![]() ![]() This gives 2rlπΔp instead of the correct answer above. ![]() However I thought that you would need to integrate the individual forces over the circumference of the semi-circle, πr, and multiply by $ \int_0^\pi \mathrm\theta$ in order to extract the vertical component of each small force. So across a small length of the circumference of the tube is dF = Δp l dx.įrom reading I know that the sum of the vertical components of these forces is is 2rlΔp as apparently you can just multiply the horizontal projection of the curved surface by the length and the pressure difference. The weight- density of water is 9,800 newtons per cubic meter. The dimensions are b 1 m and a 7 m (see figure). Find the fluid force on the vertical plate that is submerged in water. The force perpendicular to a small area of the tube surface is dF = Δp dA. Find the fluid force of a square vertical plate submerged in water, where a 6, and the weight density of water is 9800 newtons per cubic meter. Let Δp = P 1 - P 2, r be the radius of the semi-circle and l be the length of the tube. The semi-circle has radius r and the tube an internal pressure P 1 higher than the external P 2. I am attempting to extract the vertical component (in relation to diagram) of the force on this semi-circular cross-section. ![]() I have a cross section of a half-tube with a pressure gradient across it causing a force outwards. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |