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Net force on panel with water and glycerin

Short answer

Panel AB (2 m high) with water left (depths 3–5 m) and glycerin right (depths 1–3 m) yields net force 29.04 kN/m towards glycerin. Line of action 1.059 m above B from moment balance of individual fluid forces.

Find the net hydrostatic force per unit width on rectangular panel AB in the figure given below and determine its line of action. [Figure: vertical wall with water on the left and glycerin (sp. wt. = 12.36 kN/m³) on the right; panel AB is 2 m high; on the water side A is 3 m (2 m + 1 m) below the water surface and B is 5 m below it, with the wall continuing 1 m below B; the glycerin surface is 2 m below the water surface, i.e. 1 m above A.]

Work per metre width. Find the force from each liquid and its line of action, then combine them by taking moments about B.

Given: panel AB is 2 m high. Water side: A at 3 m depth, B at 5 m depth. Glycerin side: A at 1 m depth, B at 3 m depth. γw=9.81\gamma_w = 9.81 kN/m³, γg=12.36\gamma_g = 12.36 kN/m³.

 water surface ==
                 |  2 m
       glycerin  |  == glycerin surface
                 |  1 m
               A +
   water   ->    |   <-  glycerin
   (left)  ->    | 2 m <-
               B +
                 | 1 m

Force from water (left)

Fw=γwAhˉ=9.81×(2×1)×4=78.48 kNhw∗=4+1×23/122×4=4.0833 m\begin{aligned} F_w &= \gamma_w A\bar h = 9.81 \times (2\times1) \times 4 = 78.48\ \text{kN} \\ h^*_w &= 4 + \frac{1\times 2^3/12}{2 \times 4} = 4.0833\ \text{m} \end{aligned}

So it acts 5−4.0833=0.91675 - 4.0833 = 0.9167 m above B.

Force from glycerin (right)

Fg=12.36×2×2=49.44 kNhg∗=2+0.66672×2=2.1667 m\begin{aligned} F_g &= 12.36 \times 2 \times 2 = 49.44\ \text{kN} \\ h^*_g &= 2 + \frac{0.6667}{2 \times 2} = 2.1667\ \text{m} \end{aligned}

So it acts 3−2.1667=0.83333 - 2.1667 = 0.8333 m above B.

Net force

F=78.48−49.44=29.04 kN per m width (towards the glycerin side)F = 78.48 - 49.44 = 29.04\ \text{kN per m width (towards the glycerin side)}

Line of action (moments about B)

F y=78.48(0.9167)−49.44(0.8333)=71.94−41.20=30.74 kN⋅my=30.7429.04=1.059 m above B\begin{aligned} F\,y &= 78.48(0.9167) - 49.44(0.8333) \\ &= 71.94 - 41.20 = 30.74\ \text{kN·m} \\ y &= \frac{30.74}{29.04} = 1.059\ \text{m above B} \end{aligned}

Answer: net hydrostatic force =29.04= 29.04 kN per metre width, acting from the water side towards the glycerin, at 1.059 m above B (0.941 m below A, i.e. 3.941 m below the water surface).

✓ Answer checked by Claude Opus 5.5 (high effort)

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