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Water Force - Static Fluid Watch

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    A fresh water storage tank in a factory is shown in the figure above. Its top side is denoted as C, the narrow side as B and the long side as A. In order to allow regular maintenance and cleaning side A is hinged at its bottom edge and secured using a heavy-duty clasp at its top edge. L = 1.25m, H = 0.6m & W = 0.5m

    Side a has a length of 1.25m and a height of 0.6m

    (Assume the density of fresh water is 1000 kg/m3 and the acceleration due to gravity is 9.81 m/s2)

    a) Draw a side on view of side A (looking along arrow D) and show the size of the two forces acting on the tank side and how high they are located from the bottom of the tank


    I've calculated the water thrust force to be 2207.25 Newtons and it's height to be 0.4m meters from the top of the box.

    I am correct in assuming that water thrust force is the resultant force? and the second force being the equilibrium force?
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    (Original post by AutumnBeds)
    a) Draw a side on view of side A (looking along arrow D) and show the size of the two forces acting on the tank side and how high they are located from the bottom of the tank
    The question itself is poor, as there is a continually varying force along the depth of A. They're modelling it as a single force.

    There are also (at least) three forces on the tank side.

    I've calculated the water thrust force to be 2207.25 Newtons and it's height to be 0.4m meters from the top of the box.

    I am correct in assuming that water thrust force is the resultant force? and the second force being the equilibrium force?
    Can you post your calculation? (I've not checked it)

    You have the effective force on A from the water. This is actually the summation of the force at each depth, which results in both a net force and moment. They are modelling this as the net force at a given distance.

    That force must be opposed by forces at both the top and bottom hinges. These will be such that there is no net force or moment on the door. Therefore, to calculate it at the clasp, take moment around the axis of the bottom hinges.
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    (Original post by RogerOxon)
    The question itself is poor, as there is a continually varying force along the depth of A. They're modelling it as a single force.


    Can you post your calculation? (I've not checked it)

    You have the effective force on A from the water. This is actually the summation of the force at each depth, which results in both a net force and moment. They are modelling this as the net force at a given distance.

    That force must be opposed by forces at both the top and bottom hinges. These will be such that there is no net force or moment on the door. Therefore, to calculate it at the clasp, take moment around the axis of the bottom hinges.
    Workings attached for all parts I have worked through the given example for part a, however, b and c do not have any worked examples.

    For c I have calculated the overturning moment on the hinge and then the force required to keep the clasp shut.

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    (Original post by AutumnBeds)
    Workings attached for all parts I have worked through the given example for part a, however, b and c do not have any worked examples.

    For c I have calculated the overturning moment on the hinge and then the force required to keep the clasp shut.
    Sorry, but I don't have time to see where you went wrong (assuming that I'm right).

    I got 1104N (4 s.f.) and 0.4m from the top.

    I derived:

    F=\frac{\rho gh^2 A}{2}

    M=\frac{\rho gh^3 A}{3}

    \therefore d=\frac{M}{F}=\frac{2}{3}h

    A=1.25*0.5, h=0.6

    \therefore F=1104N (4.s.f.), d=0.4m

    I also get the force on the clasp as C=\frac{F}{3}=368N (3 s.f.)
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    (Original post by RogerOxon)
    Sorry, but I don't have time to see where you went wrong (assuming that I'm right).

    I got 1104N (4 s.f.) and 0.4m from the top.

    I derived:

    F=\frac{\rho gh^2 A}{2}

    M=\frac{\rho gh^3 A}{3}

    \therefore d=\frac{M}{F}=\frac{2}{3}h

    A=1.25*0.5, h=0.6

    \therefore F=1104N (4.s.f.), d=0.4m

    I also get the force on the clasp as C=\frac{F}{3}=368N (3 s.f.)
    Hi Roger,

    It seems that you have a typo for the force or the formula for the force is incorrect.
    The base SI unit for your force formula is

     \dfrac{[\rho gh^2 A]}{[2]} = \dfrac {\text{kg}}{\text{m}^3} \dfrac{\text{m}}{ \text{s}^2} \text{m}^2 \text{m}^2 = \dfrac{\text{kg} \text{ m}^2}{\text{s}^2}

    The base SI unit for force should be kg m /s2.

    I am not sure would the "extra factor of length" cause an error in your calculation.

    Ignore this post if you are confident about your calculation.
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    (Original post by Eimmanuel)
    It seems that you have a typo for the force or the formula for the force is incorrect.
    The base SI unit for your force formula is

     \dfrac{[\rho gh^2 A]}{[2]} = \dfrac {\text{kg}}{\text{m}^3} \dfrac{\text{m}}{ \text{s}^2} \text{m}^2 \text{m}^2 = \dfrac{\text{kg} \text{ m}^2}{\text{s}^2}

    The base SI unit for force should be kg m /s2.
    Good point. I'll have a look later.
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    (Original post by Eimmanuel)
     \dfrac{[\rho gh^2 A]}{[2]} = \dfrac {\text{kg}}{\text{m}^3} \dfrac{\text{m}}{ \text{s}^2} \text{m}^2 \text{m}^2 = \dfrac{\text{kg} \text{ m}^2}{\text{s}^2}

    The base SI unit for force should be kg m /s2.
    You're right. The equations should have used the width of the door, not the area of the horizontal cross section. Sorry for the confusion.

    With the 0.5m, removed, I believe that my calculations now agree with your answers.
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    (Original post by AutumnBeds)
    ...
    Hi. There seems to be a bit of confusion about the solution without a proper explanation being given.

    Firstly, note that the net force acting on the side panel due to the presence of the fluid within the tank is the product of the pressure of the fluid and the area of the side panel.

    You should know that the fluid pressure is a function of its depth:

    P_{\text{fluid}} = \rho_{\text{fluid}} g h.

    The net force acting on the side panel can then be given as a surface integral:

    F = \int_{l = 0}^{1.25} \int_{y = 0}^{0.6} P(l, h) \ \text{dl} \ \text{dh}

    F = \int_{l = 0}^{1.25} \int_{y = 0}^{0.6} \rho_{\text{fluid}} g h \ \text{dl} \ \text{dh}

    F = \left[ \left [ \dfrac{\rho_{\text{fluid}} g h^2 l}{2} \right]_{l = 0}^{1.25} \right]_{h = 0}^{0.6}

    \implies F = 2207.25 \ \text{N}.

    Since the pressure varies linearly from the top of the tank to the bottom of the tank, so does the force. The force distribution forms a triangular profile and since the centroid of a triangle lies at a distance of a third of its base length from its right angle, this is where the centre of pressure also lies and where the resultant force acts.

    There is a force of unknown magnitude acting on the side panel in the w-axis direction due to the clasp. You need to take moments about the bottom of the tank for the side panel about the l-axis. It is advisable to take moments about this location to eliminate the unknown reactions from the hinges. Note that the hinges do not generate a reaction moment as per the definition of a hinge. You know: the force and distance from the hinges of the force due to fluid pressure, and the distance from the hinges of the force from the clasp. Hence, you can solve for the only unknown of the force from the clasp.
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    (Original post by pleasedtobeatyou)
    Hi. There seems to be a bit of confusion about the solution without a proper explanation being given.

    Firstly, note that the net force acting on the side panel due to the presence of the fluid within the tank is the product of the pressure of the fluid and the area of the side panel.

    You should know that the fluid pressure is a function of its depth:

    P_{\text{fluid}} = \rho_{\text{fluid}} g h.

    The net force acting on the side panel can then be given as a surface integral:

    F = \int_{l = 0}^{1.25} \int_{y = 0}^{0.6} P(l, h) \ \text{dl} \ \text{dh}

    F = \int_{l = 0}^{1.25} \int_{y = 0}^{0.6} \rho_{\text{fluid}} g h \ \text{dl} \ \text{dh}

    F = \left[ \left [ \dfrac{\rho_{\text{fluid}} g h^2 l}{2} \right]_{l = 0}^{1.25} \right]_{h = 0}^{0.6}

    \implies F = 2207.25 \ \text{N}.

    Since the pressure varies linearly from the top of the tank to the bottom of the tank, so does the force. The force distribution forms a triangular profile and since the centroid of a triangle lies at a distance of a third of its base length from its right angle, this is where the centre of pressure also lies and where the resultant force acts.

    There is a force of unknown magnitude acting on the side panel in the w-axis direction due to the clasp. You need to take moments about the bottom of the tank for the side panel about the l-axis. It is advisable to take moments about this location to eliminate the unknown reactions from the hinges. Note that the hinges do not generate a reaction moment as per the definition of a hinge. You know: the force and distance from the hinges of the force due to fluid pressure, and the distance from the hinges of the force from the clasp. Hence, you can solve for the only unknown of the force from the clasp.
    So I was to take the force as 2207.25 and the overturning moment as 0.2mfrom the hinge.

    The resisting force on the clasp (F2) would be;

    F1D1 = F2D2

    2207.25 * 0.2 = F2 * 0.6

    F2 = (2207 * 0.2) / 0.6 = 735.75 N

    And then the force on the hinge (F3)

    2207.25 = F2 + F3

    F3 = 2207.25 - 735.75 = 1471.5 N
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    (Original post by AutumnBeds)
    So I was to take the force as 2207.25 and the overturning moment as 0.2mfrom the hinge.

    The resisting force on the clasp (F2) would be;

    F1D1 = F2D2

    2207.25 * 0.2 = F2 * 0.6

    F2 = (2207 * 0.2) / 0.6 = 735.75 N

    And then the force on the hinge (F3)

    2207.25 = F2 + F3

    F3 = 2207.25 - 735.75 = 1471.5 N
    Yep sounds good
 
 
 
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