Question

A waste treatment pond is 50 m long and 25 m wide, and has an average depth of 2 m. The density of the waste is 75.3lbm/ft^3 .Calculate the weight of the pond contents in lbf ,using single dimensional equation for your calculation.

Answer 1

Answer:

$W = 6.65 \cdot 10^{6} lbf$

Explanation:

To find the weight (W) of the pond contents first we need to use the following equation:

$W = m\cdot g$   (1)

Where m the mass and g is the gravity  

Also, we have that the mass is:

$m = \rho*V$  (2)    

Where ρ is the density and V the volume

We cand calculate the volume as follows:

$V = L*w*d$   (3)

Where L is the length, w is the wide and d is the depth  

By entering equation (2) and (3) into (1) we have:

$W = \rho*L*w*d*g$

$W = 75.3 lbm/ft^{3}*50 m*25 m*2 m*9.81 m/s^{2}$  

$W = 75.3 lbm/ft^{3}*\frac{(1 ft)^{3}}{(0.3048 m)^{3}}*\frac{0.454 kg}{1 lbm}*50 m*25 m*2 m*9.81 m/s^{2} = 2.96 \cdot 10^{7} N}*\frac{0.2248 lbf}{1 N} = 6.65\cdot 10^{6} lbf$              

Therefore, the weight of the pond is 6.65x10⁶ lbf.

I hope it helps you!