Chapter 3

Question 1

Can you derive an expression for the resultant force acting on an inclined plane surface? (hydrostatic)

Answer 1

total are of surface = A
small amount of force dF acting on surface area dA so from the definition of pressure we get:
dF=gamma.h.dA (as Pgauge=gamma.h)
using trig. we get: h=ysin(theta) => dF=gamma.y.sin(theta).dA
Fr can be found by integrating both sides:
Fr=gamma.sin(theta).integral of y.dA (overA).

Question 2

How to find an expression for the centre of the resultant force acting on an inclined plane surface? (hydrostatic).

Answer 2

Multiply both sides of Fr=gamma.sin(theta).integral of y.dA (overA). by y!
Fr.yr=gammasin(theta)integral of y squared.dA (over A)
=> yr = gamma.sin(theta).integral of y squared.dA.(overA)/Fr.

Question 3

How to calculate the resultant force acting on a curved surface? THIS IS THE SAME METHOD USED FOR THE DAM EXCEPT ACTUAL WATER WEIGHT IS PRESENT AND USED AS Y FORCE

Answer 3

Horizontal force = gamma.sin(theta).integral of y dA (over A). –> theta will equal 90 deg.
Vertical force: Fy = weight of water contained at block from curved surface to top = mg = gamma.h.A
Resultant force = sqrt(Fx^(2) + Fy^(2)).

Question 4

Loactions of horizontal and vertical forces acting on a curved surface? THIS IS THE SAME METHOD USED FOR THE DAM

Answer 4

Location of Fx = gamma.sin(theta).integral of y squared.dA.(overA)/Fx.
Location of Fy is x: x(total area of shape) = SUM of all(C.G(area of individual shapes))