2.52: A gate having the cross section shown in the figure...

Name: 2.52: A gate having the cross section shown in the figure closes an opening 1.5 m wide and 1.2 m high in a water reservoir. The gate weighs 2.2 kN, and its center of gravity is 0.3 m to the left of AC and 0.6 m above BC. Determine the horizontal reaction that is developed on the gate at C.
Uploaded: 2023-03-14T14:25:12-08:00
Duration: 8 min 22 s
Channel: Prabhat Tyagi
Description: 2.52: A gate having the cross section shown in the figure closes an opening 1.5 m wide and 1.2 m high in a water reservoir. The gate weighs 2.2 kN, and its center of gravity is 0.3 m to the left of AC and 0.6 m above BC. Determine the horizontal reaction that is developed on the gate at C.

2.52: A gate having the cross section shown in the figure closes an opening 1.5 m wide and 1.2 m high in a water reservoir. The gate weighs 2.2 kN, and its center of gravity is 0.3 m to the left of AC and 0.6 m above BC. Determine the horizontal reaction that is developed on the gate at C.

Transcript

00:01 Here in this question first we are going to calculate the hydrostatic force acting on the inclined portion ab so here f1 equals to gamma multiply by h of c multiply by a now we can write this as f1 equals to row multiply by g multiplied by h of c multiply by h of c multiply by l multiply by w or we can write this f1 equals to gamma multiply by h1 plus h2 minus 2 multiply by l multiply by w now here row is the density of water g is the acceleration due to gravity gamma is the specific weight of the water h1 is the distance from the free surface to the hinge point s2 is the distance from the hinge point a to point c l is the length of the inclined portion of gate and w is the width of the gate so here we substitute 62 .4 pound per feet cube for gamma 8 feet for h1 4 feet for s2 and square root of three whole square plus 4 whole square feet for l and 5 feet for w so now on substituting the value we get f1 equals to 62 .4 multiply by 8 plus 4 minus 2 multiply by a square root of 3 whole square plus 4 whole square multiply by 5.

01:58 So now we get f1 equals to 1560 pound now we calculate the value of theta from the triangle ac b so here we get cos theta equals to 4 divided by square root of 3 whole square plus 4 whole square so now we get cost theta equals to four divided by five now again in triangle o d a here cost theta equals to eight divided by o a so now now we get o a equals to 8 divided by cos theta now here we substitute 4 divided by 5 4 cost theta so we get oa equals to 8 divided by 4 divided by 5 so now we get oa equals to 10 feet now we calculate the value of y of c which is the distance between point o to the point of ab so we get y of c equals to oa plus a b divided by 2 so we get y of c equals to now we get y of c equals to 12 .5 field now we calculate the moment of inertia of the gate ab which is given by i of xc equals to 5 multiply by 5 to the power 3 divided by 12 so we get i of x c equals to 52 .0833 feet to the power 4.

04:53 Now here we calculate the location of point p from point o, which is given by y of r equals to y of c plus i of x, divided by y of c multiplied by l multiply by w...

Question

2.52: A gate having the cross section shown in the figure closes an opening 1.5 m wide and 1.2 m high in a water reservoir. The gate weighs 2.2 kN, and its center of gravity is 0.3 m to the left of AC and 0.6 m above BC. Determine the horizontal reaction that is developed on the gate at C.

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