Weightless horizontal bar equilibrium

In summary, the weightless horizontal bar in the given figure is in equilibrium with a reading of 4.15 kg on Scale B. The distances are D1 = 6.5 cm, D2 = 10.0 cm, and D3 = 5.0 cm, and the masses of block X and Y are 0.94 kg and 2.04 kg respectively. To determine the reading on Scale A, the total force downward must be equal to the total force upward. This results in a single variable equation, A = 4.15 - 0.94 - 2.04 = 1.17 kg.
  • #1
nora00
2
0
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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  • #2
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:
 
  • #3
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.
 
  • #4
if u just do that then A is negative?
 
  • #5
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:
 

Related to Weightless horizontal bar equilibrium

What is weightless horizontal bar equilibrium?

Weightless horizontal bar equilibrium refers to a state where a horizontal bar is supported by two or more points or forces, resulting in zero net torque and the bar remaining at rest.

What factors affect weightless horizontal bar equilibrium?

The factors that affect weightless horizontal bar equilibrium include the distance between the points of support, the weight of the bar, and the distribution of weight along the bar.

What is the formula for calculating weightless horizontal bar equilibrium?

The formula for calculating weightless horizontal bar equilibrium is M = F1 x d1 = F2 x d2, where M is the torque, F is the force, and d is the distance from the point of support to the force.

How does the center of mass affect weightless horizontal bar equilibrium?

The center of mass plays a crucial role in weightless horizontal bar equilibrium. If the center of mass is aligned with the points of support, the bar will remain in equilibrium. However, if the center of mass is outside the support points, the bar will rotate and not be in equilibrium.

What are some real-life applications of weightless horizontal bar equilibrium?

Weightless horizontal bar equilibrium is commonly used in construction and engineering to design structures such as bridges and cranes. It is also utilized in gymnastics and acrobatics, where performers balance on horizontal bars for various stunts and routines.

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