A cattle trough has a trapezoidal cross section with a height of 100 cm and horizontal sides of width 100 cm and 50 cm.

Question

A cattle trough has a trapezoidal cross section with a height of 100 cm and horizontal
sides of width 100 cm and 50 cm. Assume the length of the trough is 1000 cm. The
density of water is  = 1000 kg/m3.
(a) How much work is required to pump the water out of the trough when it is full?
(b) Determine the total force of the water on a trapezoidal face of the trough.

 

Already have that W = 20pg(1-y)xdy

and F = pg(1-y)2xdy

 

Not sure what bounds to use and how to write x in terms of y.

Answer 1

It looks like you are on the right track with this one.   However, you have a factor of 2 in your equations that I don't understand.

 

Since ρ is quoted in the SI unit (kg/m³)  and ultimately one would like to used g = 9.8 m/s² ,   it is a good idea to convert the dimensions of the trough from cm to m.  Thus the trough is 1 m deep ,  10 m long with a top width of 1 m and a bottom width of 0.5 m.

 

The integrations will be along the vertical coordinate which can be taken to be y.   For problems of this type, it is convenient to choose the zero of y to be at the top and to choose the direction of positive y to be down.  The width as a function of y is then  w =  1- 0.5h.

 

With this choice, the mass of a thin layer of water is  dM =  ρ 10 w dh =  ρ 10 (1- 0.5h) dh,  where dh is the infinitesimal thickness of the thin layer.   The work required to pump this thin layer out is  dM g h.   Thus the work required to pump out all the water is 

 

           W =  ∫ dM h    

 

           W  =   10 g ρ ∫ (1 - 0.5 h) h dh   The lower limit is zero, the upper limit is 1

 

This works out to   (10/3) g ρ

 

The pressure on the trapezoidal face  is given by  P =  ρ g (1- 0.5 y) 

 

The Force is  the integral of this expression over the area of the trapezoid. 

    Symbolically  F  =  ∫ P  dA

   Since P depends on y but not on the width variable,   one can write  dA =  w dy =  (1 - 0.5 y) dy

F = rho g integral [ y(1- 0.5y) dy = (1/3) rho g