Weightless horizontal bar equilibrium

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Homework Help Overview

The problem involves a weightless horizontal bar in equilibrium, with readings from two scales and the masses of two blocks. Participants are tasked with determining the reading on scale A based on the given information about scale B and the distances involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest taking moments of the forces on the bar to analyze equilibrium, while others question the relevance of the distances provided. There is also a concern about the possibility of a negative reading for scale A based on initial calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the approach to take moments, but there is no consensus on the calculations or the implications of the results.

Contextual Notes

Participants are working within the constraints of the problem as presented, including the given readings and distances, while questioning the assumptions made about the system's equilibrium.

nora00
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The weightless horizontal bar in the figure below is in equilibrium. Scale B reads 4.15 kg. The distances in the figure (which is not to scale) are: D1 = 6.5 cm, D2 = 10.0 cm, and D3 = 5.0 cm. The mass of block X is 0.94 kg and the mass of block Y is 2.04 kg. Determine what the reading on scale A must be.
I am not sure where to start with this problem.
 
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Welcome to PF!

nora00 said:
The weightless horizontal bar in the figure below is in equilibrium. Scale B reads 4.15 kg. The distances in the figure (which is not to scale) are: D1 = 6.5 cm, D2 = 10.0 cm, and D3 = 5.0 cm. The mass of block X is 0.94 kg and the mass of block Y is 2.04 kg. Determine what the reading on scale A must be.
I am not sure where to start with this problem.

Hi nora00! Welcome to PF! :smile:

You can assume that the bar is uniform, so its centre of mass is in the middle … and the readings on the scales are the normal forces … so …

Hint: take moments of the forces on the bar about some point. :wink:
 
That's a very strange problem. If you were not told the reading on scale B, then it would be an interesting calculation: you would need to calculate the total "moment of inertia" about some point and set it 0, the set the total force on the bar to 0 so that you would get two equations for the two scale readings. But since you are told the reading on scale B, you can just set the total force downward equal to the total force upward and solve for the single variable, the reading on scale A. In this case the distances you give are irrelevant: 0.94+ 2.04= 4.15+ A where A is the reading on scale A.
 
if u just do that then A is negative?
 
nora00 said:
if u just do that then A is negative?

Hi nora00! :smile:

Always show us your calculations, so that we can see where the problem is, and know how to help. :wink:
 

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