What is the tension in a bar submerged in water?

[foo9008]

Homework Statement

calculate the tension in the cable

Homework Equations

The Attempt at a Solution

the resultant force act on the each metal bar is equal to the tension of each bar . so FR= tension in bar
but , i have problem of finding the the area in FR , i have only (0.75/2) sin45(1000)(9.81) A . For A , i only know the one of the length , how to do this question ?

 

[SteamKing]

Homework Statement

calculate the tension in the cable

Homework Equations

The Attempt at a Solution

the resultant force act on the each metal bar is equal to the tension of each bar . so FR= tension in bar
but , i have problem of finding the the area in FR , i have only (0.75/2) sin45(1000)(9.81) A . For A , i only know the one of the length , how to do this question ?

It's not clear what FR is.

In any event, are you saying you can't work out the dimensions of this trough using trigonometry and the information given in the figure?

 

[TSny]

The Attempt at a Solution

the resultant force act on the each metal bar is equal to the tension of each bar .

I'm not sure what you mean here.

To add a little to SteamKing's comment: You are dealing with a trough with two rectangular side panels that are hinged at the bottom and a triangular panel at each end. The cable prevents the rectangular panels from rotating about the hinge. This is a rotational equilibrium problem. (unless I'm misinterpreting the problem)

 

[foo9008]

It's not clear what FR is.

In any event, are you saying you can't work out the dimensions of this trough using trigonometry and the information given in the figure?

FR = resultant force hydrostatic force acting on the metal bar

 

[foo9008]

I'm not sure what you mean here.

To add a little to SteamKing's comment: You are dealing with a trough with two rectangular side panels that are hinged at the bottom and a triangular panel at each end. The cable prevents the rectangular panels from rotating about the hinge. This is a rotational equilibrium problem. (unless I'm misinterpreting the problem)

Is there anything wrong with my working ?

 

[TSny]

i have only (0.75/2) sin45(1000)(9.81) A . For A , i only know the one of the length , how to do this question ?

I'm not sure what you are calculating here. Can you elaborate? It appears to be a certain weight of water.

Since this is a rotational equilibrium problem, you will need to consider torques rather than just forces.

You might want to consider the torque about the hinge due to fluid pressure on a narrow strip of the trough. See the yellow strip in the figure below.

 

[foo9008]

I'm not sure what you are calculating here. Can you elaborate? It appears to be a certain weight of water.

Since this is a rotational equilibrium problem, you will need to consider torques rather than just forces.

You might want to consider the torque about the hinge due to fluid pressure on a narrow strip of the trough. See the yellow strip in the figure below.

I am calculating the resultant hydrostatic pressure acting on the two metal bar ... I'm not sure how to do the calculation for the torque about the hinge due to fluid pressure on a narrow strip of the trough.

 

[TSny]

I am calculating the resultant hydrostatic pressure acting on the two metal bar ... I'm not sure how to do the calculation for the torque about the hinge due to fluid pressure on a narrow strip of the trough.

The problem doesn't mention any metal bars. My interpretation of the diagram is that it represents a cross-sectional view of the trough.

 

[foo9008]

The problem doesn't mention any metal bars. My interpretation of the diagram is that it represents a cross-sectional view of the trough.

Ok , i misunderstood it as metal bar . sorry . i changed to trough now , my calculation is still the same . FR , i have only (0.75/2) sin45(1000)(9.81) A , can you show your working ?

 

[TSny]

FR , i have only (0.75/2) sin45(1000)(9.81) A

What is FR? Please explain the reasoning behind your mathematical expression.

can you show your working ?

No. This problem is for you to work.
We are here to give a little guidance, but you must do the work.

 

[foo9008]

What is FR? Please explain the reasoning behind your mathematical expression.

No. This problem is for you to work.
We are here to give a little guidance, but you must do the work.

FR is the total resultant hydrostatic pressure force acting on the trough

 

[TSny]

FR is the total resultant hydrostatic pressure force acting on the trough

Is this the hydrostatic force on just one of the rectangular sides of the trough?

 

[foo9008]

Is this the hydrostatic force on just one of the rectangular sides of the trough?

just 1 side

 

[TSny]

OK. I think that's right. But knowing the force on one side is not very helpful for this problem. The side of the trough is free to rotate about the hinged edge. The water pressure tends to make the side rotate about the hinge while the cable prevents the side from rotating.

What is the basic mechanical condition that must be satisfied in order for an object not to rotate about some axis?

 

[foo9008]

OK. I think that's right. But knowing the force on one side is not very helpful for this problem. The side of the trough is free to rotate about the hinged edge. The water pressure tends to make the side rotate about the hinge while the cable prevents the side from rotating.

What is the basic mechanical condition that must be satisfied in order for an object not to rotate about some axis?

ok , can you show which is the direction of tension . I coudn't figure out

 

[foo9008]

OK. I think that's right. But knowing the force on one side is not very helpful for this problem. The side of the trough is free to rotate about the hinged edge. The water pressure tends to make the side rotate about the hinge while the cable prevents the side from rotating.

What is the basic mechanical condition that must be satisfied in order for an object not to rotate about some axis?

moment anticlockwiese= moment clockwise, but how to find the pressuree center ? it's given by the formula of yp= yc +(Ixx) /ycA , while Ixx has the formula of a(b^3)/12 , since we can't find the area of this question , is it possible to find the answer for this question ?

 

[TSny]

moment anticlockwiese= moment clockwise,

Yes.

but how to find the pressuree center ? it's given by the formula of yp= yc +(Ixx) /ycA , while Ixx has the formula of a(b^3)/12

I'm not familiar with these expressions. When giving formulas that are not "well-known", you should define all symbols.

we can't find the area of this question

The area of one side of the trough can be calculated easily from the information given in the statement of the problem and the diagram given in the problem.

is it possible to find the answer for this question ?

Yes. You can figure out the moment due to the fluid pressure by considering the moment on a thin strip (see post #7) and then adding the moments for all such strips. Or, if you were given the formula that you posted above, you could use it. I don't know if the formula is valid since I don't know what the symbols represent.

 

[foo9008]

Yes.

I'm not familiar with these expressions. When giving formulas that are not "well-known", you should define all symbols.
The area of one side of the trough can be calculated easily from the information given in the statement of the problem and the diagram given in the problem.
Yes. You can figure out the moment due to the fluid pressure by considering the moment on a thin strip (see post #7) and then adding the moments for all such strips. Or, if you were given the formula that you posted above, you could use it. I don't know if the formula is valid since I don't know what the symbols represent.

what is the meaning of the trough is 6m long ? which side ? can you point it out ? i could only see the length of trough is 0.75m long , the 'width' of the trough is unknown , how to do this question ?

 

[TSny]

The distance from one triangular end to the other triangular end is 6 m.

 

[foo9008]

The distance from one triangular end to the other triangular end is 6 m.

the pressure act on the metal trough, right? how to find the area of the metal trough ? i have only 0.75m

 

[haruspex]

the pressure act on the metal trough, right? how to find the area of the metal trough ? i have only 0.75m

Since the pressure is not uniform, knowing the area of a side of the trough is not helpful. But the fact that you ask makes me think you do not understand the physical arrangement. TSny's diagram should have been more than adequate.
The trough is formed from two plates. Each plate has width 0.75m, as shown, and a length of 6m, into the page.

 

[foo9008]

Since the pressure is not uniform, knowing the area of a side of the trough is not helpful. But the fact that you ask makes me think you do not understand the physical arrangement. TSny's diagram should have been more than adequate.
The trough is formed from two plates. Each plate has width 0.75m, as shown, and a length of 6m, into the page.

for the center of pressure , by using the formula of yp = yc + Ixx / (ycA) = 0.375 +(0.75)(6^3) / (12 x 0.375 x 0.75 x6 ) = 8.375 m the answer is rather weird , since the length of trough is 0.75m

yp = center of pressure where the resultant force acts on, yc = centroid of trough

 

[foo9008]

for the center of pressure , by using the formula of yp = yc + Ixx / (ycA) = 0.375 +(0.75)(6^3) / (12 x 0.375 x 0.75 x6 ) = 8.375 m the answer is rather weird , since the length of trough is 0.75m

yp = center of pressure where the resultant force acts on, yc = centroid of trough

or the a is 6m , b = 0.75m ? how to determine a and b ? I'm confused

 

[TSny]

or the a is 6m , b = 0.75m ?

Yes.

how to determine a and b ?

That information should have been given by whatever source you used to get the formula.

 

[foo9008]

Yes.That information should have been given by whatever source you used to get the formula.

why a=6 , b = 0.75 ? since the trough is slanted , so i 'turn ' it into perfect horizontal . so , when calculating the second moment of inertia about x-axis , i have the 6m 'cut thru ' x-axis , so that's why my b = 6m

 

[TSny]

when calculating the second moment of inertia about x-axis , i have the 6m 'cut thru ' x-axis , so that's why my b = 6m

Why do you say that $b$ = 6 m rather than $a$ = 6 m?

 

[haruspex]

yp = yc + Ixx / (ycA)

Centre of pressure is defined by an integral which allows for an arbitrary pressure distribution. The formula you quote is only applicable in some specific pressure distribution. It might not be appropriate in this problem. Can you quote exactly what all the variables represent and what physical circumstance the formula applies to?

 

[TSny]

See pages 6 and 7 here: http://www.sfu.ca/~mbahrami/ENSC 283/Notes/Fluid Statics.pdf

 

[foo9008]

Why do you say that $b$ = 6 m rather than $a$ = 6 m?

here's what i mean in the previous post

 

[TSny]

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

 

[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 ?