Weightless horizontal bar equilibrium

AI Thread Summary
The discussion revolves around a weightless horizontal bar in equilibrium, with scale B reading 4.15 kg, and the masses of blocks X and Y being 0.94 kg and 2.04 kg, respectively. Participants suggest using the principle of moments to analyze the forces acting on the bar, emphasizing that the total downward force must equal the total upward force. The calculation simplifies to the equation 0.94 + 2.04 = 4.15 + A, where A represents the reading on scale A. Concerns are raised about the possibility of A being negative, indicating a potential miscalculation. Participants encourage sharing calculations for better assistance in resolving the issue.
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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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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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