Answer to Question #269093 in Classical Mechanics for joseph

Question

1. The 5-m-wide wall is in the form of a parabola. If the depth of the water is h = 4 m, determine the magnitude and direction of the resultant force on the wall.

2. The Tainter gate for a water channel is 1.5 m wide and in the closed position, as shown. Determine the magnitude of the resultant force of the water acting on the gate. Also, what is the smallest torque T that must be applied to open the gate if its weight is 30 kN and its center of gravity is at G.

Answer 1

1.

The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the wall.

$w_{A} = \rho _{w}gh_{A}b = 196.2 ~ \frac Nm.$ Thus,

$​ F_h=\frac 12w_{A}h_{A} = 392.4~k N .$ It acts at

$\breve{y} = \frac{1}{3}h_{A} = \frac{4}{3}~m .$

The vertical component of the resultant force is equal to the weight of the column of water above surface AB of the wall. The volume of this column of water is

$V = \frac{1}{3} ahb = 26.67 ~m^{3}.$ Thus,

$F_{v} = \rho _{w}gV = 261.6 ~kN.$ It acts at

$\breve{x} = \frac{3}{10}a =\frac{6}{5} ~m .$ The magnitude of the resultant force is

$F_{R} = 2\sqrt{F_{h}^{2} + F_{v}^{ 2}} = 472 ~kN.$ And its direction is

$\theta = \arctan\frac{F_{v}}{F_{h}} ​ = 33.7°.$

2.

The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the gate.

$w_{1} = \rho_{w}gh_{1}b = 29.43~\frac{kN}m,$

$w_{2} = \rho _{w}gh_{2}b =48.35~\frac{kN}m.$ Then

$(F_{h})_{1} = 37.83 ~kN,\\ (F_{h})_{2} =12.16~ kN,$

$F _ h ​ =(F _ h ​ ) _ 1 ​ +(F _ h ​ )_2=50~kN.$ Also,

$\breve{y}_{1} = 0.64 m ,$ $\breve{y}_{2} = 0.86 ~m .$

The vertical component of the resultant force is equal to the weight of the imaginary column of water above the gate but acts upward. The volume of this column of water is

$V = 2.02 ~m^{3}.$

$F_{v} = \rho _{w}g V = 19.83~ kN.$

$\bar{r} = 1.31m ,\\ \breve{x}_{2} = 1.23 m , \\ \breve{x}_{1} = 1.77 m , \\ \breve{x}_{3} = 1.02 m .$

Thus, $F_v$ ​acts at

$\bar{x} = 1.75~ m .$

The magnitude of the resultant force is

$F_{R} = \sqrt{F_{h}^{2} + F_{v}^{2}}= 53.8~ kN.$

$\sum M_{0} = 0,$

$T=42.3~kN\cdot m.$

The resultant force acting on the give must act normal to its surface. Therefore it produces moment about the pin.