Question 9.3: Gate AB has an evenly distributed mass of 180 kg and is 1.2 ...
Gate AB has an evenly distributed mass of 180 kg and is 1.2 m wide “into the page” as illustrated in Figure 9.17. The gate is hinged at A and rests on a smooth tank floor at B. For what water depth h will the force at point B be zero?
Given: Gate dimensions and mass; fluids on either side.
Find: Water depth h required for R_B = 0.
Assume: No relative motion of fluid elements (hydrostatics); fluids have constant, uniform density, and it is appropriate to evaluate densities at 20˚C. Looking up these values, we find that water’s density is 998 kg/m³ and that glycerin’s is 1260 kg/m³.
We start with a free-body diagram (FBD) of gate AB (Figure 9.18).
We intend to apply the equations of equilibrium to the gate to ensure the proper relationships between forces and to meet the constraint that R_B must equal zero. To do this, we first need to evaluate all the forces in the FBD. The weight of the gate is simply mg, or W = (180 kg)(9.81 m/s²) = 1766 N, and this force acts at the centroid of the gate (Figure 9.19).
Next, we must find the forces on the gate due to fluid pressure and where they act. We begin with the glycerin. Since we would prefer not to have to find the reaction forces at A, we plan to sum moments about this point. We need to know the depth of the centroid of the gate, h_c , measured from the surface of the glycerin. The gate is a rectangle, as shown in Figure 9.19, and due to its symmetry its centroid is simply 0.433 m down from the hinge at A. The depth of the centroid is thus 2 m – 0.433 m = 1.567 m below the glycerin surface:
F_{glycerin}= ρ_{_{glycerin}}gh_cA= ( 1260 \frac{\textrm{kg}}{\textrm{m}^3})( 9.81 \frac{\textrm{m}}{\textrm{s}^2})(1.567 \textrm{ m})(1.2 \textrm{ m}^2)
= 23.2 kN.
This force acts at
y_{_R}=y_c + \frac{I_{xc}}{y_cA}=y_c+\frac{I_{xc} \sin \theta}{h_cA} ,
and since we intend to sum moments about A, we would like to know the moment arm from F_{glycerin} to point A. Thus, we are most concerned with how
much deeper y_{_R} is than y_c , as y_c is clearly 0.5 m from point A:
y_{_R}-y_c=\frac{I_{xc}\sin \theta}{h_cA}=\frac{\frac{1}{12}bh^3 \sin \theta }{h_cA}=\frac{\frac{1}{2}(1.2 \textrm{ m})(1 \textrm{ m})^3 \sin 60 ° }{(1.567 \textrm{ m})(1.2 \textrm{ m}^2)}=0.0461 m.
We now know that F_{glycerin} = 23.2 kN has a moment arm of 0.5461 m relative to point A.
What remains to be found is the force due to the water on the other side of the gate. Both the magnitude of this force and its moment arm (where it acts) depend on the depth of water, h. For the moment, we leave both these values in terms of the depth of the centroid of the gate, h_c , measured from the water surface – the depth h = h_c + 0.433 m:
F_{water}= ρ_{_{water}}gh_cA \\= (998 \frac{\textrm{kg}}{\textrm{m}^3})( 9.81 \frac{\textrm{m}}{\textrm{s}^2})h_c(1.2 \textrm{ m}^2)\\=(11.75 h_c) kN.
And this force acts at y_R , where
y_{_R}-y_c=\frac{I_{xc}\sin \theta}{h_cA}=\frac{\frac{1}{12}bh^3 \sin \theta }{h_cA}=\frac{\frac{1}{12}(1.2 \textrm{ m})(1 \textrm{ m})^3 \sin 60 ° }{h_c (1.2 \textrm{ m}^2)}=\frac{0.0722}{h_c} .
So F_{water}=11.75h_c kN has a moment arm of (0.5 + \frac{0.0722}{h_c}) m relative to point A (Figure 9.20).
We choose to sum moments about point A to avoid having to solve for the hinge reaction forces and require that the gate be in equilibrium:
\sum{M_A} = 0 = (23,200 \textrm{ N})(0.5461 \textrm{ m}) + (1766 \textrm{ N})(0.5 \cos 60°) – (11,750 h_c )(0.5 + \frac{0.0722}{h_c}) .
Solving this expression for h_c , we find that h_c = 2.09 m. The depth of the water is then h = h_c + 0.433 m = 2.52 m.
