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Physical problem for Nonlinear EquationsGeneral Engineering

By Joel Ellis,2014-06-02 20:44
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Physical problem for Nonlinear EquationsGeneral Engineering

Chapter 03.00A

Physical Problem for Nonlinear Equations

    General Engineering

Problem Statement

    You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are asked to find the depth to which the ball will get submerged when floating in water (see Figure 1).

     Figure 1 Depth to which the ball is submerged in water

    Solution

    According to Newton’s third law of motion, every action has an equal and opposite reaction. In this case, the weight of the ball is balanced by the buoyancy force (Figure 2).

    Weight of ball = Buoyancy force (1)

    The weight of the ball is given by

     Weight of ball = (Volume of ball) (Density of ball) (Acceleration due to gravity)

    4?3;;;;R(g = (2) (b3)

    where

    R = radius of ball , (m)

    3 = density of ball ;;, (kg/m

    03.00A.1

    03.00A.2 Chapter 03.00A

    2 = acceleration due to gravity . ;;gm/s

    1The buoyancy force is given by

     Buoyancy force = Weight of water displaced

     = (Volume of ball under water) (Density of water)

     (Acceleration due to gravity)

    x?2 = (3) xR(g(w3)

    where

     = depth to which ball is submerged, x

     = density of water. (w

    Weight of ball

    Buoyancy force

     Figure 2 Free Body Diagram showing the forces acting on the ball immersed in water

Now substituting Equations (2) and (3) in Equation (1),

    4x?32 R(gxR(g(bw33)

    x32 4(3()(RxRbw3

    323 4R(3xR(x(0bww

    (323b 4R3xRx0(w

    323 (4) 4R3xRx0b

    where

     the specific gravity of the ball, is given by b

    (b (5) b(w

    Given

     , R5.5cm0.055 m

     , and 0.6b

    substituting in Equation (4), we get

     The derivation of the volume of the ball submerged under water is given in the appendix.

    Physical Problem of Nonlinear Equations: General Engineering 03.00A.3

    323 4(0.055)(0.6)3x(0.055)x0

    423 (6) 3.993100.165xx0

    The above equation is a nonlinear equation. Solving it would give us the value of ‘’, that is, xthe depth to which the ball is submerged under water.

Appendix A

    Derivation of the formula for the volume of a ball submerged under water.

    How do you find that the volume of the ball submerged under water as given by

    23;;hrh (7) V3

    where

    = radius of the ball, r

    = height of the ball to which the ball is submerged. h

    From calculus,

    r

     (8) VAdx?rh

    where is the cross-sectioned area at a distance from the center of the sphere. The lower Ax

    limit of integration is as that is where the water line is xrh

     O

     x r x

     A B

     h

     Figure 3 Deriving the equation for volume of ball under water

    rand the upper limit is as that is the bottom of the sphere. So, what is the A at any location

    . x

    From Figure 3, for a location , x

    , OBx

    , OAr

    then

    22 ABOAOB

    22 . (9) rx

    Band AB is the radius of the area at . So at location is x

    03.00A.4 Chapter 03.00A

    222 (10) ;;A(AB)rx

    so

    r22 ;;Vrxdx?rhr3?x2( rx(3)rh

    33~???rr;;h22(( ;;rrrrh?((33?))??

    2;;3hrh (11) .3

    NONLINEAR EQUATIONS

    Topic Physical problem for nonlinear equations for general engineering

    Summary A physical problem of finding the depth to which a ball would float in

    water is modeled as a nonlinear equation.

    Major General Engineering

    Authors Autar Kaw

    Date June 2, 2012

    Web Site http://numericalmethods.eng.usf.edu

     The derivation of the volume of the ball submerged under water is given in the appendix.

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