Chapter 9 Exercise 9A

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1 Chater 9 Exercise 9A Q ,000 = 0 kg/m Q ,000 = 0.75 Q.. (i) 1 m has mass 1.6 1,000,000 = 1,600,000 grammes (ii) = 1,600 kg Density = 1,600 kg/m 1,600 1,000 = 1.6 Q.. Volume = = m Mass Density = Volume 1.0 = = 9,000 kg/m Relative density = 9,000 1,000 = 9 Q. 5. Volume = Mass Density = = 1 1 m Mass = Volume Density = 1 2, = 175 kg 1 Q. 6. s v sv _ s = = 0.9 Q. 9. Volume = = 0.02 m eight = Vrg = (0.02)(1,000 g) = 2g N Q. 10. Volume = r 2 h = (0.0) 2 (0.1) = m eight = Vrg = ( )(00)g = g N Q. 11. (i) V = 1 h(r2 + Rr + r 2 ) = 1 (1)( ) = 1,026 cm = m eight = Vrg = ( )(1,000)g = 1.026g N (ii) V = 1 h(r2 + Rr + r 2 ) = 1 (6)( ) = 25 cm = m eight = Vrg = ( )(9)(g) = 0.21g N Mass (iii) Density = Volume = = = 990 Exercise 9 ecific gravity = 990 1,000 = 0.99 Q. 7. s v sv Q.. s v sv x 0.9x x x _ s = = x = x x = x x = 60 ml Q. 1. (i) Pressure = hrg = (2)(1,000)g = 2,000g N/m 2 (ii) Thrust = Pressure Area = 2,000g (2) 2 =,000g N eight = Vrg = (2) (1,000)g =,000g N Q. 2. (i) Pressure = hrg = (0.11)()g = 9.5g N/m 2 (ii) Thrust = Pressure Area = (9.5g) (0.05) 2 = 0.275g N (iii) eight = Vrg = r 2 hrg = (0.05) 2 (0.11)()g = g N 1

2 Q.. Thrust = Pressure Area = (hrg)(r 2 ) = (0.1)rg((0.2) 2 ) = rg eight = Vrg = 1 h(r2 + Rr + r 2 )rg (b) x 17 x 17 Q.. (a) = 1 (0.1){(0.05)2 + (0.05)(0.02) + (0.02) 2 }rg = g The ratio is, therefore, : 1 Pressure at = Pressure at h(1,000)g = ()g h = 6. cm Difference = 6. = 1.2 cm (b) Pressure under mercury = Pressure under oil h(9)g = ()g h = 7.16 cm Pressure at = Pressure at x ()g + (17 x)(1,600)g = 17(1,000)g x = 16. cm Q. 6. Pressure = hrg = ()(1,000)g =,000g N/m 2 Thrust = Pressure Area = (,000g)(2 2) = 12,000g N (i) 2 2 h = h = 1 m (ii) P = hrg = ( 1 )(1,000)(g) 2g N/m2 (iii) T = P A = (2g)() = 1,000g N Q. 7. (i) R 2 h = r (16)h = (27) h = 2 1 cm = m (ii) P = hrg = (0.0225)(1,000)g = 22.5g N/m 2 (iii) Thrust = P A = (22.5g)( (16)) = 60 N Q. 5. (a) 17 Pressure at = Pressure at h(1,600)g = 17(1,000)g h = 1.25 cm Difference = = cm Q.. (i) R 2 h = r (6)h = (27) h = 1 cm = 0.01 m (ii) P = hrg = (0.01)(900)g = 9g N/m 2 (iii) T = P A = (9g)( (0.06) 2 ) = 0.02 g N Q dyne cm 2 = 10 5 Newtons 10 = 0.1 N/m2 m2 Q. 10. (i) Thrust = Pressure Area = (hrg)(r 2 ) = (2)(1,2)(9.)( 22 7 ) (1.)2 = 1,920 N = 1.92 kn (ii) Volume = M r = 7 = 0. m 2,0 at x = Increase in deth r 2 x = 0. x = 0. r 2 2

3 Increase in ressure = hrg = 0. r 2 (1,2)g Increase in thrust = P A = 0. r 2 (1,2)g(r2 ) =,675 N =.675 kn Exercise 9C (i) H2 O = I 5 = = 10 (ii) = I L = () L L = 0. (iii) In liuid, L = 0.75 Aarent t. Q. 1. ouyancy = 12 = = = 12 = Q. 2. 5% 65% = Vrg = ( ) V (1,000)g = 6Vg ut = r = 6 = 0.65 Q.. 0.9, as in the last uestion Q.. = 1 15 = = = 1 = 6 = 1 16 = 2 w 2 () L = 2 Q. 5. (i) = = 0 6 = 5 Aarent weight = 0 5 = 25 N (ii) = (0.9)(5) =.5 Aarent weight = 0.5 = 25.5 N Q. 6. ( ) ( = ) = 160 Aarent weight = = 0 N Q. 7. Note: H2 O = ouyancy in water AIR H 2 O = ouyancy in liuid I = Immersed weight = ecific gravity of object L = ecific gravity of liuid 5 = H2 O 5 OIL = 6 Aarent t. = Aarent t. = I L Aarent t. = 0.75 () = 6.25 N Q.. ouyancy in water, = 0 60 = 20 ouyancy in oil, = 0 6 = (20) L = 0. Q. 9. (i) V = r = (0.) = 0.06 m OR (ii) = rvg Q. 10. V = r2 h = 2,0 ( 9 2 ) g = 90g N 9 2 m (iii) A. t. = eight ouyancy, = I = 90g 90g 2.5 = 5g N r = 0., h = 0.7 (i) V = 22 (0.) 2 (0.7) 7 V = 0 m OR V = m (ii) = rvg =,000 ( 0 ) 9. = 5,17. N

4 (iii) H2 O = I = 517. = 66.6 A. t. = eight ouyancy = = N Q. 11. Let = the weight of the body. Let = ouyancy in water. 1 = ouyancy in liuid 1 = (0.) 2 = 0. imilarily, = = 5( 0.) ( 0.75) = = 1 QED Exercise 9D Q. 1. of volume under water = r = 1,000 = 7 = Vrg = (0.1)(7)g = 75g = = 75g = 100g = + T 100g = 75g + T T = 25g Q. 2. Its secific gravity is 0., as in uestion 1 Let V = its volume, xv = volume under the liuid = Vrg = V(00)g = 00Vg = xv(1,200)g = 1,200xVg xv ince it is in euilibrium, = 00Vg = 1,200xVg x = of its mass, or %, is above the surface. Q.. (i) ( ) of the immersed art. where = weight = (1.1) ( 0.9 ) = 0.99 ince it is in euilibrium = = 0.99 = 0.99 (ii) 0.99 of its mass will be below, as in uestion 2. Q.. Its secific gravity is, as in uestion 1. Mg = 20g Let M = the mass of the glass = s L ( ) = (0.)( 20g 0.75 ) = 21 1 g = Mg + 20g 21 1 g = Mg + 20g M = 1 1 kg Q. 5. = 60 0 = 0 = 0 = 60 = 12. It is lead

5 Q. 6. xv (iii) = 12.5g, = g = T + T = 12.5 =.5g =. N Let V = its volume, xv = volume under the sea. = weight of liuid dislaced = (xv)(1,00)g = weight of the object = V(900)g ince =, xv(1,00)g = V(900g) x = 0.7 Answer: 7% Q. 7. V = (0.)(0.6)(0.) = m Q.. = Vrg R = (0.192)(2,0)g = 0g = (0.192)(1,000)g = 192g + R = 192g + R = 0g (i) V = M r R = 2g = 2,22. N T = ,0 (ii) M = Vr = m = 0.005(00) = kg Q. 9. (i) V = (0.1)(0.2)(0.0) (ii) = m M = Vr a w 1 = r r = 625 = 625 1,000 = 5 5 th of its length is submerged Deth = 5 20 = 12.5 cm Q. 10. = 0 5 = 5 = 5 = 0 = Q. 11. Total area = (0 60) + 2(0 0) + 2(60 0) 0 60 = 16,000 cm 2 = 1.6 m 2 0 Volume = eight = Vrg = m = (0.002)(,000)g = 19.2g Let X = the deth of the tank in the lake. = weight of liuid dislaced = ( X)(1,000)g = 0Xg = 0Xg = 19.2g X = 0.0 m = cm 5

6 Q. 12. Pressure at 5 cm = hg = (0.05)(900)g = 5g Pressure at 9.5 cm = (0.05)(900)g + (0.05)(1,000)g = 90g Pressure at 9.5 cm = Twice ressure at 5 cm Pressure at cm = (0.0)(900)g = 27g 5 times ressure at cm = 15g Let x = the deth (in metres) Pressure = (0.05)(900)g + (x 0.05)(1,000)g = 15g 5g + 1,000g g = 15g x = 0.1 m = 1 cm 0.05 x Q. 1. Let A = Atmosheric ressure Pressure at 1 m = 2 Pressure at 2 m A + 1(1,000)g = 2(A + 2(1,000)g) A = 10,000g = 9,000 N/m 2 Q. 1. (i) ouyancy = weight of liuid dislaced (ii) Pressure = hrg (iii) = ( 2 (1) ) (1,000)g = 2,000 g N = (2)(1,000)g = 2,000g Thrust = P A = Fu Fd = (2,000g)((1) 2 ) = 2,000g N 2,000 g = 2,000g Fd Fd =,000 g N Q. 15. (i) ouyancy = weight of dislaced liuid = ( 1 (2)2 (6) ) (900)g = 7,200 g N (ii) Pressure = hrg = (7)(900)g = 6,00g N Thrust = P A = (6,00g)((2) 2 ) = 25,200 g N (iii) = Fu Fd 7,200 g = 25,200 g Fd Fd = 1,000 g N Q. 16. h R ince r 2 = 1 (R2 ), R = 2r Volume = 1 h{(2r)2 + (2r)r + r 2 } = 7 hr2 eight = Vrg = 7 hr2 rg 6

7 Pressure at base = hrg Thrust = P A = (hrg)((2r) 2 ) = hr 2 rg Ratio, Thrust : eight = hr 2 rg : 7 hr2 rg = 12 : 7 Q. 17. (i) P = hrg = ( 1 2 )(1,000)g = 0g N/m2 (ii) T = P A = (0g) ( ( 1 ) 2 ) = g N (iii) = Vrg (ii) (iii) 0.01 Oil ater Mercury Pressure at = Pressure at (0.6)(1,000)g = (0.01)rg + (0.6)(1,600)g r = 01.6 kg/m ater 0.6 h Oil = 1 ( 1 2 ) { ( 1 2 ) 2 + ( 1 2 ) ( 1 ) + ( 1 ) 2 } (1,000)g = g eight = 7 T 1 Q. 1. Relative density of the wood = = 0.75 Volume = M r = 7 = 1 10g 15 m Q. 20. (0.6)(1,000g) = h(01.6)g h =.7 cm = 7 mm r r r g = g + 10g = 60g = eight of liuid 60g = ( 1 15 ) rg r = = 900 s = 0.9 Q. 21. (i) Pressure = hrg = rrg Thrust = P A = rrg(r 2 ) = r rg N (ii) = Vrg = ( 1_ r2 (r))rg = r rg = Fu Fd r rg = Fu r rg Fu = 2r rg N 1 ater Q. 19. (i) M = rv = 1,600 ( 1 2) = 6. tonnes = 6,00 kg 2 Mercury = Vrg = (v 1 + v 2 )(7,00)g = 7,00(v 1 + v 2 )g 1 = v 1 (1,000)g = 1,000v 1 g 2 = v 2 (1,600)g = 1,600v 2 g 7

8 ince the body is in euilibrium = 1,000v 1 g + 1,600v 2 g = 7,00(v 1 + v 2 )g Exercise 9E Q. 1. (i) Q. 2. (i) 1 5,00v 2 = 6,00v 1 v 1 v2 = 5 6 = 29 1 F + s = Taking moments about : 2 () = F(5) F = 2 5 (ii) = T = 5 = (ii) 1 T + 2 = 2 (Taking moments about t) t ater 2 (2) = T() T = 1 ( = ) (iii) = 2 = = 2 F Q.. Q.. T Oil T2 ater = 1 = = ( ) 2 = (0.9) ( ) _ = = T T 2 T 1 + T 2 = (Taking moments about the lower end) (1) (5) = (1) + (5) + T 2 () T 2 = 5 T 1 = = 1 20 x 2 x x 2 1 x 2 1 x 2 (1 x) Let = the weight of the rod. Let x = the length of the submerged art. = = x 0.6 = 25x x + F = 9 2 (Taking moments about ). (1 x) ( x x 2 ) ( = F x x ) (1 x) ( 1 2 ) ( = F 1 x 2 ) F = (1 x) 2 x F

9 Q. 5. Putting this result into euation 1 gives: 25x (1 x) + = 9 (2 x) 9(2 x)... Multily by 25x(2 x) + 9(1 x) = 9(2 x) 25x 2 x + 9 = 0 (5x 1)(5x 9) = 0 x = 1 5 m (x = 9 m is too long) 5 Answer: 20 cm is submerged. T _ = 1 = 2 6 = 12 ( V s ) = 0. ( 1_ T T 2 = T 1 + T 2 = T 2 2 ) = 6 15 (Taking moments about the lower end) () + T 2 () = 2 (1) + 2 () T 2 = T 1 = = Oil ater Q. 6. (i) Let x and y be the horizontal and vertical comonents of the reaction at, x = 0 since no other forces act along the vertical. Therefore, the reaction at is vertical. e shall henceforth call it R. (ii) = 10 = = 10s R = 2 (Taking moments about ) 7 () = R(65) 10 R = 7 1 Putting this into euation 1 gives: = 10 = 6 1 = 9 60 = 1 20 y x 20 9

Assignment Set 2 - Solutions Due: Wednesday October 6; 1:00 pm

Assignment Set 2 - Solutions Due: Wednesday October 6; 1:00 pm Assignments CE 312 Fluid Mechanics (Fall 21) Assignment Set 2 - s Due: Wednesday October 6; 1: m Question A (7 marks) Consider the situation in the Figure where you see a gate laced under an angle θ=1