mm. Participants recommend using the equilibrium equations, specifically the sum of moments and forces, to determine the reaction force at support 2. A diagram is essential for visualizing the forces and moments involved. Additionally, the impact of an extra 5G force on the mass at support 2 is discussed, emphasizing the need to account for this in the calculations.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of moment calculations in physics
  • Familiarity with reaction forces in structural analysis
  • Ability to draw and interpret free-body diagrams
NEXT STEPS
  • Study the concept of static equilibrium in detail
  • Learn how to calculate moments about a point in a beam
  • Explore the effects of additional forces on static structures
  • Practice drawing free-body diagrams for various support configurations
USEFUL FOR

Engineers, physics students, and anyone involved in structural analysis or mechanics will benefit from this discussion, particularly those working with beam support systems and static load calculations.

abe_cooldude
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Hi all,

I am working on a problem that I can't even seem to figure out where to start. I have a horizontal beam of known weight (1100lbs, or 4893N). It is 1714.5mm long, and is supported by two vertical supports, which are at known location along the beam (Support 1 @ 200mm; support 2 @ 1150mm). I am trying to figure out how much mass (in Newtons) is carried by support 2, and then I want to add additional 5G of force to the mass supported by support 2 straight down in vertical direction.

I know the starting point would be:
sum of moments = sum of forces = 0
\SigmaM = \SigmaF = 0

I know I do something with the reaction forces at the support, but not sure.

Any help to get started would be appreciated.

Thanks!
 
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abe_cooldude: Yes, start by performing a summation of moment about support 1, then solve for the reaction force at support 2.

Is the additional 5 g acceleration applied only to the mass supported by support 2? Or is an additional 5 g also applied to the mass supported by support 1? If the latter, then just multiply the beam weight by 6, then solve for the reaction force at support 2 again, as you did before.
 
First, draw a diagram! Like you say, for the system to remain in equilibrium the forces (so the system doesn't accelerate) and the moments (so the system doesn't rotate) need to sum to 0.

Like nvn said I would start by summing moments about support 1 (remembering to consider the reactive force at support 2 as an unknown). If the horizontal beam is uniform it's entire weight will act at the centre of gravity, which will be in the middle. A moment is the magnitude of force * perpendicular distance from line of action of force.

Obviously for forces Reaction 1 + Reaction 2 = Weight of beam

Not entirely sure about the second part of your question, but hopefully you can work it out now :) Hope this helps.

2013-010.jpg
 

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