Rod Suspended in Fluid

1. Jan 22, 2012

Rapier

1. The problem statement, all variables and given/known data

A metal alloy rod is submerged 15 cm below the surface of a fresh water pool by steel cables tied 10cm from each end. It has a length of 110 cm, a mass of 1.5 kg and a uniform square cross sectional area of 5 cm2. Because its density is not uniform, its center of mass is located 30 from the left end. The total depth of the water is 45 cm.

a) What is the force of tension on the left cable?

b) What is the force of tension on the right cable?

2. Relevant equations

T = F*d
Fb = ρgV
F = ma
Ftension = -Fbuoy + Fgrav + Fwaterpress

3. The attempt at a solution

I have three 5 forces acting on the rod.
- The pressure of the water pressing down on the rod
- The force of gravity acting down on the rod (@ center of gravity 30cm from the left end)
- The buoyant force acting up on the rod (@ the same point as the gravitational force)
- Tension from the wire on the left end
- Tension from the wire on the right end

Since there is no rotation on the rod the sum of the torques must be zero. Since there is no motion on the rod, the sum of the forces must also be zero.

I believe I need to include the torques in the solution someplace but I'm just not seeing it.

2. Jan 22, 2012

ehild

The pressure of the water is taken into account by the buoyant force.

According to Pascal's Law, the hydrostatic pressure P at depth D in the fluid results in a normal force of magnitude PA on a surface area A independently of its orientation. If you have a block submerged in water it experiences a downward force Fd=ρgD1A +Po at is top surface and an upward force Fu=Fd=ρgD2A+Po at the base. The buoyant force is the resultant of these forces. If the base area is A and the height of the block is h, the resultant is
FB=ρgA(D2-D1)=ρgAh = ρgV where V is the immersed volume.

The force of gravity acts at the CM, but the bouyant force acts at the CM of the displaced fluid. As the rod has uniform cross section, the CM of the displaced water is at the middle of the rod.

ehild

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3. Jan 23, 2012

Rapier

I see.

So if I use the right wire as my pivot the sum of the torques is equal to:

0 = Ftension-left*(.9m) + Fbuoyancy*(.45m) - Fgravity*(.7m) + Ftension-right*(0m)

Fbuoyancy = A*L*ρ*g
Fbuoyancy = (5e-4 m^2)*(1.1m)*(1e3 kg/m^3)*g
Fbuoyancy = 5.394 N

Fgravity = mg
Fgravity = (1.5 kg)g
Fgravity = 14.71 N

Ftension-left*(.9m) = -(5.394 N)*(.45m) + (14.71 N)*(.7m)
Ftension-left = 8.7441 N

However, it tells me my answer is incorrect.

EDIT: I miskeyed the equation in my calculator. 8.7441 N is the correct answer.

Last edited: Jan 23, 2012
4. Jan 23, 2012

ehild

If you use a point along the rod as pivot the pieces of rod at the opposite sides exert opposite torques. You can not assume that the whole mass is concentrated in the CM. As the mass distribution is not known you do not know where is it concentrated. Maybe, it is in the 10 cm length part outside the pivot.

Write up the torque equation with respect to one end of the rod, and also the sum of the forces which has to be zero. You get two equations for the unknown tensions.

ehild

5. Jan 23, 2012

Rapier

I chose the point of the right wire as my pivot so that the level arm at that point would be zero and its torque would be zero.

Actually, I believe I can. The problem specifically states that the center of mass is 30cm from the left end.

EDIT: I had mislabeled my diagram and had an incorrect distance on one of the torques. I fixed the distance to the center of mass and got the correct answer.

My previous response was the equation for the torques pivoting around the point where the wire on the right side is attached. Once I have used the torque equation to determine the tension on the left wire I can set the forces equal to zero and calculate the right.

Last edited: Jan 23, 2012
6. Jan 23, 2012

Rapier

Tension on the left: 8.7441 N
Tension on the right: .5719 N

Thanks for all the help!

7. Jan 23, 2012

ehild

Ops! You were right. The place of the CM was explicitly given. Good work!

ehild

Last edited: Jan 23, 2012
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