2.32 The symmetric trough shown in Figure P2.32 is used to...

2.32 The symmetric trough shown in Figure P2.32 is used to hold water. Along the length of the trough, steel wires are attached to support the sides at distances $w$ apart. Find the magnitude of the resultant force exerted by the water on each side, and the magnitude of the tension in the steel wire, neglecting the weight of the trough. (5 pts) $P_a$ Steel wire $\frac{h}{3}$ Water $90^\circ$ $P_a$ $\frac{2h}{3}$ Hinge Figure P2.32

Transcript

00:01 Hello everyone in this problem we have given a diagram in this manner so here this one is h and here the force is here the force that is f of h now this distance is h by 2 now if i resolve the forces here the tension will be in the side and its weight that will be in the downward direction so here it is denoted by w now here we have given the value of theta that is equal to 45 degree also the length that is denoted by l is given as 0 .75 meter and the value of the row is thousand kilogram per meter cube now firstly we are calculating the value of height that is h that will be equals to l into sine of theta so here the l is 0 .75 into sign of 45 degree so so here we got the value that is 0 .530 meter.

01:23 That is the value of height that is equal to 0 .530 meter.

01:31 Also the value of the b that is this much will be equals to l into course of theta.

01:44 So here l is 0 .75 course of 45 degrees.

01:50 So here after solving we get 0 .530 meter.

01:55 So we have got b is equal to 0 .530.

01:59 Also we have given the value of the width that is w six meter now we are calculating the value of the horizontal force that will be equals to pressure into area minus that will be equals to where pressure is row into g into h of the c into a where h of c is this much that is h by two so here it will be row g into h by 2 into its area now here row is thousand into g which is 9 .8 into h where h is 0 .530 divided by 2 into area where the value of area is b into w that is 0 .530 into 6 so here after multiplying we get 826 .7 newton also the value of the vertical force that is f of b can be equals to its weight so here weight will be density into volume into g so here density into g where volume is w into b into height that is h by 2 so here density is thousand into g that is 9 .8 into w which is 6 into b which is 0 .30 into 0 .530 by 2.

03:51 So here we get 8267 newton.

03:56 Here this value is 8 ,267 neuter.

04:09 Now we know that at this table condition the total moment about any point is 0...

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