Mechanical Engineering: Distributed Loads on Beams (11 of 12) Find Distributed Load on Beam Ex. 10

In this video I will find (using the table method) distance to the centroid, and Force total at the centroid of a distributed load on a beam.

Next video in this series can be seen at:

[Spectrojamz]

Exquisitely done...
Concept understood✅
Thanks Sir.

[rickjljr]

I know this is an old video but I'm confused. The diagram labels the vertex as (6, 800), a parabola that opens downward. The formula for this is given in one of the comments below.
The diagram also says the formula of the curve is x=ky^2 which is a parabola that opens to the right with a vertex of (0, 0) (conflict???). The formula for this parabola is y=302.77*SQRT(x) but it passes through the points given at (0.982, 300) and (6.982, 800) (harder to integrate).
The third parabola given is y=204.12*SQRT(x) (used in the calculations) which could be the one shown on the board but it is not clear. The vertex at (0, 0) for this does not seem right.

[shubhamkukreja]

So if function is y=ky^2, rather than ky^2+c then we take the rectangle area

[koredetobiloba]

Sir what if we consider a horizontal strip when I did that I got the 1000N not 2000N

[saeedullahyasinzai]

It is better you make a colour print of these Q’s like software type of demonstration will be better for students to download it.

[diflection]

12600 is not the total moment at A is 12900 instead