Calculate the force on the side of this water storage box

[didaw]

Homework Statement

A hinged water storage is shown below. Its top side is denoted as C, the narrow side as B and the long side as A. Assume water density is 1000 kg/m3 and gravity is 9.81 m/s2
Side A is hinged at the bottom edge and secured using a clasp at its top edge.
Dimensions are: L = 1m, H = 0.5m & W = 0.5m
a) Calculate the size and location of the water thrust force on side A

Relevant Equations

Area = B x D
1st moment of area = Area x 𝑦̅

Area = 1 x 0.5= 0.5
1st moment of area = 0.5 x (0.5/2) = 0.125 m
total force = p x g x 1 moment = 1000 x 9.81 x 0.125 = 1226.25N
Force = 1226.25
im pretty sure all that is right
i think the second moment = Ix = bh^3 / 3 = 1 x 0.5^3 / 3 = 0.0416
then 2nd moment of area / 1st moment of area 0.0416/0.125=0.3328
so would that mean that the location of the thrust force of 1226.25 would be 0.3328 m from the top of the box?

 

[kuruman]

Problem Statement: A hinged water storage is shown below.

Where "below"?

 

[didaw]

picture of the box

 

[didaw]

Where "below"?

ive just posted it

 

[haruspex]

would that mean that the location of the thrust force of 1226.25 would be 0.3328 m from the top of the box?

Consider the top and bottom halves of the panel. Which half is subject to the greater force?

Good style is to keep everything symbolic, only plugging in numbers at the end. This has many advantages. One is greater accuracy, another is greater insight into the way things work.

 

[didaw]

Consider the top and bottom halves of the panel. Which half is subject to the greater force?

Good style is to keep everything symbolic, only plugging in numbers at the end. This has many advantages. One is greater accuracy, another is greater insight into the way things work.

the bottom will have the most force on it since their is more water mass above it but i have the force 0.1672 from the bottom of the box surely that sounds about right? unless i have used the wrong formula to work out moment two ' Ix = bh^3 / 3 ' should i have used Ixc=bh^3/12

 

[didaw]

Consider the top and bottom halves of the panel. Which half is subject to the greater force?

Good style is to keep everything symbolic, only plugging in numbers at the end. This has many advantages. One is greater accuracy, another is greater insight into the way things work.

is this what i should have done
IG=bd^3/12 = 1 x 0.5^3 / 12 = 0.0104
H= IG / Ah + h = 0.0104 / (1 x 0.5) x 0.25 + 0.25 = 0.3332m
so the force would be 0.3332m from the top and 0.1668 from the bottom

 

[haruspex]

is this what i should have done
IG=bd^3/12 = 1 x 0.5^3 / 12 = 0.0104
H= IG / Ah + h = 0.0104 / (1 x 0.5) x 0.25 + 0.25 = 0.3332m
so the force would be 0.3332m from the top and 0.1668 from the bottom

Sorry, for some reason I thought you had found it to be 1/3 of the way from the top, but you wrote 1/3 of a metre, so 2/3 from the top.
(If you keep everything symbolic you will find it is exactly 2/3 from the top.)

 

[haruspex]

so i have gone wrong on both attempts since it should be 0.166666666666?

It is one third of the way from the bottom.
You are probably expected to give the answer in decimals of a meter, but it is a bit awkward here because the standard rule is that you only quote as many significant figures as you were given in the data, so here you round 0.16666... to 0.2m, and that would probably be marked wrong.
I suggest you answer either 0.17m from the bottom or 0.33m from the top. The wording not seem to specify which way to measure it.

 

[didaw]

Sorry, for some reason I thought you had found it to be 1/3 of the way from the top, but you wrote 1/3 of a metre, so 2/3 from the top.
(If you keep everything symbolic you will find it is exactly 2/3 from the top.)

so both attempt's are right because i rounded some numbers to 3 dp if i used the full numbers i would get 0.16 recurring

 

[haruspex]

so both attempt's are right because i rounded some numbers to 3 dp if i used the full numbers i would get 0.16 recurring

Yes.

 

[didaw]

Yes.

thank you, i didn't realize that if i used the full numbers it would get me to the right answer

 

[didaw]

Yes.

i don't suppose you can help me out on the next 2 questions? the same data applies and the picture.

A hinged water storage is shown below. Its top side is denoted as C, the narrow side as B and the long side as A. Assume water density is 1000 kg/m3 and gravity is 9.81 m/s2

Side A is hinged at the bottom edge and secured using a clasp at its top edge.

Dimensions are: L = 1m, H = 0.5m & W = 0.5m

Draw a side on view of side A (looking along arrow D shown) 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

Calculate the required resisting force at the clasp to keep the panel shut.

my working out for all 3 questions are their i am just unsure what the other force would be on the side of the box

 

[haruspex]

unsure what the other force would be on the side of the box

It can't just be the force of water pressure. There must be a combination of forces to achieve equilibrium. There's the force of the clasp, but that is at the top, so there'd still be a net torque. With just those two forces, which way would the panel rotate? What prevents that?

Your numbers look right, but please, please get into the habit of working entirely symbolically. Where you are given numbers, create variables to represent them. You should find that the force at the clasp is exactly one third of the force from the water.

 

[didaw]

It can't just be the force of water pressure. There must be a combination of forces to achieve equilibrium. There's the force of the clasp, but that is at the top, so there'd still be a net torque. With just those two forces, which way would the panel rotate? What prevents that?

Your numbers look right, but please, please get into the habit of working entirely symbolically. Where you are given numbers, create variables to represent them. You should find that the force at the clasp is exactly one third of the force from the water.

so would it be a force the other side of a of 1226.25 to make it equal? i will do next time i didn't realize that it was 1/3 when i started this equation but now i do, thanks

 

[didaw]

It can't just be the force of water pressure. There must be a combination of forces to achieve equilibrium. There's the force of the clasp, but that is at the top, so there'd still be a net torque. With just those two forces, which way would the panel rotate? What prevents that?

Your numbers look right, but please, please get into the habit of working entirely symbolically. Where you are given numbers, create variables to represent them. You should find that the force at the clasp is exactly one third of the force from the water.

 

[haruspex]

so would it be a force the other side of a of 1226.25 to make it equal?

There must be that total of force the other way, but where is it coming from? Is it all one force?

 

[didaw]

There must be that total of force the other way, but where is it coming from? Is it all one force?

im not to sure, would the force come from the bottom of the tank? the force would equal the amount of pressure the water is pushing on it and that would be different for every point of the side

 

[haruspex]

would the force come from the bottom of the tank?

Yes, from the hinge(s). Using balance of forces and torques, you can deduce the force at the hinges and the force at the clasp.

the force would equal the amount of pressure the water is pushing on it and that would be different for every point of the side

The forces balancing the water pressure do not need to be spread out across the area to match the distribution of force from the water. The panel is stiff. It suffices to have three forces, in this case, two at the bottom and one at the top. There could be more in each case, by having more hinges and clasps, but then it becomes impossible to say how much force each individual such supplies.

 

[didaw]

so with using a side view like the question says i will only have to do a load at the top and bottom? What method would i need to use to find out what the loads are? or since its a 3rd of the way up would i need to 1226.25/3=408.75 for the bottom and 1226.25-408.75=817.5 for the top?

or do you think that it would be suffice just leaving it like the image that i have posted up?

Thanks.

 

[haruspex]

since its a 3rd of the way up would i need to 1226.25/3=408.75 for the bottom and 1226.25-408.75=817.5 for the top?

Along the right lines, but not quite.
Take moments about some point. Any point will do, but some will make it simpler than others.

 

[didaw]

Along the right lines, but not quite.
Take moments about some point. Any point will do, but some will make it simpler than others.

The only think I can think of is to used point loads but I don't know how to work that out with only one force and no provided UDL. Is that the right method to use?

 

[haruspex]

I don't know how to work that out with only one force

You have three forces, all horizontal: the overall force from the water (which you found in the first part), the force at the catch and the force at the hinge.
Take moments about, say, the hinge. What moments do the forces exert about that point?

 

[didaw]

You have three forces, all horizontal: the overall force from the water (which you found in the first part), the force at the catch and the force at the hinge.
Take moments about, say, the hinge. What moments do the forces exert about that point?

The force is 410 on the hinge since that is the required pressure to open the hinge?

 

[haruspex]

The force is 410 on the hinge since that is the required pressure to open the hinge?

That is what you wrote in post #20, but it is wrong and I see no adequate explanation of how you arrive at it.
The correct method is to write the torque balance equation about some axis. Do you understand how to find the torque a force exerts about an axis?

 

[didaw]

Looking at my notes it looks like it doesn't provide information on how to work it out. However it has one example of it just not anything to determine how it was worked out

 

[haruspex]

Looking at my notes it looks like it doesn't provide information on how to work it out. However it has one example of it just not anything to determine how it was worked out

Are you saying you have not been taught how to find the torque exerted by a force about an axis?

 

[didaw]

Are you saying you have not been taught how to find the torque exerted by a force about an axis?

No, I will have a look into it a bit later though. The closest thing I have been taught is about UDL point loads, conversion of moments, equilibrium conditions, moments of force but I have a 20 page document covering all of them and a few others so only a few things are in a good amount of detail

 

[haruspex]

No, I will have a look into it a bit later though. The closest thing I have been taught is about UDL point loads, conversion of moments, equilibrium conditions, moments of force but I have a 20 page document covering all of them and a few others so only a few things are in a good amount of detail

You might find https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ useful.

 

[didaw]

You might find https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ useful.

Thank you