WORK and HYDROSTATIC FORCE

Calculus is used in many disciplines. Physics and Engineering are the principal ones. Our focus here is on the application of the calculus to the specific subjects of Work & Hydrostatics.

Work is the product of the force required to move an object and the distance that the object moves as a consequence of that force. The Calculus is required to compute work when the force depends or varies with the distance. The examples that follow illustrate this.

Example # 1(a): A worker hauls a bucket of water over a distance of 20 ft at a constant rate of 2 ft/sec. The bucket weighs 5 lb and the water in the bucket weighs 17 lb. The rope's weight is negligible and the bucket does not leak. Determine the work, "  ", that the worker must expend.

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The effort or force, "  ", that the worker must exert does not depend on the position, "  ". Therefore, the work, "  ", can be computed without the calculus.

The total force, "  ", is the sum of the bucket weight, "  ", and the water weight, "  ".

Example # 1(b): A worker hauls a bucket of water over a distance of 20 ft at a constant rate of 2 ft/sec. The bucket weighs 5 lb and the water in the bucket weighs 17 lb. The rope weighs 0.5 lb/ft and the bucket leaks at the 0.2 lb/sec. Determine the work, "  ", that the worker must expend.

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The effort or force, "  ", that the worker must exert does depend on the position, "  ". Therefore, the work, "  ", must be computed with the calculus.

The total force, "  ", is the sum of the bucket weight, "  ", the water weight, "  "

and the rope weight, "  ".

Although the bucket weight is a constant, the water weight and the rope weight both depend on the distance, "  ", that the bucket has been lifted.

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Since the force varies with "  ", the work expended also depends on "  ". Let the worker haul the bucket an infinitesimal distance, "  ", from the position, "  ". This is the work, "  ", required to move the bucket that infinitesimal distance.

The total work, "  ", is the "sum" of all of those infinitesimal amounts of work from the bottom to the top.

We use Maple to compute this definite integral.

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Example # 2: Determine the work required to place a 1500 lb satellite into an orbit 1000 miles above the earth's surface. Assume that the radius of the earth is 4000 miles and that the attractive gravitational force, "  " exerted by the earth on a mass weighing, " ", at the earth's surface at a distance: " x " miles from the center of the earth is given by the following expression.

Since the force varies with "  ", the work expended also depends on "  ".

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The total work, "  ", is the "sum" of all of those infinitesimal amounts of work from the earth's surface to the orbital height.

We use Maple to compute this definite integral.

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Example # 3: Determine the work required to pump all of the water from a half-filled vertical, right circular cylindrical tank with the dimensions shown in the figure below. Assume that the water's density, .

Take a cross-sectional slice of water of thickness, " dx ", at a depth of " x " ft below the top of the tank. The weight, "  ", of that slice is the product of its volume, " ", and its density, "  ".

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The slice of water must be moved a

distance, " x ", to take it to the top of the tank. The work, " dW ", required to accomplish that equals the product of the slice's weight and the vertical distance it is moved.

The total work, "  ", is the "sum" of all of those infinitesimal amounts of work.

We use Maple to compute this definite integral.

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Example # 4: Determine the work required to pump all of the oil from a full horizontal, right circular cylindrical tank with the dimensions shown in the figure below. Assume that the oil's density, .

Take a cross-sectional slice of water of thickness, " dx ", at a depth of " x " ft below the top of the discharge pipe. The weight, "  ", of that slice is the product of its volume, " ", and its density, "  ".

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The volume, "  ", is the product of the cross-sectional area, "  ", of the slice and its thickness, "  ".

The area, "  ", is the product of the oil tank length, " 12 ft ", and the length, "  ", of the cord of the circular cross-section of the oil tank that looks like this.

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The slice of water must be moved a distance,

" x ", to take it to the top of the tank. The work,

" dW ", required to accomplish that equals the product of the slice's weight and the vertical distance it is moved.

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The total work, "  ", is the "sum" of all of those infinitesimal amounts of work.

We use Maple to compute this definite integral.

The Hydrostatic Force on the surface of an object is the product of the pressure produced by the fluid and the surface area of the object. The Calculus is required to compute the hydrostatic force when the pressure varies. The examples that follow illustrate this.

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Example # 5: Determine the total hydrostatic force acting on the dam. Assume that the water has a density: "  ".

The pressure, "  ", varies directly with the depth, " x ".

At a depth, " x ", the pressure is constant everywhere on the rectangular strip of

width, "dx", and length, "L". The force, "dF ", produced by the water on that strip is the product of the pressure acting on that strip and the area, "dA", of that strip.

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The total force, "  ", is the "sum" of all of those infinitesimal forces.

Example # 6: Determine the total hydrostatic force acting on the semi-circular gate at the bottom of the dam. Assume that the water has a

density: .

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The pressure, "  ", varies directly with the depth, " x ".

At a depth, " x ", the pressure is constant everywhere on the rectangular strip of

width, "dx", and length, " ". The force, "dF ", produced by the water on that strip is the product of the pressure acting on that strip and the area, "dA", of that strip.

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The total force, "  ", is the "sum" of all of those infinitesimal forces.

We use Maple to compute this definite integral.

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Example # 7: Determine the total hydrostatic force acting on the ends of the trough. Assume that the fluid has a density:  and a depth: h.

The pressure, "  ", varies directly with the depth, "  ".

At a depth, "  ", the pressure is constant everywhere on the rectangular strip of

width, "", and length, " ". The force, "dF ", produced by the water on that strip is the product of the pressure acting on that strip and the area, "dA", of that strip.

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The total force, "  ", is the "sum" of all of those infinitesimal forces.

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