What is the force required to release the lock on the water storage window?

[Smartornot]

Force in water storage, please help!

[PLAIN]http://img13.imageshack.us/i/storageo.jpg/Hey!

I am dealing with a an old construction/physics assignment and really need some help.

The assignment is as follows:

To make sure that a waterstorage is not flooded, one of the walls have a window with the measures (metres): width x height = b x h = 3 x 3

In the lower part of the window, there is a lock built in. This lock is unlocked and releases the water when the waterlevel reached height d.

The assignment is to calculate the force RA (kN), that arises when the water level reaches the height d. This force is the force that "tells" the lock to release and to let water push out of the storage. Water has the "weight" 10 kN/m^3.

My start of the solution: The force that pushes from the water is q= Yv*x. According to Newton's third law, RB and RA are the same size as q. If b*h= 9, this would mean that q is = 9 * 10 = 90 kN.

But I have a feeling this is wrong because I need to implement the x somehow. Can someone please help? I really need help with this and would be grateful for any reply!


Please see the attached picture below to get a better idea.

http://img13.imageshack.us/i/storageo.jpg/

 

[Doc Al]

The assignment is to calculate the force RA (kN), that arises when the water level reaches the height d. This force is the force that "tells" the lock to release and to let water push out of the storage. Water has the "weight" 10 kN/m^3.

If I understand the problem correctly, you want to calculate the force on the lock (that's R_A). Hint: That force must exert a torque on the window to balance the torque due to the water on the window. Figure out the torque exerted by the water.

My start of the solution: The force that pushes from the water is q= Yv*x. According to Newton's third law, RB and RA are the same size as q. If b*h= 9, this would mean that q is = 9 * 10 = 90 kN.

I don't understand this. (Explain your notation, for one thing.)

But I have a feeling this is wrong because I need to implement the x somehow.

Sure. The water pressure increases with x, so the lower part of the window feels a greater force per area than the top part--and an even greater torque.

 

[Smartornot]

Thank you for your reply. I have calculated the force on the window by using an integral. How does the force of the water on the window relate to RA? Is it the same force? Would really appreciate an answer.

 

[Doc Al]

I have calculated the force on the window by using an integral. How does the force of the water on the window relate to RA? Is it the same force? Would really appreciate an answer.

Since the window is in equilibrium, the force from the water must equal the sum of RA and RB.

To find RA alone, consider the torque about hinge B. (Use a slightly different integral to get the torque due to the water, instead of just the force due to the water.)

 

[Smartornot]

Sorry, my mother tongue isn't english. I thought torque was the same as force. I googled and it's what I call the "moment of force". Torque = force x the perpendicular distance.

But in my initial integral, I feel that I have already taken the distance into consideration, please see formula:

http://img823.imageshack.us/i/integralstorage.png/

How does this integral need to be changed in order for it to be torque due to the water?

I know I need to study moment of force more, and I will, but this assignment is due soon ... thanks for your help!

 

[Doc Al]

I thought torque was the same as force. I googled and it's what I call the "moment of force". Torque = force x the perpendicular distance.

Right. Torque and force are related, but different.

But in my initial integral, I feel that I have already taken the distance into consideration, please see formula:

http://img823.imageshack.us/i/integralstorage.png/

Yes, to find the total force you took distance into consideration. But your integral is for total force, dF = ρx bdx. (I assume that ρx is the pressure and that bdx is the area. If ρ is the mass density, then the pressure should be ρgx.)

How does this integral need to be changed in order for it to be torque due to the water?

You need the torque, which is "distance from hinge"*dF. Express that distance in terms of x, multiply, and integrate.

You'll then compare that to the torque from RA about the hinge. That will allow you to solve for RA.

 

[Smartornot]

Aha I am starting to understand the procedure, but i don't understand what the hinge is, where is it located?

 

[Doc Al]

Aha I am starting to understand the procedure, but i don't understand what the hinge is, where is it located?

At the top of the window, at point B. Think of the window as a door, hinged at the top. There's a lock at point A holding it shut. But if the force there is too much, the lock releases and the window opens.