Question 1 |
A sector gate is provided on a spillway as shown in the figure. Assuming g = 10\: \: m/s^{2} the resultant force per meter length (expressed in kN/m) on the gate will be _______ .
127 | |
125 | |
22.7 | |
87.5 |
Question 1 Explanation:
\begin{aligned} F_{x} &=\rho g \bar{h} A_{v} \\ &=\left(10^{3}\right)(10)\left(\frac{5}{2}\right)(5 \times 1)\\ &=125 \mathrm{kN}\text{ per unit width}\\ F_{y} &=\rho g \forall \\ \text { where } \forall &=\frac{\pi(5)^{2}}{6}-\left(\frac{1}{2} \times 5 \times 5 \cos 30^{\circ}\right)=2.264 \mathrm{m}^{3} \\ F_{y} &=(10)^{3}(10)(2.264)=22.64 \mathrm{kN} \end{aligned}
Resultant force per unit width,
\begin{aligned} F_{R}&=\sqrt{F_{x}^{2}+F_{y}^{2}}\\ &=\sqrt{125^{2}+22.64^{2}}=127.03 \mathrm{kN} \end{aligned}
Question 2 |
A triangular gate with a base width of 2 m and a height of 1.5 m lies in a vertical plane. The top vertex of the gate is 1.5 m below the surface of a tank which contains oil of specific gravity 0.8. Considering the density of water and acceleration due to gravity to be 1000 kg/m^{3} and 9.81 m/s^{2}, respectively, the hydrostatic force (in kN) exerted by the oil on the gate is__________.
22.55 | |
29.43 | |
24.58 | |
26.43 |
Question 2 Explanation:
\begin{aligned} \bar{h} &=1.5+1.5\left(\frac{2}{3}\right)=2.5 \mathrm{m} \\ F_{R} &=\rho g \bar{h} A \\ &=(800)(9.81)(2.5)\left(\frac{1}{2} \times 2 \times 1.5\right) \\ &=29430 \mathrm{N}=29.43 \mathrm{kN} \end{aligned}
Question 3 |
The necessary and sufficient condition for a surface to be called as a free surface is
no stress should be acting on it | |
tensile stress acting on it must be zero | |
shear stress acting on it must be zero | |
no point on it should be under any stress |
Question 4 |
Cross-section of an object (having same section normal to the paper) submerged
into a fluid consists of a square of sides 2 m and triangle as shown in the figure.
The object is hinged at point P that is one meter below the fluid free surface. If
the object is to be kept in the position as shown in the figure, the value of 'x'
should be
2 \sqrt{3} | |
4 \sqrt{3} | |
4m | |
8m |
Question 4 Explanation:
The vertical force on the surface bounded by square and triangle would be respectively,
F_{1}=4(S-1) \gamma_{w} for square edge and acting at 1 m from P
F_{2}=x(S-1) \gamma_{w} for inclined edge of triangle and acting at x / 3 from P
Taking moments of both the forces about P, we get,
\begin{aligned} F_{ 1} \times 1 &=F_{2} \times \frac{x}{3} \\ \Rightarrow \quad 4(\mathrm{S}-1) \gamma_{w} &=x(S-1) \gamma_{w} \times \frac{x}{3} \\ \Rightarrow \quad x^{2} &=12 \\ \Rightarrow \quad x &=2 \sqrt{3} \mathrm{m} \end{aligned}
F_{1}=4(S-1) \gamma_{w} for square edge and acting at 1 m from P
F_{2}=x(S-1) \gamma_{w} for inclined edge of triangle and acting at x / 3 from P
Taking moments of both the forces about P, we get,
\begin{aligned} F_{ 1} \times 1 &=F_{2} \times \frac{x}{3} \\ \Rightarrow \quad 4(\mathrm{S}-1) \gamma_{w} &=x(S-1) \gamma_{w} \times \frac{x}{3} \\ \Rightarrow \quad x^{2} &=12 \\ \Rightarrow \quad x &=2 \sqrt{3} \mathrm{m} \end{aligned}
Question 5 |
The force 'F' required at equilibrium on the semi-cylindrical gate shown below is
9.81kN | |
0kN | |
19.62kN | |
none of the above |
Question 5 Explanation:
Depth of water (H)=2 m
Consider unit width, of cylinder,
\begin{aligned} F_{H} &=\rho g \bar{h} A_{h} \\ &=1000 \times 9.81 \times\left(\frac{2}{2}\right) \times(2 \times 1) \\ &=19.62 \mathrm{kN} / \mathrm{m} \text { width } \end{aligned}
Vertical component,
\begin{aligned} F_{v}&=\rho g(\text { volume })\\ &=1000 \times 9.81\left(\frac{\pi R^{2}}{2} \times 1\right) \\ &=15.41 \mathrm{kN} / \mathrm{m} \text { width } \end{aligned}
location of center of pressure of F_{h}
h^{*}=\frac{2}{3} H=\frac{4}{3} m
location of center of pressure of F_{v}
\therefore \quad \bar{x}=\frac{4 R}{3 \pi}=\frac{4 \times 1}{3 \pi}=0.424 \mathrm{m}
Moment about hinge,
\begin{aligned} \therefore \quad F_{H} \times\left(h^{*}-\bar{h}\right)&=F_{V} \times \bar{x}+F \times 1 \\ \therefore 19.62 \times\left(\frac{4}{3}-1\right)&=15.41 \times 0.424+F \times 1 \\ \therefore \quad F&=0 \mathrm{kN} \end{aligned}
Consider unit width, of cylinder,
\begin{aligned} F_{H} &=\rho g \bar{h} A_{h} \\ &=1000 \times 9.81 \times\left(\frac{2}{2}\right) \times(2 \times 1) \\ &=19.62 \mathrm{kN} / \mathrm{m} \text { width } \end{aligned}
Vertical component,
\begin{aligned} F_{v}&=\rho g(\text { volume })\\ &=1000 \times 9.81\left(\frac{\pi R^{2}}{2} \times 1\right) \\ &=15.41 \mathrm{kN} / \mathrm{m} \text { width } \end{aligned}
location of center of pressure of F_{h}
h^{*}=\frac{2}{3} H=\frac{4}{3} m
location of center of pressure of F_{v}
\therefore \quad \bar{x}=\frac{4 R}{3 \pi}=\frac{4 \times 1}{3 \pi}=0.424 \mathrm{m}
Moment about hinge,
\begin{aligned} \therefore \quad F_{H} \times\left(h^{*}-\bar{h}\right)&=F_{V} \times \bar{x}+F \times 1 \\ \therefore 19.62 \times\left(\frac{4}{3}-1\right)&=15.41 \times 0.424+F \times 1 \\ \therefore \quad F&=0 \mathrm{kN} \end{aligned}
There are 5 questions to complete.