To solve this problem, we’ll follow these steps:

1. Force of Water on the Triangular End

Given:

• The trough has width $w = 2 \, \text{ft}$, length $l = 18 \, \text{ft}$, and height $h = 5 \, \text{ft}$.
• The triangular end is an isosceles right triangle (assuming this from common trough problems) with base $w = 2 \, \text{ft}$ and height $h = 5 \, \text{ft}$.
• The density of water $\rho = 62.4 \, \text{lb/ft}^3$.

The force due to water pressure at depth $y$ is given by:

$dF = \text{(pressure at depth } y) \times \text{(area of the strip at depth } y)$

• Pressure at depth $y$ = $\rho \cdot g \cdot y = 62.4 \cdot y \, \text{lb/ft}^2$.
• The width of the strip at depth $y$ is proportional to the height, as the triangle is isosceles. Thus, at depth $y$, the width is:

$\text{Width at } y = \frac{2}{5} y \, \text{ft}.$

• The area of a strip is the width at $y$ multiplied by a small height $dy$:

$dA = \left(\frac{2}{5} y\right) dy.$

Thus, the infinitesimal force $dF$ is:

$dF = 62.4 \cdot y \cdot \left(\frac{2}{5} y \right) dy.$

Now, integrate from 0 to 5 to find the total force on the triangular end:

$F = \int_0^5 62.4 \cdot \frac{2}{5} y^2 \, dy.$

We can compute this integral to find the total force.

2. Work to Pump Water Over the Top

To find the work required to pump the water over the top of the trough:

• The volume of a horizontal strip of water at depth $y$ is $dV = \left(\frac{2}{5} y\right) \cdot l \cdot dy = \left(\frac{2}{5} y \cdot 18 \cdot dy\right)$.
• The weight of this water is $dW = \rho \cdot dV = 62.4 \cdot \frac{2}{5} y \cdot 18 \cdot dy$.
• The distance the water needs to be lifted is $5 - y$ (the height of the trough minus the depth).

The infinitesimal work is then:

$dW_{\text{pump}} = 62.4 \cdot \frac{2}{5} y \cdot 18 \cdot (5 - y) dy.$

Now, integrate this from 0 to 5 to find the total work:

$W = \int_0^5 62.4 \cdot \frac{2}{5} y \cdot 18 \cdot (5 - y) \, dy.$

We'll compute this integral to find the total work.

3. Convert Force and Work into SI Units

• To convert force from pounds (lb) to newtons (N):

  $1 \, \text{lb} = 4.44822 \, \text{N}.$

• To convert work from foot-pounds (ft-lb) to joules (J):

  $1 \, \text{ft-lb} = 1.35582 \, \text{J}.$

Let’s compute the total force, total work, and then convert them to SI units.### Results:

1. Force on the triangular end:

   • Total force $F = 1040 \, \text{lb}$.
   • In newtons: $F = 4626.15 \, \text{N}$.

2. Work to pump all the water over the top:

   • Total work $W = 9360 \, \text{ft-lb}$.
   • In joules: $W = 12690.48 \, \text{J}$.

Would you like any further details or explanations on these calculations?

5 Related Questions:

1. How does the shape of the trough affect the force distribution on its walls?
2. What would happen to the force and work calculations if the trough had a different density of liquid?
3. How can we generalize these calculations for non-isosceles triangular ends?
4. How does water pressure vary with depth in different container shapes?
5. How does changing the height of the trough influence the total work needed to pump the water?

Tip: Always check the units carefully when dealing with force and energy, especially when switching between imperial and SI systems.