What is the tension in a bar submerged in water?

[foo9008]

Yes. You just need to convince yourself that the dimension parallel to the x-axis corresponds to $a$ in your expression $ab^3/12$.

so, i don't have to 'turn' the plane to becomeperfectly horizontal , just left the object as it is and the 0.75m cut thru x-axis will do ?

 

[TSny]

so, i don't have to 'turn' the plane to becomeperfectly horizontal

RIght, you don't need to turn the plane.

just left the object as it is and the 0.75m cut thru x-axis will do ?

Not sure what you are saying. The 0.75 m dimension of the rectangle should be parallel to the y-axis. Therefore, the 0.75 m side will "cut thru x-axis", if that's what you mean.

 

[foo9008]

RIght, you don't need to turn the plane.

Not sure what you are saying. The 0.75 m dimension of the rectangle should be parallel to the y-axis. Therefore, the 0.75 m side will "cut thru x-axis", if that's what you mean.

so the final ans is
anticlockwise moment = clockwise moment
T(0.75sin45) = (0.75/2) sin45 x9.81x1000x6x0.75x( 0.75-0.5)
T= 5518N
( 0.75-0.5) represent the distance from the FR to the hinge

so , yp = yc + Ixx / (ycA)
= 0.375 + 6(0.75)^3) / (12 x 0.375 x 0.75 x6) = 0.5m

 

[TSny]

so the final ans is
anticlockwise moment = clockwise moment
T(0.75sin45) = (0.75/2) sin45 x9.81x1000x6x0.75x( 0.75-0.5)
T= 5518N
( 0.75-0.5) represent the distance from the FR to the hinge

so , yp = yc + Ixx / (ycA)
= 0.375 + 6(0.75)^3) / (12 x 0.375 x 0.75 x6) = 0.5m

Yes, I think that's correct.

 

[foo9008]

RIght, you don't need to turn the plane.

Not sure what you are saying. The 0.75 m dimension of the rectangle should be parallel to the y-axis. Therefore, the 0.75 m side will "cut thru x-axis", if that's what you mean.

can you draw out the x-axis on the diagram ? IMO, both 6m and 0.75 m cut thru the x-axis , so , i am not sure which one cut thru

 

[TSny]

You have the formulas yp= yc +(Ixx) /ycA , Ixx = a(b^3)/12 for one of the rectangular sides of the trough. The x and y axes lie in the plane of the rectangle with the origin at the centroid. The x-axis is horizontal while the y-axis is perpendicular to the x-axis and increases toward the bottom of the trough.

 

[foo9008]

You have the formulas yp= yc +(Ixx) /ycA , Ixx = a(b^3)/12 for one of the rectangular sides of the trough. The x and y axes lie in the plane of the rectangle with the origin at the centroid. The x-axis is horizontal while the y-axis is perpendicular to the x-axis and increases toward the bottom of the trough.

how do u know that the x-axis is pointing outwards ? can't it be pointing left or right?

 

[TSny]

The x-axis does not point outwards. Both the x and y axes lie in the plane of the rectangular side. I drew the x-axis as pointing to the right as you look perpendicularly at one of the rectangular sides (into the trough). Pointing to the left would be just as good.

The 6m-long edges of the trough run parallel to the x axis.

The z axis (not shown) would be perpendicular to the rectangular side.

Edit: Actually, in order to correspond to your formula yp = yc + Ixx /(yc A), the origin of the xy system should be at the top edge of one of the rectangular sides. See figure below. The x-axis runs along the top edge that is 6 m long.

yc is the y-coordinate of the centroid in this coordinate system.

 

[foo9008]

The x-axis does not point outwards. Both the x and y axes lie in the plane of the rectangular side. I drew the x-axis as pointing to the right as you look perpendicularly at one of the rectangular sides (into the trough). Pointing to the left would be just as good.

The 6m-long edges of the trough run parallel to the x axis.

The z axis (not shown) would be perpendicular to the rectangular side.

Edit: Actually, in order to correspond to your formula yp = yc + Ixx /(yc A), the origin of the xy system should be at the top edge of one of the rectangular sides. See figure below. The x-axis runs along the top edge that is 6 m long.

yc is the y-coordinate of the centroid in this coordinate system.

so , the x-axis will depends on the orientation of the plane ?
why the x and y-axis shouldn't act at centroid of the plane of the rectangular door?

 

[TSny]

so , the x-axis will depends on the orientation of the plane ?
why the x and y-axis shouldn't act at centroid of the plane of the rectangular door?

If you put the origin at the centroid, then what would be the value of y_c? The formula y_p = y_c + I_xx /(y_c A) would then not make sense.

You posted a picture that shows the y-axis for a general plane surface submerged in a fluid. I reproduced it in the first figure below. I highlighted the y-axis in red. This corresponds to a cross-sectional view of the plane surface. The brown region is the cross section of the plane surface. Note that the origin of the coordinate system is at the surface of the fluid. The x-axis would be running out of the page through the origin (along the surface of the fluid). In this picture, the plane surface is well submerged beneath the surface of the fluid. For the trough in your problem, the surface of the water is at the top of the trough as shown in the second figure below.

 

[foo9008]

You have the formulas yp= yc +(Ixx) /ycA , Ixx = a(b^3)/12 for one of the rectangular sides of the trough. The x and y axes lie in the plane of the rectangle with the origin at the centroid. The x-axis is horizontal while the y-axis is perpendicular to the x-axis and increases toward the bottom of the trough.

for the plane surface , how to know that which is y and x -axis? I am confused..

 

[foo9008]

how to determine the y and x-axis of a surface ? is there any method ?

 

[TSny]

Can you be more specific about why you are having difficulty with setting up the axes? The axes are shown in the figures in posts 23, 38, and 40 of this thread.

The submerged planar surface lies in a plane. The x-y coordinate system is chosen such that the submerged planar surface lies in the x-y plane of the coordinate system with the x-axis running along the surface of the water. You can choose the origin of the coordinate system arbitrarily as long as the the previously stated conditions are met. Note that if the planar surface is tilted with respect to the vertical, then the x-y coordinate plane will also be tilted.

In your problem, you are concerned with each of the rectangular sides of the trough. They are tilted at a 45 degree angle to the vertical.

Consider a rectangle that is initially oriented vertically and submerged in water until the surface of the water is at the top edge of the rectangle, as shown in the attachment below. The figure on the left shows a particular choice of the x and y axes where the y-axis passes through the center of the rectangle. The right figure shows the y-axis shifted arbitrarily to the left. Either choice of position of the y-axis is OK.

Now imagine rotating the vertical rectangle about the x-axis so that it is no longer vertical. You would also rotate the x-y plane of the coordinate system so that the x-y plane still contains the rectangle. That is, the x-y plane and the rectangle are both tilted the same relative to the vertical.

You might find this video helpful:

 

[foo9008]

Can you be more specific about why you are having difficulty with setting up the axes? The axes are shown in the figures in posts 23, 38, and 40 of this thread.

The submerged planar surface lies in a plane. The x-y coordinate system is chosen such that the submerged planar surface lies in the x-y plane of the coordinate system with the x-axis running along the surface of the water. You can choose the origin of the coordinate system arbitrarily as long as the the previously stated conditions are met. Note that if the planar surface is tilted with respect to the vertical, then the x-y coordinate plane will also be tilted.

In your problem, you are concerned with each of the rectangular sides of the trough. They are tilted at a 45 degree angle to the vertical.

Consider a rectangle that is initially oriented vertically and submerged in water until the surface of the water is at the top edge of the rectangle, as shown in the attachment below. The figure on the left shows a particular choice of the x and y axes where the y-axis passes through the center of the rectangle. The right figure shows the y-axis shifted arbitrarily to the left. Either choice of position of the y-axis is OK.

Now imagine rotating the vertical rectangle about the x-axis so that it is no longer vertical. You would also rotate the x-y plane of the coordinate system so that the x-y plane still contains the rectangle. That is, the x-y plane and the rectangle are both tilted the same relative to the vertical.

You might find this video helpful:

well , i know to choose the surface already . I am just not sure for the surface , which one should be x , and which one should be y-axis . How to choose x and y -axis in a plane surface ? is there any specific method?

 

[TSny]

I am just not sure for the surface , which one should be x , and which one should be y-axis . How to choose x and y -axis in a plane surface ? is there any specific method?

I'm not sure what "one" refers to when you ask, "which one should be x"?

 

[foo9008]

I'm not sure what "one" refers to when you ask, "which one should be x"?

for the 2d surface submerged in water , which one should be the x-axis ? which one should be the y-axis ?

 

[TSny]

Are you saying that you understand how the axes are oriented, but you don't know which of the axes is the x axis?

 

[foo9008]

Are you saying that you understand how the axes are oriented, but you don't know which of the axes is the x axis?

yes

 

[TSny]

The axis that runs along the surface of the water is the x axis.

 

[foo9008]

The axis that runs along the surface of the water is the x axis.

no matter how the plane is tilted , The axis that runs along the surface of the water is the x-axis ?
this will always true ?

 

[TSny]

Yes. The formula that you gave for y_p assumes that the x-axis runs along the surface of the water.

For any orientation of the plane surface, imagine extending that plane in all directions. The extended plane will intersect the surface of the water along a straight line. That straight line corresponds to the x axis.

 

[foo9008]

Yes. The formula that you gave for y_p assumes that the x-axis runs along the surface of the water.

For any orientation of the plane surface, imagine extending that plane in all directions. The extended plane will intersect the surface of the water along a straight line. That straight line corresponds to the x axis.

what do you mean ?

 

[TSny]

what do you mean ?

The "extended plane" is just the infinite plane that the planar surface lies in. Take the door in your room. It is oriented vertically. That is, it is oriented in a vertical plane. If you imagined all four edges of the door becoming bigger and bigger until the edges are infinitely long, the door would become an infinite plane. That infinite plane is the infinite plane in which the original door was lying.

Likewise, if you took a planar surface of any shape then you can imagine the infinite plane in which the surface lies. If the planar surface is submerged in water, then the infinite plane in which the planar surface lies will intersect the surface of the water (unless the planar surface is horizontal).

In the case of the trough in your problem, the top edge of a rectangular side is already at the surface of the water. (No need for "extending the plane".) So, for that rectangular side, the top edge is the x-axis. See the picture below which shows the x-axis for the yellow rectangular side.

 

[TSny]

One thing that might be confusing is that the quantity $I_{xx}$ in the formula for $y_p$ represents the second moment of the area of the planar region about an axis that is parallel to the x-axis and that runs through the centroid of the planar area. It is not the second moment about the x-axis at the surface of the water.

 

[foo9008]

Yes. The formula that you gave for y_p assumes that the x-axis runs along the surface of the water.

For any orientation of the plane surface, imagine extending that plane in all directions. The extended plane will intersect the surface of the water along a straight line. That straight line corresponds to the x axis.

when you extend the line , the straight line that intersect with the water surface should be y-aixs , am right ? since we have yp ( center of pressure) , yc ( center of centroid) ...

 

[TSny]

when you extend the line

I was extending a planar region, not a line.

the straight line that intersect with the water surface should be y-aixs , am right ?

The extended plane intersects the surface of the water. The surface of the water lies in a horizontal plane. So, the intersection of the extended plane with the surface of the water corresponds to the intersection of two planes. Two planes intersect in a straight line. This straight line is the x axis.

The y-axis is perpendicular to the x-axis and the y-axis lies in the extended plane of the submerged planar surface. Yes, the y-axis will intersect the surface of the water at a point (the origin of the x-y coordinate system). As you move from the origin along the y axis, you move deeper into the water.

The first minute of the video that I linked to in post #43 explains the orientation of the x and y axes and the location of the origin. Can you pinpoint which part of the video that is not clear to you?

 

[foo9008]

I was extending a planar region, not a line.

The extended plane intersects the surface of the water. The surface of the water lies in a horizontal plane. So, the intersection of the extended plane with the surface of the water corresponds to the intersection of two planes. Two planes intersect in a straight line. This straight line is the x axis.

The y-axis is perpendicular to the x-axis and the y-axis lies in the extended plane of the submerged planar surface. Yes, the y-axis will intersect the surface of the water at a point (the origin of the x-y coordinate system). As you move from the origin along the y axis, you move deeper into the water.

The first minute of the video that I linked to in post #43 explains the orientation of the x and y axes and the location of the origin. Can you pinpoint which part of the video that is not clear to you?

ok , i think i gt the idea now . so , y-axis is the axis that bascially increases with the depth ?

 

[TSny]

ok , i think i gt the idea now . so , y-axis is the axis that bascially increases with the depth ?

Yes, that's right.