Hydrostatic force on a plane surface

[sdaysley]

Homework Statement

im required to find the width of the strip of integration across the triangular lamina w(s) and then integrate to find the area

I am given w(s) = x_right(s) - x_left(s) between 0<s<λh and λh<s<h

Homework Equations

The Attempt at a Solution

i have found the equation i think for the first inequality using similar triangles

w(s) = (d-b)s/λh + ds/h (do i need to find anonther equation for w(s) with for the second inequality?)

due to the orientation of the triangle I am not sure whether i have to split the region up into two sections when integrating to find the area

i would really appreciate it if someone could point me in the right direction i don't want any solutions

thanks in advance

 

[LawrenceC]

You could split it into two sections and integrate over y, then sum the results. You have to find the equations of the lines that form the three sides of the triangle. Your limits of integration should go from -h to -lambda*h and from -lambda*h to 0 for the second integral.

 

[pongo38]

your function "w(s) = (d-b)s/λh + ds/h" can't be correct on 2 accounts.
In the first term, why is d-b used and not b-d?. In the second term the dimensions are not correct. Some length is missing?
"im not sure whether i have to split the region up" If the functions are different, they must be done separately.

 

[sdaysley]

thank you both for your replies, yes your right i copied it up wrong i meant to put (b-d)

how do you man some legnth is missing? a h canceled when calculating w(s). that's fine i was just unsure about the integration but i think that may help thanks again

 

[pongo38]

The term ds/h is dimensionless, whereas w(s) has units of length. (EDIT) Sorry, I misread the d as a delta. (second edit) Doesn't ds/h give the length xl(s). But that isn't exactly the length you want to include in w(s).

 

[sdaysley]

The term ds/h is dimensionless, whereas w(s) has units of length. (EDIT) Sorry, I misread the d as a delta. (second edit) Doesn't ds/h give the length xl(s). But that isn't exactly the length you want to include in w(s).

erm for the first inequality w(s) would be given by xr(s) - xl(s)? so it would be (b-d)s/λh + h - ds/h - h = (b-d)s/h - ds/h. this would be a dimension of width as the h and λh would cancel with the s to leave b-d and d?

 

[pongo38]

I agree your first term, but the second term ds/h is not right. Can you do similar triangles on the triangles Odh and s xls h. ( I think there is a 1-lambda in there somewhere)