View Full Version : Application to physics

hollysti

05-12-2007, 02:50 PM

A cylinder whose cross-section is an isolceles triangle with sides 5 in, 5 in, and 6 in has a height H = 20 in and weight 5 Ib. After being pushed down so that its top is level with the surfaceof the water and released, the cylinder bobs, with axis vertical, in water whose weight per unit volume is 62.5 Ib/ft^3. Solve for the motion y(t) where y(t) is the position relative to equilibrium. Determine, and state clearly, the period and amplitude of the motion. HINT: Draw a picture.

I really don't have a clue on this one, so help would be greatly appreciated. Thanks!

The cross section of a cylinder is a circle. What do you mean by cross-section?

hollysti

05-12-2007, 04:25 PM

I think the base is the isosceles triangle. This is the problem as it was given to me, so I am not sure, but that is what I was assuming as I read the problem.

skeeter

05-13-2007, 09:04 AM

perhaps a prism rather than a cylinder?

hollysti

05-13-2007, 12:09 PM

Well maybe, but I am just quoting the problem to you. That is exactly what the problem says.

wjm11

05-14-2007, 01:31 AM

A cylinder whose cross-section is an isosceles triangle with sides 5 in, 5 in, and 6 in has a height H = 20 in and weight 5 Ib. After being pushed down so that its top is level with the surface of the water and released, the cylinder bobs, with axis vertical, in water whose weight per unit volume is 62.5 Ib/ft^3. Solve for the motion y(t) where y(t) is the position relative to equilibrium. Determine, and state clearly, the period and amplitude of the motion. HINT: Draw a picture.

Assume a prism, not a cylinder. You need this for is the volume, from which you can calculate the density of the object. The isosceles triangular cross-section allows for an easy calculation of triangle height of 4 inches. Therefore area is 12 in^2 and volume is 240 in^3, leading to a density of 5 lb/240 in^3 = .02083 pci = 36 pcf. As this is less than the 62.5 pcf of water, this object will float.

At equilibrium it will displace 5 lb of water (per Archimedes). Five lb of water occupies a volume of .08 ft^3 or 138.24 in^3. From this it can be determined that the object’s equilibrium position is about 8.48 inches above the waterline and 11.52 inches below. Therefore it has been pushed down 8.48” initially. When released it will oscillate between 8.48” below its equilibrium position and 8.48” above (amplitude). I think.

Please check my assumptions and calculations.

Hope this helps.

hollysti

05-15-2007, 02:13 PM

Is there any way that you could post what work you did to find those values? I am not quite getting how you were able to come to those conclusions... Or if you maybe just want to scan any hand written work I could give you an email address... Thank you very much.

tkhunny

05-15-2007, 02:24 PM

More appropriately, you could show your work so we can see how you are getting what you are getting.

hollysti

05-15-2007, 03:38 PM

Assume a prism, not a cylinder. You need this for is the volume, from which you can calculate the density of the object. The isosceles triangular cross-section allows for an easy calculation of triangle height of 4 inches. Therefore area is 12 in^2 and volume is 240 in^3, leading to a density of 5 lb/240 in^3 = .02083 pci = 36 pcf. As this is less than the 62.5 pcf of water, this object will float.

At equilibrium it will displace 5 lb of water (per Archimedes). Five lb of water occupies a volume of .08 ft^3 or 138.24 in^3. From this it can be determined that the object’s equilibrium position is about 8.48 inches above the waterline and 11.52 inches below. Therefore it has been pushed down 8.48” initially. When released it will oscillate between 8.48” below its equilibrium position and 8.48” above (amplitude). I think.

Please check my assumptions and calculations.

Hope this helps.

Ok, I understand the whole first paragraph... I guess I was confused about the 12 in^2 at first glance, but I see that you were calculating the area of the triangle, not the whole prism. Now what I am not understanding is how to get the volume of 5 pounds of water. I have looked online and haven't really found anything that can help. After that, I haven't really looked at the rest of the problem yet, but i am not sure that I will understand the oscillation part. But if you could help me with this next step, then I will keep going from there and see how it works. Thanks!

Now what I am not understanding is how to get the volume of 5 pounds of water.

density*volume = mass

The density of water varies but is generally assumed to be 1 g/cm^3.

hollysti

05-15-2007, 05:18 PM

ooooooohhhhh... sorry, that was a really silly question. I will keep working through the problem now. Thank you!

hollysti

05-15-2007, 05:55 PM

OK, one last question... Any ideas on how to come up with the period of motion? Thank you all very much for your help! I actually understand this problem :wink:

skeeter

05-15-2007, 06:52 PM

determine the maximum net force exerted on the prism ... when the prism is pushed down to its maximum displacement, max F<sub>net</sub> = buoyant force - weight

now determine the magnitude of a<sub>max</sub> using

max F<sub>net</sub>/m = a<sub>max</sub>

once you have a<sub>max</sub>, use the following formulas to determine the period of SHM.

\L a_{max} = A\omega^2

\L T = \frac{2\pi}{\omega}

hollysti

05-15-2007, 07:22 PM

Sorry, but what is the boyant force in this problem and what is A?

skeeter

05-15-2007, 07:32 PM

The buoyant force is the weight of the water displaced by the object.

A is the variable that represents the maximum displacement from equilibrium in SHM.

hollysti

05-15-2007, 07:46 PM

ok, so the boyant force would be 5 lb and A would be 8.48 inches. What is the weight? I originally thought the weight was 5 lb.

skeeter

05-15-2007, 08:56 PM

the buoyant force is not 5 lbs ... it is the weight of the water displaced by the prism.

hollysti

05-15-2007, 09:37 PM

At equilibrium it will displace 5 lb of water (per Archimedes). Five lb of water occupies a volume of .08 ft^3 or 138.24 in^3.

I thought the weight of the displaced water was equal to the weight of the prism.

skeeter

05-15-2007, 09:53 PM

not when you do this ...

After being pushed down so that its top is level with the surface of the water and released, the cylinder bobs ...

you are trying to find the maximum buoyant force that is achieved when the above description happens, not the buoyant force at equilibrium. at equilibrium, the buoyant force equals the weight, but the net force = 0.

you haven't studied any physics, have you?

hollysti

05-15-2007, 10:33 PM

I haven't had any phisics, and I have almost (amazingly) made it through an entire semester of applied mathematics. Could you tell me what I need to do to get the maximum buoyant force?

skeeter

05-16-2007, 08:37 AM

maximum buoyant force = weight of water displaced by the entire prism

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