# Homework Help: Gravity force term

1. Jul 5, 2012

### swmmr1928

This is not a problem statement this is not homework this is not a textbook exercise. This is my own question about a formula in a textbook. I was given an infraction because I did not post this here.
1. The problem statement, all variables and given/known data

Source: Fluid Mechanics 2nd ed. - P. Kundu, I. Cohen.pdf

I am trying to understand the force balance, specifically the force due to gravity. Why is the (1/2) present?

2. Relevant equations

(1/2) * ρ * g * dx * dz

3. The attempt at a solution

I tried my own force balance and I agree with the book except for the gravity term:

ƩF in vertical = 0 = +p2dx - Fgravity - p1ds*cos(θ)
Fgravity = m*a = m*g = ρ*V*g
m=V*ρ
ƩF in vertical = 0 = +p2dx - ρ*V*g - p1ds*cos(θ)

2. Jul 5, 2012

### swmmr1928

Oh I have received an email regarding the thread I started in the general physics section. Another user replied:

"m=ρ*A*d <=> (since d=1, for its of unit thickness)
m=ρ*(dxdz/2) = 1/2 ρ dx dz (since A is that of a triangle) and hence
F= -1/2 ρ dx dz g"

So is this correct: ?
V = s*A
A=1/2*x*y
V=1/2*x*y*s
s=1
V=1/2*x*y

How are you able to suddenly invoke differential length?

3. Jul 5, 2012

### Staff: Mentor

It's a triangle, so its volume is 1/2 that of a rectangular element. Area = 1/2dxdz.

4. Jul 6, 2012

### swmmr1928

How are you able to suddenly invoke differential length?

5. Jul 6, 2012

### Staff: Mentor

What's sudden about it? The infinitesimal fluid element in your diagram has a triangular cross section with sides dx & dz.