How Do You Calculate Forces in a Supported Bar with Uneven Loads?

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Homework Help Overview

The problem involves calculating the forces acting on a bar supported at two points (A and B) with an uneven load applied. The bar has a weight of 500 N and an additional weight of 1500 N is hung from it. The dimensions of the bar and the positions of the weights are provided, but there seems to be confusion regarding the calculations of the forces FA and FB.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equilibrium of moments and attempt to set up equations based on the forces and distances involved. There are questions about the correct values to use in the calculations, particularly regarding the total weight and the signs of the moments.

Discussion Status

There is ongoing clarification regarding the values used in the calculations, with some participants suggesting corrections to the initial data. Multiple interpretations of the problem setup are being explored, and there is no explicit consensus on the correct forces at this time.

Contextual Notes

Participants note that the drawing provided is unclear, which may be contributing to the confusion. There are also side discussions about the properties of uniform bars and the relationship between forces and the center of mass.

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Homework Statement



We have a bar with the weight of 500 N which stands in A and B.In this bar an object which weights 1500 N has been hung.The length of the bar AB is 2 meters and AD is 0.8 meters.Find FA and FB just like in the picture.

http://s1171.photobucket.com/albums/r549/Elaia06/?action=view&current=sdsd.jpg


The Attempt at a Solution



The resultant momentum is zero so

FA*0+1900*0.8+500*1+2*FB=0
But FB in my book is 850 N.This answer does not fit my answer.
 
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Elaia06 said:
FA*0+1900*0.8+500*1+2*FB=0
(1) Why 1900?
(2) Distinguish clockwise and counterclockwise moments: they have different signs.
 
it is 1900 sorry , I have written it the wrong way in the data.
FA*0-1900*0.8-500*1+2*FB=0
Still FB is not 860..
 
Your drawing is not clear enough.
Maybe you make the drawing, take a snapshot and insert image.
 
Elaia06 said:
it is 1900 sorry , I have written it the wrong way in the data.
FA*0-1900*0.8-500*1+2*FB=0
Still FB is not 860..
Are you sure? When I solve for FB using your original data (weight of object = 1500N) I get FB = 850 N. Which matches the answer in your book (at least according to your first post).
 
i have written it wrong ,i was sleepy last night :P
It should be 860 N, the Fb.
Yes I get 850 N too.The drawing is just like in my book.
 
Not related to this exercise :Ah and by the way, If we have a bar with two forces acting in the beginning and the end of it, what is the relationship of the weight of the bar with the forces? I mean, is the distance from a force to the point of weight equal to the distance of the other force with the weight or not?
 
Elaia06 said:
Not related to this exercise :Ah and by the way, If we have a bar with two forces acting in the beginning and the end of it, what is the relationship of the weight of the bar with the forces? I mean, is the distance from a force to the point of weight equal to the distance of the other force with the weight or not?
If the bar is uniform, the weight acts at its center. The support forces at each end will (of course) be equidistant from the center.
 
what is a uniform bar? Sorry but english is not my native language :P
 
  • #10
Elaia06 said:
what is a uniform bar?
A bar whose weight is evenly distributed along its length.
 
  • #11
Yes this is an uniform bar, thanks :)
 

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