Please answer the following questions using the examples as references.

1.

Calculate the fluid force on one side of the plate using the coordinate system shown below. Assume the density is 62.4 lb/ft's. The fluid force on one side of the plate is 334361088 lb. y ( ft ) Surface of pool 10." * >x ( ft) Depth -----y= -5 lyl (x,y) - 10The viewing portion of the rectangular glass window in a fish tank is 81 inches wide and runs from 0.5 inch below the water's surface to 39.5 inches below the surface. Find the fluid force against this portion of the window. The weight-density of seawater is 64 lb/ft's - . . The fluid force against the window is |Ib.We model pumping from spherical containers the way we do from other containers, with the axis of "'-.\"' integration along the vertical axis of the sphere. Use the figure shown to the right to nd how much work it 0 6 12 m takes to empty a full hemispherical water reservoir of yl' % . radius 6 m by pumping the water to a height of 12 m \"\"1;- above the top ofthe reservoir. Water weighs \\% ' _.. 9800 Nim3 ~ g 3'." 361,-\" F(y) : w - dV 9800::(36 3:2] cly The distance through which F(y) must act to lift the slab to the level of 12 m above the top of the reservoir is (12 7 5!), so the work done lifting the slab is approximatelyr the following. their = 9800::(36 3:2] (12 v)dv Recall from the diagram that the top ofthe reservoir is at y: 0. Since the radius ofthe reservoir is 6, the height at the bottom must be at y = 6. Thus, the limits of integration are from y = 6 to v = O. Integrate the expression for mail from y: 6 to y: 0. 0 W = I 9800:: (36 (12 y)dy -6 We model pumping from spherical containers the way we do from other containers, with the axis of integration along the vertical axis of the sphere. Use the figure shown to the right to find how much work it 6 12 m takes to empty a full hemispherical water reservoir of radius 6 m by pumping the water to a height of 12 m above the top of the reservoir. Water weighs BYE -y 9800 N/ms Av V36 -V2 . . . W = 9800x (36 - y2) (12 - y)dy 0 = 9800m (432 - 36y - 12)2 + y3) dy - 6 9800m 432y - 18y2 - 4y3 + - 6 Evaluate the integral. W = 9800x 432y - 18y2 - 4y3 + y 4 4 - 6 ~ 63, 176, 171.63We model pumping from spherical containers the way we do from other containers: with the axis of \"a integration along the vertical aXis ot the sphere. Use the figure shown to the right to find how much work it , , El m takes to empty a full hemispherical water reservoir of y radius 6 m by pumping the water to a heightotS m '\\ above the top ofthe reservoir. Water weighs ' 9800 Nlma. The amount of work required to pump the water to a height of 8 m above the top of the reservoir is approXimately 45 44 -10'5 (Do not round until the final answer Then round to the nearest hundredth as needed ) A force of 10 N will stretch a rubber band 10 cm (0. 10 m). Assuming that Hooke's law applies, how far will an 8-N force stretch the rubber band? How much work does it take to stretch the rubber band this far? How far will an 8-N force stretch the rubber band? 0.08 m (Simplify your answer.) How much work does it take to stretch the rubber band this far? (Simplify your answer.)The graph of a force function (in newtons) is given. How much work is done by the force in moving an object 10 m? F (N) 24 20- 16- 12- x (m) 10 The amount of work done by the force in moving an object 10 m is J. (Type a whole number.)Calculate the fluid force on one side of the plate using the coordinate system shown The fluid (weight of denSityr w) force against the side of a vertical plate With below. Assume the density is 52.4 ibrii3. D the bottom at y = a and with the top at y = b ist - (strip depth) - L{y) - dy: t 4 a where L(y) is the length of the horizontal strip of area. 3' (ft) The depth of a horizontal element in the given coordinate system is , y. The length ofthe strip (element of area} y ft below the surface is 2x Surface ofpgol With (x,y) the coordinates ofthe right end ofthe strip. The equation, y : x) of the line containing the right side of the submerged plate is y =x 6. H Since y=x6: x=y+5and L(y)=2x=2{y+6). b F: WI 2yly+6ldy a The limits of integration are the ycoordinates of the bottom and top of the plate, 6 and l respectively. :4160 lb. Calculate the fluid force on one side of a semicircular plate of radius 5 ft that rests vertically on its diameter at the bottom of a X pool filled with water to a depth of 6 ft. Assume the weight-density of water is 62.4 lb/fts. . . . The fluid force on one side of the plate is (Round the final answer to the nearest tenth as needed. Round all intermediate values to the nearest thousandth as needed.)Calculate the fluid force on one side of a semicircular plate of radius 16 ft that rests vertically on its diameter at the bottom ofa X pool filled with water to a depth of 23 ft. Assume '3 the weightdensity of water is 62.4 lbi'tt3_ b F =wI(23 y}[21/(16}2 5:2 ] dy a The limits of integration are the ycoordinates of the bottom and top of the plate, or O and 16. respectively, Substitute 62.4 lbltt3 for w 16 F = 524I(23y)[21/(16)2 y2]dy 0 After distributing (23 y) and factoring out the constant: 2: the equation can be rewritten as the following 16 F = 124.8IZ3q/(1BJZy2 y1/(16)2y2dy 0 0 = N - du 256 256 ~ 1365.333 Now substitute these values to find the fluid force. 16 16 F = 124.8 231(16)2 -y2 dy - |yv(16)2->2dy = 124.8[F, - F2] 124.8 4624.424 - 1365.333] ~ 406734.6The viewing portion of the rectangular glass window in a fish tank is 87 inches wide and runs from 1.5 inches below the water's surface to 28.5 inches below the surface. Find the fluid force against this portion of the window. The weight-density of seawater is 64 lb/ft . . . Suppose that a plate submerged vertically in fluid of weight-density w runs from y = a to y = b on the y-axis. Let L(y) be the length of the horizontal strip of the plate at level y. The force exerted by the fluid against one side of the plate is given by the following definite integral. F = w . (strip depth) . L(y) dy The weight-density w of seawater is given as 64 lb/ft". All the other values are in inches. In order to use the formula, the units must be compatible. Either convert the weight-density to cubic inches or all the other values to feet. This solution will use inches To convert pounds per cubic foot to pounds per cubic inch, substitute 12 inches for 1 foot and simplify. 64 lb 64 lb f+ 3 (12 in) 3 1 = 27 1b/ in Use the coordinate system in the figure on the right to determine the depth to the bottom of a horizontal strip in y (in) terms of y. 28.5 1 surface 27 strip depth = (28.5 -y) in X (in) -43.5 43.527 F = w . (strip depth) . L(y) dy 28.5-y a 27 x (in) -43.5 43.5 27 . (28.5 -y) . 87 dy 0 Use the constant multiple rule to simplify the integral. Then integrate. 27 27 29 27 * (28.5 -y) . 87 dy = 9 (28.5 -y) dy 29 y2 727 = 9 28.5y - 2 0 Now evaluate the integral for the fluid force. 29 29 9 28.5y 2 9 = 1305 lb The fluid force against the window is 1305 pounds