Fluid-Mechanics-and-Hydraulics.pdf · version 1 PDF

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A(0)B(2)El. 2El. 0ABhIf 6 m 3 of soil weight 47 kn , calculate the following.a) specific weighty =47 6 = 7 kN/m 3 b) Density:p =7833 9 = 798 kg/m 3Problem 2 - HydraulicsFor the open tank with piezometers attached on the side, contains two different liquids. ➀ Find the elevation of the liquid in piez...

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Problem 1 - Hydraulics If 6 m3 of soil weight 47 kn , calculate the following. a)

specific weight

47 y = 6 = 7.833 kN/m3 b)

c)

Density:

specific gravity:

7.833 sp.gr. = 9.81 = 0.80

7833 p = 9.81 = 798 kg/m3

Problem 2 - Hydraulics For the open tank with piezometers attached on the side, contains two different liquids. ➀ Find the elevation of the liquid in piezometer A. ➁ Find the pressure at the bottom of the tank. ➂ Find the elevation of the liquid in piezometer B. Solution: A B ➀ Elevation of liquid at piezometer A = 2m. El. 2.0m ➁ Pressure at the bottom of the tank: Pbottom = 9.81(0.72)(2 - 0.3) + 9.81(2.36)(0.3) Pbottom = 18.95 kPa ➂ Elevation of the liquid at piezometer B: 0 + 9.81(0.72)(1.7) + 9.81(2.36)(0.3) - 2.36(h)(9.81) = 0 h = 0.819 Elevation 0 + 0.819 = 0.819 m.

A(0.72) h

El. 0.3m

B(2.36)

Problem 3 - Hydraulics Oil of sp.gr. 0.750 flows through the nozzle shown and deflects the mercury in the U-tube gage. ➀ Determine the value of h if the pressure at A is 142 kPa. ➁ What is the pressure at B. ➂ If the diam. at A is 200 mm, compute the velocity at A if oil flows at a rate of 36000 liters per minute. Solution: ➀ Value of h: 142 + 9.81(0.75)(h + 2.75) - 13.6(9.81)(h) = 0 142 - 126.0585h + 20.233 = 0 h = 1.29 m.

A 2.75

B

➁

Pressure at B: PB = 142 + (9.81)(0.75)(2.75 + 1.29) PB = 177.72 kPa

➂

Velocity at A: Q = 36000 liters/minute Q = 0.60 m3/s Q = AV 0.60 =

π (0.2)2 V 4

V = 19.10 m/s

Hg(13.6)

D C

h

Problem 4 - Hydraulics A cylinder glass tubing 2.8 cm. inside diameter and 90 cm long with one end closed is immersed vertically with the open end down into a tank of cleaning solvent (sp.gr. = 0.73) until only 5 cm. of its length remain above the liquid surface. If the barometric pressure is 1 kg/cm2 and neglecting vapor pressure, ➀ How high will the fluid rise in the tube? ➁ Compute the height of air inside the glass. ➂ What force required to maintain equilibrium. Solution: ➀ Height of fluid rise in the tube: P1 = 1 kg/cm2 P2 = P1 + wh m3 1000(0.73)(h) Kg ) ( P2 = 1 + m3 1003 cm3 (100)3 P2 = 1 + 0.00073h kg/cm2

2.8cm 5cm 90cm h x

Cleaning solvent (S=0.73)

π V1 = 4 (2.8) 2 (90) V1 = 554.18 cm3 π V2 = 4 (2.8) 2 (h + 5) V2 = 6.158 (h + 5)

Using Boyle's Law: P 1V 1 = P 2V 2 1(554.18) = (1 + 0.00073h)(6.158)(h + 5) 554.18 6.158 = (1 + 0.00073h) (h + 5) 89.99 = h + 0.00073h2 + 5 + 0.00365h 89.99 = 0.00073h2 + 1.00365h + 5 h2 + 1374.86h - 116424.657 = 0 h = 80 cm. x = 90 - 5 - 80 x = 5 cm.

➁ Height of air inside the glass: h + 5 = 85 cm. ➂ Force required equilibrium: ρA = γ h A

to

π(0.028) 2 F = 9810 0.80 4 F = 4.83 N

maintain

Problem 5 - Hydraulics The 8 ft. diam. cylinder weighs 500 lb. and rests on the bottom of a tank that is 3 ft. long. Water and oil are poured into the left and right portions of the tank to depths of 2 ft. and 4 ft. respectively. ➀ Find the horizontal component of the force that will kept the cylinder touching the tank at B. ➁ Find the vertical component of the force that will push up the cylinder. ③ Compute the force that will keep the cylinder touching the tank at B. Solution: ➀ Horizontal component of the force that will kept the cylinder touching the tank at B: P1 = γw h A 500 lb P1 = 62.4 (1) (2) (3) Oil (0.75) P1 = 374 lb. 4 2

H2O P1

P2 = γw h A P2 = 62.4 (2) (4) (3) (0.75) P2 = 1123 lb. Ph = P2 - P1 H2O Ph = 1123 - 374 2 Ph = 749 lb. ➁ Vertical component of the force that will push up the cylinder:

π(4) 2 Fv2 = 62.4 4 (3)(0.75) = 1764 lb.

π(4) 2 (60 2(4)Sin 60˚ Fv1 = 62.4 [ 360 ] (3) 2 Fv1 = 920 Fv = 1764 + 920 Fv = 2684 lb. ③ Force that will keep the cylinder touching the tank at B: FB + 500 = 2684 FB = 2184 lb. (downward)

60˚

2 2

P2

4

B Fv1

Fv2

500 lb

4 4 60˚

2 2

4

B Fv1

Oil (0.75)

Fv2

Problem 6 - Hydraulics The 6 ft. diameter cylinder weighs 5000 lb. and is 5 ft. long. ➀ ➁ ③

Determine the upward force due to the effect of oil in the left side. Compute the horizontal reaction at A. Compute the vertical reaction at B.

➀

Solution: Upward force due to the effect of oil in the left side: E

Pv = γw V

C

Oil D (0.80)

2

(π)(3) (5) Pv = 62.4(0.8) 2

A

B

Pv = 3529 lb. (upward)

➁ Horizontal reaction at A: Ph = γw h A Ph = 62.4 (.80)(3)(6)(5) Ph = 4493 lb. RA = 4493 lb. to the left.

5

E

C

D

6

A

Ph

③ Vertical reaction at B: RB + Pv = 5000 RB = 5000 - 3529 RB = 1471 lb. (upward)

B Pv RB

RA

Problem 7 - Hydraulics From the figure shown. ➀ Compute the horizontal component of the hydrostatic force. ➁ Compute the vertical component of the hydrostatic force. ③ Compute the location of the vertical component horizontally from B. Solution: ➀ Horizontal component of the hydrostatic force: Ph = γw h A Ph = 62.4 (3)(6)(1) Ph = 1123 lb.

Hinge 1

6

A

6’

C

4’ Ph 1/3(6)=2

6’ B

R

➁ Vertical component of the hydrostatic force: (π)(6) 2 (1) Pv = 62.4 4 Pv = 1764 lb. ③ Location of the vertical component horizontally from B: 4r x = 3π

4(6) x = 3π x = 2.55 ft. from B

Pv

x

Problem 8 - Hydraulics A wooden storage vat, 20 ft. outside diam. is filled with 24 ft. of brine sp.gr. = 1.06. 1 The wood staves are bound by flat steel bands 2 in. wide by 4 inch thick, whose allowable stress is 16,000 psi. ➀ ➁ ➂

What is the bursting pressure? What is the tensile force of steel bands. What is the spacing of the bands near the bottom of the vat, neglecting any initial stress?

Solution: ➀ Bursting pressure: 2T = F 2T = P D(S)

20’

T 24’

1/4” D T

sp.gr.=1.06

2” 1/4”

P = γw h P = 62.4 (1.06)(24) P = 1587.5 psf.

T F S

T D

➁ Tensile force of steel bands: T = Ss A ⎛ 1⎞ T = 16000 (2) ⎜⎜4⎟⎟ ⎝ ⎠ T = 8000 lb. ➂ Spacing of the bands near the bottom of the vat: 2T = P D S 2(8000) = 1587.5 (20) S S = 0.504 ft. S = 6.05 inches

Problem 9 - Hydraulics A concrete dam retaining water is shown. If the specific weight of concrete is 23.5 kN/m3. ➀ ➁ ➂

Find the factor of safety against sliding. Find the factor of safety against overturning if the coeff. of friction is 0.48. Find the max. and min. pressure intensity

Solution: ➀ Factor of safety against sliding: Considering 1 meter strip: ___ P=γhA P = 9.79 (3)(6)(1) P = 176.20 kN

2m

7m 6m

4m 2

2

3

2(7) w1 = 2 (1)(23.5) w1 = 164.5 kN w2 = 2 (7)(1)(23.5) w2= 329 kN Ry = 164.5 + 329 Ry = 493.5 kN Factor of safety against sliding: uRy F.S. = P 0.48 (493.5) F.S. = 176.20 F.S.= 1.34

W2

7m

6m

W1 P 2 1.333 x R y 2

2

Pmin

Pmax 1.73 e Ry

Problem 9 – Hydraulics (Continuation) ➁

Factor of safety against overturning. O.M. = P(2) O.M. = 176.20(2) O.M. = 352.4 kN.m. R.M. = w1 (1.333) + w2 (3) R.M. = 164.5 (1.333) + 329(3) R.M. = 1206 kN.m. R.M. F.S. = O.M. 1206 F.S. = 352.4 F.S. = 3.42

➂

The max. and min. pressure intensity: ___ Ry x = R.M. - O.M. 1206 - 352.4 x= 493.5 x = 1.73 e = 2 - 1.73 e = 0.27 Ry ⎛ 6e ⎞ Pmin. = B ⎜⎜1 - B ⎟⎟ ⎝ ⎠ 493.5 6(0.27) Pmin. = [1 4 ] 4 Pmin. = 73.4 kN/m2. R ⎛⎜ 6e⎞⎟ Pmax. = B ⎜1 + B ⎟ ⎝ ⎠ 6(0.27) 493.5 Pmax. = 4 [1 + 4 ] Pmax. = 173.34 kN/m2.

Problem 10 - Hydraulics A prismatic object 200 mm thick by 200 mm wide by 400 mm long is weighed in water at a depth of 500 mm and found to be 50 N. ➀ ➁ ➂

Find its weight in air. Find its specific gravity. Find its specific weight.

Solution: ➀ Weight in air: W = BF + 50 BF = 0.2(0.2)(0.4)(9810) BF = 156.96 N W = 156.96 + 50 W = 206.96 N ➁ Specific gravity: 206.96 Sp.gr. = 156.96 Sp.gr. = 1.32 ➂

Specific weight: Sp.wt. 9.81 = 1.32 Sp.wt. = 12.94 kN/m3 Check: 206.96 3 0.2(0.2)(0.4) = 12935 N/m = 12.94 kN/m3

Problem 11 - Hydraulics ➀ ➁ ➂

What fraction of the volume of a solid object of sp.gr. 7.3 floats above the surface of a container of mercury? If the volume of the object below the liquid surface is 0.014 m3, what is the wt. of the object. What load applied vertically that would cause the object to be fully submerged?

Solution: ➀ Fraction of vol. of a solid object above the mercury: W = BF (7.3)(V)(9.81) = V1 (9.81)(13.6) V = 1.863 V1 V1 = 0.536 V V2 = V - 0.536 V V2 = 0.464 V V2 V = 0.464

V2 V1

Mercury (13.6)

➁ Wt. of object: V1 = 0.014 V = 1.863(0.014) V = 0.026 m3 W = 0.026(9.81)(7.3) W = 1.86 kN ➂

Load to cause the object to submerged: P = V2 (9.81)(13.6) P = 0.464(0.026)(9.81)(13.6) P = 1.61 kN

Problem 12 - Hydraulics A hydrometer weighs 0.002 kg has a stem at the upper end, which is 3 mm in diameter. How much deeper will it float in oil (sp.gr. = 0.78) than in alcohol having sp.gr. of 0.82? Solution: In alcohol: (sp.gr. = 0.82) BF = W (9810)(0.82)V1 = 0.002(9.81) V1 = 2.44 x 10-6 m3 V1 = 2440 mm3 In oil (sp.gr. = 0.78) BF = W 9810(0.78)V2 = 0.002(9.81) V2 = 2.56 x 10-6 m3 V2 = 2560 mm3 V = V1 – V2 V = 2560 – 2440 V = 120 mm3 π 2 (3) h = 120 4 h = 16.98 mm

h

w.s.

Alcohol (sp. gr. = 0.82)

w.s.

Oil (sp. gr. = 0.78)

Problem 13 - Hydraulics A piece of wood of sp.gr. 0.651 is 3 in. square and 5 ft. long. ➀

➁ ➂

What is the volume of lead having a unit weight of 700 pcf that should be fastened at one end of the stick so that it will float upright in 1 ft. out of water? Determine the weight of the lead? Determine the total weight of lead and the wood.

Solution: ➀ Volume of lead: w1 + w2 = BF1 + BF2 ⎛ 3 ⎞⎛ 3 ⎞ 0.651 ⎜⎜ ⎟⎟⎜⎜12⎟⎟ (5)(62.4) + 700 V ⎝ 12⎠ ⎝ ⎠ ⎛ 3 ⎞⎛ 3 ⎞ = ⎜⎜12⎟⎟⎜⎜12⎟⎟(4) 62.4 + 62.4 V ⎝ ⎠⎝ ⎠ V = 0.00456 ft3. ➁ Weight of lead: w=VxD w = 0.00456 (700) w = 3.19 lb. ➂ Total weight: 3 3 = 3.19 + 0.651 ( 12 )( 12 )(5)(62.4) = 15.88 lb.

Problem 14 - Hydraulics

W1 3”

3” 1’

BF1

4’

W2

Lead

BF2

A piece of wood floats in water with 50 mm projecting above the water surface. When placed in glycerine of sp.gr. 1.35, the block projects 75 mm above the liquid surface. ➀ ➁ ➂

Find the height of the piece of wood. Find the sp.gr. of wood. Find the weight of the wood if it has a cross sectional area of 200 mm x 200 mm.

Solution: ➀ Height of wood: W = 9.81(A)(h - 0.05) S h (9.81)(A) = 9.81A (h - 0.05) S h = h - 0.05 W = 9.81(1.35) A (h - 0.075) S h (9.81) A = 9.81(1.35) A (h - 0.075) S h = 1.35(h - 0.075) h - 0.05 = 1.35 (h - 0.075) 0.35h = 0.05125 h = 0.146 m.

0.05

h-0.05

water

0.075

➁ Sp.gr. of wood: S h = h - 0.05 S (0.146) = 0.146 - 0.05 S = 0.658 ➂

Wt. of wood: W = 9.81(0.658)(0.146)(0.2)(0.2) W = 0.038 kN W = 38 N

h-0.075

Glycerine

Problem 15 - Hydraulics An open horizontal tank 2 m high, 2 m. wide and 4 m. long is full of water. ➀ ➁ ➂ ➀

➁

➂

How much water is spilled out when the tank is accelerated horizontally at 2.45 m/sec2 in a direction parallel with its longest side? What is the force acting on the side with the greatest depth? Compute the required accelerating force. Solution: Volume of water spilled Water spilled out: θ a x=1 tan θ = g θ 2 1 2.45 tan θ = 9.81 4 1 tan θ = 4 2m x 1 1m F1 F2 4 =4 x = 1 m. 4 1(4)(2) Vol. of water spilled = 2 2m Vol. of water spilled = 4 cu. m. 1m F1 F2 Force acting on the side with the greatest depth. F1 = γw h1 A1 F1 = 9.81 (1)(2)(2) F1= 39.24 kN Required accelerating force: F2 = γw h2 A2 1 F2 = 9.81 (2 ) (1)(2) F2= 9.81 Accelerating force = 39.24 - 9.81 Accelerating force = 29.43 kN Check: F=ma W F = (g ) a [2(2)(4) - (4)] (9.81) (2.45) = 29.4 kN F= 9.81

Problem 16 - Hydraulics A 75 mm diameter pipe, 2 meter long is just filled with oil, specific gravity is 0.855 and then capped. Placed in a horizontal position, it is rotated at 27.5 rad/sec. about a vertical axis 0.5 meter from one end. ➀ What is the pressure in kPa at the far end of the pipe? ➁ What is the pressure in kPa at the other end of the pipe? ➂ What is the pressure at the midpoint of the pipe? Solution: ➀ Pressure in kPa at the far end of the pipe: P2 = γw h h = y - y1 ω2 r2 ω y= 2g rad ω = 27.5 sec h y (27.5)2 (2.5)2 y = 2(9.81) h1 1 2 y2 y = 240.9 m. y1 ω2 r12 y1 = 2g r1 0.5m 2m (27.5)2 (0.5)2 r=2.5m y1 = 2(9.81) r =1.5 m. y1 = 9.64 h = 240.90 - 9.64 = 231.26 m. P2 = γw h P2 = 9.81(0.822) (231.26) = 1864.8 kPa ➁ Pressure in kPa at the other end of the pipe: P1 = 0 ➂ Pressure at the midpoint of the pipe: ω2 r2 y2 = 2g (27.5)2 (1.5)2 y2 = 2(9.81) y2 = 86.73 Pm = γw h1 Pm = 9.81(0.822)(86.73 - 9.64) = 621.6 kPa

Problem 17 - Hydraulics A turbine is rated at 450 KN when the flow of water through it is 0.609 m3/s. Assuming an efficiency of 87%, what head is acting on the turbine? Solution: Power = QWE x Efficiency 450 = 0.609(9.81)E(0.87) E = 86.6 m.

Problem 18 - Hydraulics Oil of sp.gr. of 0.75 is flowing through a 150 mm pipe under a pressure of 103 kPa. If the total energy relative to a datum plane 2.40 m. below the center of the pipe is 17.9 m. kN/kN, determine the flow of oil. Solution: Energy =

V12 2g

+

P1 w

+Z

V2 103 + 2.40 17.9 = + 2(9.81) 0.75(9.81) V = 5.43 m/s Q = AV π (0.15)2 (5.43) 4 Q = 0.096 m3 / ss

Q=

Problem 19 - Hydraulics A 150 mm diameter jet of water is discharge from a nozzle into the air. The velocity of the jet is 36 m/s. Find the power in the jet? Solution: V2 E= 2g (36) 2 = 66.06 E= 2(9.81) Power = QwE Q = AV π Q = (0.15)2 (36) = 0.636 m3 / s 4 Power = 0.636(9.81)(66.06) Power = 412 kN

Problem 20 - Hydraulics For laminar conditions, what size of pipe will deliver 0.0057 m3/s of oil having viscosity of v = 6.09x10-6 m2/s. Solution: Q = AV π 2 dV 4 0.00228 V= πd2 Rendds no. for laminar flow = 2000 Vd R= V ⎛ 0.0228 ⎞ (d) 2000 = ⎜ 2 ⎟ ⎝ πd ⎠ 6.09x10−6 d = 0.596 m. 0.0057 =

Problem 21 - Hydraulics A 1m. diameter new cast iron pipe C = 130 is 845 m. long and has a head loss of 1.11 m. Find the discharge capacity of the pipe according to Hazen Williams Formula? Solution: 10.64LQ1.85 HL = CD4.87 10.64(845)Q1.85 1.11 = 130(1)4.87 Q1.85 = 0.016 Q = 1.01 m3 / s

Problem 22 - Hydraulics A barrel containing water weighs 1.260 kN. What will be the reading on the scale if a 50 mm by 50 mm piece of wood is held vertically in the water to a depth of 0.60 m. Solution: BF = 0.05(0.05)(0.60)(9.81) BF = 0.015 kN Scale reading = 0.015 + 1.260 = 1.270 kN

Problem 23 - Hydraulics A piece of wood of sp.gr. 0.651 is 80 mm square and 1.5 m. long. How many Newtons of lead weighing 110 kN/m3 must be fastened at one end of the stick so that it will float upright with 0.3 m. out of water? Solution: W1 = (0.08)2(1.5)(0.651)(9.81) W1 = 0.061 kN BF1 = (0.08)2(1.2)(9.81) BF1 = 0.075 kN W2 = V2 D2 W2 = 110 V2 BF2 = 9.81V2 W1 + W2 = BF1 + BF2 0.061 + 110V2 = 0.075 + 9.81V2 100.19V2 = 0.014 V2 = 0.00014 m3 W2 = 110(0.00014) = 0.0154 kN = 15.4 N

0.08 0.08 0.3

1.2

W1

W2

BF1

BF2

Problem 24 - Hydraulics To what depth will a 2.4 m. diameter log 4.6 m. long and sp.gr. of 0.425 sink in fresh water? Solution:

r =1.2

r =1.2 w.s.

w.s.

h

h

4.6 W = π(1.2)2 (4.6)(0.425)(9.81) W = 1.922655(4.6)(9.81) Area of segment: (shaded section) ⎛ Sin 2θ ⎞ A = r2 ⎜ θ 2 ⎟⎠ ⎝ Try θ = 83.1667˚ θ = 1.4515 rad Sin 2θ = 0.11813 2 A = (1.2) 2 ⎡⎣1.4515 - 0.11813 ⎦⎤ A = 1.92 BF = A(4.6)(9.81) BF = W 1.92(4.6)(9.81) = 1.92265(4.8)(9.81) 1.92 = 1.92265

almost equal

Use ø = 83.1667˚ Depth of floatation = r - r Cos θ h = 1.2 - 1.2 Cos 83.1667˚ h = 1.057

Problem 25 - Hydraulics A barge with a flat bottom and square ends has a draft of 1.8 m. when fully loaded and floating in an upright position. The barge is 7.6 m. wide and 12.8 m. long and a height of 3 m. The center of gravity of the barge is 0.30 m. above the water surface if the barge is stable. ➀ ➁ ➂

Determine the distance of the metacenter above the center of buoyancy. Determine the distance of the metacenter above the barge center of gravity. What is the righting moment in water when the angle of heel is 12˚? M

➀

Solution: Distance of the metacenter above the center of buoyancy I MBo = V 12.8(7.6) 3 I= 12 I = 468.24

MBo w.s.

C.G.

0.3 m 0.9 m

3m 1.8 m Bo

0.9 m

7.6 m

V = 7.6(12.8)(1.8) V = 175.10 468.24 175.10 MBo = 2.67 m.

MBo =

12.8 m

Distance of the metacenter is 2.67 m. above the center of buoyancy. 7.6 m

➁

Distance of the metacenter above the center of gravity of the barge Distance = 2.67 – 0.3 – 0.9 = 1.47 m.

1.47

M W

➂

Righting moment in water when the angle of heel is 12˚ x = 1.47 Sin 12˚ x = 0.3056 BF = W BF = 7.6(1.8)(12.8)(9.81) BF = 1717.77 kN Righting moment = BF(x) Righting moment = 1717.77(0.3056) Righting moment = 584.95 kN.m.

1.2 1.8

12˚ C.G.

x 0.3

Bo

0.9

7.6

BF

Problem 26 - Hydraulics A tank is 1.5 m. square and contains 1.0 m. of water. How high must its sides be if no water is to be spilled when the acceleration is 4 m/s2 parallel to a pair of sides? Solution: a tan θ = g

y

 

y tan θ = 0.75 y 4 = 9.81 0.75 y = 0.306 m. Its sides must be = 0.306 + 1 = 1.306 m. high (deep)

1.5 m 1m

1.5 m

Problem 27 - Hydraulics A rectangular tank 6 m. long by 1.8 m. deep and 2.10 m. wide is filled with water accelerated in the direction of its length at a rate of 1.52 m/sec2. How many liters of water are spilled out? Solution: a tan θ = g h a = 6 g 6(1.52) 9.81 h = 0.93 Vol. spilled out: 0.93(6) V= (2.10) 2 V = 5.859 m3

h 

h=

V = 5859 liters

6m

1.8 m

Problem 28 - Hydraulics

A rectangular tank 6 m. long by 1.8 m. deep is 2.1 m. wide contains 0.90 m. of water. Find the unbalanced force necessary to accelerate the liquid mass in the direction of the tank length to accelerate 2.45 m/s2.

Solution: F = ma m=

W g

m=

(6)(2.1)(0.90) (9810) 9.81

m = 11340 kg F = 2.45(11340) F = 27783 N

Problem 29 - Hydraulics A cubic tank is filled with 1.5 m. of oil, sp.gr. 0.752. ➀ Find the force acting on the side of the tank when the acceleration is 4.9 m/s2 vertically upward. ➁ Find the force acting on the side of the tank when the acceleration is 4.9 m/s2 vertically downward. ➂ Determine the pressure at the bottom of the tank when the acceleration is 9.81 m/s2 vertically downward.

a=4.9 m/s2

Solution: ➀ Force acting on the side of the tank when the acceleration is 4.9 m/s2...