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