Question 1

A 1.2-m high by 2.8-m wide, vertical rectangular gate is used to control the flow of water from a reservoir to a grinding mill. A frictionless hinge runs along the bottom of the gate. The
Minimum force acting normal to and at the very top of the gate needed to hold the gate shut if
The water is at the very top of the gate, is nearest:

A) 6590 N
B) 9800 N
C) 10 000 N
D) 13 300 N How high above the hinge must the water be for the gate
To just open?

6590 N

Accepted Answer

$6590 \mathrm {~N}$
The pressure distribution on the gate is linear, increasing from zero at the top to $p = \gamma h = 9810 \times 1.2 = 11772 \mathrm {~Pa}$ at the bottom. Hence, the average pressure over the gate is 5886 Pa providing a force of $5886 \times 1.2 \times 2.8 = 19780 \mathrm {~N}$ acting on the gate. This force acts through the centroid of the triangular pressure distribution (see Fig. $2.9 ) , 0.8 \mathrm {~m}$ from the top. The force on the top of the gate $( 1.2 \mathrm {~m}$ above the hinge and $0.8 \mathrm {~m}$ above the force) is found by taking moments about the hinge:
$$1.2 P = F \times d = 19780 \times 0.4 . \quad \therefore P = \underline { 6590 \mathrm {~N} }$$

Question 2

A horizontal 80-cm-diameter hatch is located on the top of a submersible designed for a diver to escape. If the submersible is 100 m below the surface of the ocean, estimate the minimum
Force required to open the hatch? $\text { Assume } S _ { \text {saltwater } } = 1.02 \text {. }$

A) 300 kN
B) 400 kN
C) 500 kN
D) 600 kN

500 kN

Accepted Answer

$500 \mathrm { kN }$
The pressure at a depth of $100 \mathrm {~m}$ is $p = \psi h = ( 9810 \times 1.02 ) \times 100 = 10 ^ { 6 } \mathrm {~N}$. The force due to the saltwater acts at the centroid of the circular hatch since it is horizontal. Moments about the hinge, which is assumed to be on the circumference of the circular hatch with the force $F$ on the opposite side, provides
$$0.8 P = F \times 0.4 = 10 ^ { 6 } \times 0.4 . \quad \therefore P = 500000 \mathrm {~N} \text { or } 500 \mathrm { kN }$$

Question 3

The force P needed to hold the 2-m-radius circular arc gate in the position shown is nearest:&#10; &#10;A) 58 kN&#10;B) 49 kN &#10;C) 39 kN&#10;D) 27 kN

Accepted Answer

$49 \mathrm { kN }$
Since the force due to the pressure on each area element of the circular arc acts through the center of the arc, we can move the horizontal and vertical components to that center of the arc. Sketch a free-body diagram of the water above the gate, as in Fig. 2.11b. (Combine $F _ { H }$ and $F _ { V }$ to make $\mathbf { F }$, then $\mathbf { F }$ will pass through the center of the arc at which point it can be decomposed into $F _ { H }$ and $F _ { V }$.) They produce the same moments when located at the center as if they were located where they actually act. Take moments about the hinge so that only $P$ and $F _ { H }$ produce moments:
11edd9e5_0f18_92b9_bca1_492c8b35e91e_TB8827_11 
For practice, maintain $F _ { H }$ and $F _ { V }$ in their normal positions and solve for $P$. We use Eq. 2.4.28 to help find the distance $d _ { H }$ above the hinge where $F _ { H }$ acts:
$$d _ { H } = 2 - y _ { p } = 2 - \left( \bar { y } + \frac { \bar { I } } { A \bar { y } } \right) = 2 - 1 - \frac { 3 \times 2 ^ { 3 } / 12 } { ( 3 \times 2 ) \times 1 } = 0.667 \mathrm {~m}$$
$F _ { V }$ acts through the center of gravity of the quarter circle. The distance $d _ { V }$ from the hinge to $F _ { V }$ is :
$$d _ { V } = \frac { 4 r } { 3 \pi } = \frac { 4 \times 2 } { 3 \pi } = 0.849 \mathrm {~m}$$
Finally, moments about the hinge results in
$$\begin{array} { c } 
2.4 P = 0.667 \times [ 9810 \times 1 \times 6 ] + 0.849 \times \left[ 9810 \times \left( \pi \times 2 ^ { 2 } / 4 \right) \times 3 \right] \
\therefore P = \underline { 49060 \mathrm {~N} }
\end{array}$$
Obviously, it's much simpler to move $F _ { H }$ and $F _ { V }$ to the center of the arc.
12

Question 4

The pressure in the oil (S = 0.86) pipe is 12 kPa. The pressure&#10;A) &#8722;34.8 kPa in the water pipe is nearest:&#10; &#10;B) &#8722;28.8 kPa&#10;C) &#8722;18.8 kPa&#10;D) &#8722;13.1 kPa

Accepted Answer

$- 13.1 \mathrm { kPa }$
Place " $a$ " and " $b$ " as shown. Then $p _ { a } = p _ { b }$ since $a$ and $b$ are at the same elevation in the same fluid. The manometer then tells us that
11edd9e4_f0a8_0fe8_bca1_0d4fe27355ad_TB8827_11
$$\begin{array} { c } 
p _ { a } = p _ { b } \
p _ { \text {oil } } + ( 0.2 + 0.08 ) \times 9810 \times 0.86 = p _ { \text {water } } + 0.08 \times 9810 + 0.2 \times 9810 \times 13.6 \
\therefore p _ { \text {water } } = 12000 + 2362 - 785 - 26683 = - 13110 \mathrm {~Pa} \text { or } - 13.11 \mathrm { kPa }
\end{array}$$

Question 5

An object requires a force of 30 N to hold it under water. The object weighs 90 N in air. Its density is nearest: (A)

850 \mathrm {~kg} / \mathrm { m } ^ { 3 }

(B)

800 \mathrm {~kg} / \mathrm { m } ^ { 3 }

(C)

750 \mathrm {~kg} / \mathrm { m } ^ { 3 }

(D)

700 \mathrm {~kg} / \mathrm { m } ^ { 3 }

Accepted Answer

$750 \mathrm {~kg} / \mathrm { m } ^ { 3 }$
A free-body diagram would show $30 \mathrm {~N}$ and $90 \mathrm {~N}$ acting down and the buoyant force $F _ { B }$ acting up (see Eq. 2.4.36):
11edd9e5_3bab_669a_bca1_513529ee91a2_TB8827_11
11edd9e5_4b3e_379b_bca1_4150e353eb26_TB8827_11

Question 6

A gate automatically opens when the water gets too high.  &#10;A) 0.6 m&#10;B) 1.2 m&#10;C) 1.8 m&#10;D) 2.4 m

Accepted Answer

$1.2 \mathrm {~m}$
The force of the water on the vertical gate acts at the hinge just as the gate is about to open. If the force moves above the hinge, the gate will open. The distance $0.6 \mathrm {~m}$ below the hinge means that the water surface is $2 \times 0.6 = \underline { 1.2 \mathrm {~m} }$ above the hinge for the gate to open. See Fig. 2.10.

Deck 4: The Integral Forms of the Fundamental Laws