Thanks. Yes That's a "typo". The value of y_cp is correct. I will make a note of this correction in the description. That's probably better than deleting this video. Thanks again.
A salvage ship is attempting to raise an iron anchor off the ocean floor. The anchor has a mass of 560 kg and the density of iron is 7 800 kg/ m3 .The cable used to lift the anchor can support a weight of 5 000 N. before breaking. Use 1 025 kg/m3 as the density of seawater. (a) Can the cable support the anchor while it is completely submerged? (b) Can the cable support the anchor when it is completely out of the water? (c) What percentage of the anchor will be out of the water when the cable breaks?
Quick question does the force due to pressure depend on the whole shape of the submerged object or only on the surface in contact with the liquid. Like if a flat plane surface of slant length L is submerged, will the force be same as a slant trapezoid of length L also.
The total force due to pressure depends upon the depth of the centroid of the surface (which might be different for a trapezoid), the surface area. I hope that helps.
Can you please explain why the yp or the hp is different for vertical surfaces that protrude from the surface or whose tops are levelled with the surface compared to when the surface is below water?
I'm not sure I understand your question. The height of the centroid (h_cg) is the distance measured from the liquid surface. The centre of pressure (y_cp) is the distance below the centroid. A portion of the surface that protrudes above the liquid has no effect, because there is no hydrostatic pressure above the free surface.
Basic statics: Once you have the horizontal force at B, then set the sum of the forces in the x-direction equals zero to get F_Ax and set the sum of the forces in the y-direction equals zero to get F_Ay. Actually, F_Ay is zero by observation.
ycp is measured from the center of gravity, not the free surface. It's the the distance from the h_cg, with the negative sign indicating its below the cg. So, the distance from the free surface to the center of pressure is h_cg + abs(y_cp). I hope that helps.
The second moment of area of a rectangle about a horizontal axis through its centroid is the width*height^3/12. Google "second moment of area of a rectangle"
In the video description: Minor Correction: There is a "typo" at 6:46. h_cg=2.8m (not 1.8m) . But the value of y_cp and the problem answers are all ok.
In 7 minutes I have learned more about Fluid Mechanics than I have in an entire semester so far with my professor. Thank you.
Thanks. Happy to help.
thx for the video-
little mistake at 06:48, hcg=2,8m !
Thanks. Yes That's a "typo". The value of y_cp is correct. I will make a note of this correction in the description. That's probably better than deleting this video. Thanks again.
true
OMG no way i just three hours until i just saw your comment.
thanks for the video though was very well explained, i understood it immediately as opposed to my lecturer that rambled for 2 hours straight
me attemping to learn this at 3 am and my exam is in 5hours 😪😪
Did you pass? Boutta take my final in 8 hours
A salvage ship is attempting to raise an iron anchor off the ocean floor. The anchor has a mass of 560 kg and the density of iron is 7 800 kg/ m3 .The cable used to lift the anchor can support a weight of 5 000 N. before breaking. Use 1 025 kg/m3 as the density of seawater. (a) Can the cable support the anchor while it is completely submerged? (b) Can the cable support the anchor when it is completely out of the water? (c) What percentage of the anchor will be out of the water when the cable breaks?
That's an easy question to answer, if you do the necessary learning. Here's a recommended video top help: ua-cam.com/video/MJnYZ6s-LsQ/v-deo.html
Quick question does the force due to pressure depend on the whole shape of the submerged object or only on the surface in contact with the liquid. Like if a flat plane surface of slant length L is submerged, will the force be same as a slant trapezoid of length L also.
The total force due to pressure depends upon the depth of the centroid of the surface (which might be different for a trapezoid), the surface area. I hope that helps.
Very clear explanation...thanks🙏
where did the 12 come from? the one you divided with to get Ixx
The second moment of area of a rectangle about it's centroid is width*(height)^3/12. See en.wikipedia.org/wiki/List_of_second_moments_of_area
thanks . please share the link for the first culculation videos.thanks
All the videos are available at my website www.drdavidnaylor.net I hope that helps.
Can you please explain why the yp or the hp is different for vertical surfaces that protrude from the surface or whose tops are levelled with the surface compared to when the surface is below water?
I'm not sure I understand your question. The height of the centroid (h_cg) is the distance measured from the liquid surface. The centre of pressure (y_cp) is the distance below the centroid. A portion of the surface that protrudes above the liquid has no effect, because there is no hydrostatic pressure above the free surface.
Very well explained
Thanks. Glad to hear it was helpful.
Thanks for this, very clear explanation
Glad to hear it was helpful.
Well, I was expecting you to use the prism way to resolve it too. :)
Yes. That works too.
At the end, Why did we write 1.8 intead of 2.8 ?
See the comment in the video description.
Thanks ✋🏻
How would I go about finding the reaction forces at points A and B?
Basic statics: Once you have the horizontal force at B, then set the sum of the forces in the x-direction equals zero to get F_Ax and set the sum of the forces in the y-direction equals zero to get F_Ay. Actually, F_Ay is zero by observation.
well explained, thank you.
Glad to hear it was helpful
For the last question why is the equation not : ycp = Ixx/hA +h
ycp is measured from the center of gravity, not the free surface. It's the the distance from the h_cg, with the negative sign indicating its below the cg. So, the distance from the free surface to the center of pressure is h_cg + abs(y_cp). I hope that helps.
02:05 , free surface has Pa, a little correction i think??
No correction is needed because Pa also acts on the right side of the gate. So, the effect of atmospheric pressure cancels out.
I'm confused, why is the value of Ixx divided by 12?
The second moment of area of a rectangle about a horizontal axis through its centroid is the width*height^3/12. Google "second moment of area of a rectangle"
What is the distance 1.8 meters??
In the video description: Minor Correction: There is a "typo" at 6:46. h_cg=2.8m (not 1.8m) . But the value of y_cp and the problem answers are all ok.
I was looking for the correction. Thanks for the correction