Why does the area centroid formula for bar y have a factor of 0.5?

  • Thread starter Thread starter chrisyuen
  • Start date Start date
  • Tags Tags
    Friction Statics
AI Thread Summary
The discussion centers on understanding why the area centroid formula for bar y includes a factor of 0.5. The formula presented is \(\bar{y}=\frac{1}{A} \int y\ dx dy = \frac{1}{A} \int 0.5\ y^2\ dx\), where the participant clarifies that the symbols used are division signs rather than addition. This indicates a misunderstanding of the notation in the formula. The factor of 0.5 arises from integrating the function related to the area centroid. Overall, the conversation aims to clarify the mathematical representation of the area centroid calculation.
chrisyuen
Messages
55
Reaction score
0

Homework Statement



DSCF0002.jpg


Homework Equations



Nil

The Attempt at a Solution



Why does the bar y have a factor of 0.5 for the forumla of area centroid?

Thank you very much!
 
Physics news on Phys.org
Am I right that the + is actually an devided by?

\bar{y}=\frac{1}{A} \int y\ dx dy = \frac{1}{A} \int 0.5\ y^2\ dx

Is that ok?
 
Those are not plus signs. They are in fact, this symbol here:

\div​

So you are right.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How to Determine Centroid of Gate Hydrostatics
Replies
2
Views
3K
Finding Second Moment of Inertia for Shaded Area Through Centroid?
Replies
3
Views
1K
How do you "read" this formula?
Replies
2
Views
1K
Find Centroids: Determine Ix'y' for Area - a=120mm, b=290mm, d=30mm, h=390mm
Replies
20
Views
3K
Centroid of a uniform shape, using area
Replies
9
Views
2K
What Am I Missing About the Centroid Equation for a Circular Segment?
Replies
2
Views
5K
Centroid of an Area - Finding X & Y Coordinates
Replies
11
Views
2K
What is the process for finding the moment of inertia for a semicircle?
Replies
6
Views
2K
Why is 2/3 required in the calculation for the equilibrium of rigid bodies?
Replies
12
Views
2K
Find the position of the centroid of the shaded area
Replies
17
Views
5K

Hot Threads

Recent Insights

Back
Top