Recent content by iSimon

Here is the main idea on how to solve. If we transpose nodes of the cylinder to the right side we will receive figure which will be composed from 2 identical elipses on top and bottom and two identical rectangulars in middle. And the thing is to add Areas of separate parts...

I struggled with replying here because its hard to explain what my teacher wants me to do but again, I am not really able to do it myself. And I don't quite understand her demands. Maybe you guys are going to help me. This is what she said: I should try to project lines from top and bottom of...

I am really sorry, I just consulted this task. My lecturer told me it's wrong again. Gave me some explanation which to be honest is quite helpful but I will need to draw it. I will manage to do it later when I get home :)

The force that is pointing at curved surface is upward and right. The force is placed somewhere on the curved surface, up and right. Its perpendicular to the axis of cylinder.

So the force acting on the curved surface is 27140 N + 13572 *sin30 = 33926 N?

To determine vertical force I will have to multiply 13572 * sin30 degree. For horizontal its 13573 * cos30 degree.

If weight is a force where mass is multiplied with g, then weight is equal to 27140 kg*m/s^2. 1kg*m/s^2 = 1N

Volume of cylinder is V=3.14*(0.6^2) * 2.4 = 2.714m^3 Density of water is 1000kg/m^3 mass of water will be the the same as volume of cylinder so 2.714kg also. And to be honest my english isn't that good, mass and weight? Isnt that the same? :/So it should be: N= g*m*sin30= 2.714 kg * sin30 * 10...

The force is directed upwards and its value is equal to the weight of our cylinder. Which is ummm, 13572 * sin30 degree?

I think the 2nd one may seem more complicated but calculating integral is beyond my reach. We didnt have integrals yet.

And how to calculate the force on the curved surface?

The lower one has greater value, and the resultant will be 13 572 N.

Magnitudes at each face is: F1 = 10179 kg/(m^2 * s^2) F2 = 23751 kg/(m^2 * s^2) And the net of these two forces is... I think that the bottom Force is pointed up and left. And the top force is down and right.

Area is equal to 1.131 m^2. Direction of pressure force on each face will be perpendicular to the axis of cylinder. And the force can be split into two forces each like I did in the 2nd picture. Oh bloody hell. Average pressure is 15000 kg/m^3. I can't really imagine how the forces vectors...

Ummm I have problem with units now. po is F/s where F = P*s (s is the area of base) F1= p1*s=qgh1*s ; F2= p2*s=qgh2*s F1 = 10179 kg/(m^2 * s^2) F2 = 23751 kg/(m^2 * s^2) V = (h2-h1)*s=(2.1-0.9)*1.131=1.3572 F = qgV= 1000kg/m^3 * 10 kg/s^2 * 1.3572m^2 = 13572 I made somewhere a mistake...