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# Fluid Dynamics Help watch

1. This is from an old exam paper but I can't seem to work it out, I know for a fact that it needs moment of inertia to work out but i keep getting ridiculous things I can't reduce any more: (9810h-9810)/(3h-3). If someone can take a look and let me know if they can work it out that'd be great.
2. (Original post by tongrer)
This is from an old exam paper but I can't seem to work it out, I know for a fact that it needs moment of inertia to work out but i keep getting ridiculous things I can't reduce any more: (9810h-9810)/(3h-3). If someone can take a look and let me know if they can work it out that'd be great.
I don't see that you need to consider moment of inertia.

At the instant of opening, the moment of just balances the moment due to the water pressure.

The water pressure increases downwards, but linearly, so we can compute its effect as being due to its average value at the depth halfway up the door (which is 1m from the hinge). The area of the door is so the average force on the door is

So we have:

From there, it's rearrange-and-calculator time.
3. the issue with this is that its assuming that the water is acting at a point half way down the gate where as in reality you need to calculate the moment of inertia to determine the point at which the water acts
4. What answer do you have at the minute? Do you have the actual answer? I get 9.84m from a quick doodle.

Which bit are you struggling with?
5. (Original post by tongrer)
the issue with this is that its assuming that the water is acting at a point half way down the gate where as in reality you need to calculate the moment of inertia to determine the point at which the water acts
Which moment of inertia will you calculate? That of the gate? That doesn't seem to be possible, as you don't have its mass. What will you do with the moment of inertia when you have it?

It's not impossible that I'm being stupid, but I can't see a role for moment of inertia in this question. Can you explain how you intend to solve the problem using moment of inertia, please?
6. Basically I can't work out the answer, I need the method for my exam monday.

CRN 31896 Engineering Thermodynamics and Fluid Mechanics E1
9
8 A gate 1 m wide and 2 m high is hinged at its upper edge, as shown in Figure 8. Find the force F
required to open the gate.

Fig. 8
yc = 4 m
Pc = 1000*9.81*4 = 39240 Pag
Fh = Pc*2*1 = 78480 N
Ixx/ycA = 1*8/12 *(1/4)*(1/2) = 0.08333 m
ycp = yc + Ixx/ycA = 4.0833 m
Moment about hinge = (ycp – 3)*78480
= 1.08333*78480 = 85020 Nm
F = 85020/1.8 = 47233.3 N
= 47.233 kN

I have this as a worked example for a similar question where the height is 5m and you need to find the force but can't seem to backward engineer it?
7. I assumed that Yc = (h-1) but this gives me a really long formula to solve with h^3 etc so i don't think they expect me to do this in the exam as it took my mathematician friend an hour to solve earlier
8. (Original post by atsruser)
the average force on the door is

So we have:

There's an error in this: the calculation should be:

Sorry.
9. (Original post by tongrer)
I assumed that Yc = (h-1) but this gives me a really long formula to solve with h^3 etc so i don't think they expect me to do this in the exam as it took my mathematician friend an hour to solve earlier
What answer is given in the book?

Can you put up all of your argument and working? Which physical principles are you using to find the solution?

It's not really possible to comment on what you've done without seeing it.
10. You need to calculate the torque about the hinge as a function of h and then invert it to have a height as a function of torque, you also know what the torque is.

Calculating the torque:
Spoiler:
Show

Integrate pressure * vertical distance from the hinge over the surface of the gate

Knowing what the torque is:
Spoiler:
Show

50 KN * 1.8

EDIT:

something subtle that is easy to overlook:
Spoiler:
Show

if the water is lower than the gate then the integral should be over the surface of contact

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Updated: August 19, 2012
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