Maximum load platform can withstand

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Homework Statement


A platform is supported by a rectangular plastic tank below . When there's no external load act on it , it oscillate on the water surface with amplitude 0.05m . When it's in equilibrium , the water reaches 0.07m of the height of tank submerged in water.
P/s : the tank has a height of 0.35m , and base area of 0.5 (m^2) . Assume it is undamped oscillation
what is the maximum load it can withstand so that the platform still can float on water ?


Homework Equations




The Attempt at a Solution


when it's in equlibrium , the tank submerged 0.06m , the amplitude of oscillaltion is 0.05m , so 0.35m - 0.06m -0.05m = 0.24m ? so the maximum load = buoyant force = (0.24)(1000)(9.81) N , am i right ? The weight of platfrom is then subtracted from (0.24)(1000)(9.81) N to get the max load it can withstand...
is my working correct ?
 

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  • #2
678
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or it should be (0.25)(1000)(9.81) ?
 
  • #3
SteamKing
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or it should be (0.25)(1000)(9.81) ?
What does Archimedes' Principle tell you the buoyant force should be?
 
  • #4
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What does Archimedes' Principle tell you the buoyant force should be?
weight of onject displaced = buoyant force
 
  • #5
SteamKing
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weight of onject displaced = buoyant force
And how do you calculate the weight of the water displaced by the object?
 
  • #6
678
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And how do you calculate the weight of the water displaced by the object?
volume displaced x rho x g
 
  • #7
SteamKing
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volume displaced x rho x g
And that calculation is, for this platform?
 
  • #8
678
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And that calculation is, for this platform?
for the volume displaced of the tank
 
  • #9
SteamKing
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for the volume displaced of the tank
You're getting stuck in an endless loop, here.

What is the actual calculation, you know, using numbers?
 
  • #10
678
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You're getting stuck in an endless loop, here.

What is the actual calculation, you know, using numbers?
pls refer to post 1 for the numbers
 
  • #11
678
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You're getting stuck in an endless loop, here.

What is the actual calculation, you know, using numbers?
why am i wrong ?
 
  • #12
SteamKing
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why am i wrong ?
That's why I'm asking you to re-examine your original calculation, which you seem extremely reluctant to do.
 
  • #13
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That's why I'm asking you to re-examine your original calculation, which you seem extremely reluctant to do.
so , maximum buoyant force = 0.35(1000)(9.81) ???
the maximum weigh it can support is 0.35(1000)(9.81) - (0.06)(10000(9.81) ?
 
  • #14
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That's why I'm asking you to re-examine your original calculation, which you seem extremely reluctant to do.
is it correct now ?
 
  • #15
SteamKing
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is it correct now ?
You're still omitting a very important piece of information from your calculations.

Why don't you check the dimensions of the numbers you do have and see if they are consistent?
 
  • #16
678
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You're still omitting a very important piece of information from your calculations.

Why don't you check the dimensions of the numbers you do have and see if they are consistent?
You mean the amplitude of oscillation of the tank?
 
  • #17
SteamKing
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You mean the amplitude of oscillation of the tank?
No, I mean all the terms in your calculation of the buoyant force.

IIRC, you just did some problems in dimensional analysis. Now, you can put that work to use in analyzing the units of your buoyant force calculation.
 
  • #18
678
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You mean the amplitude of oscillation of the tank?
No, I mean all the terms in your calculation of the buoyant force.

IIRC, you just did some problems in dimensional analysis. Now, you can put that work to use in analyzing the units of your buoyant force calculation.
sorry , it should be (0.35x0.24)(1000)(9.81) N

i am not sure about the value of height of tank immersed in water(0.24) , can you clarify on this ?
 
  • #19
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No, I mean all the terms in your calculation of the buoyant force.

IIRC, you just did some problems in dimensional analysis. Now, you can put that work to use in analyzing the units of your buoyant force calculation.
can you explain about what is the maximum height of tank submerged in water ?
 
  • #20
SteamKing
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sorry , it should be (0.35x0.24)(1000)(9.81) N

i am not sure about the value of height of tank immersed in water(0.24) , can you clarify on this ?
You still are not getting this.

According to the OP, the tank has a base area of 0.5 m2.

Don't you see how this is an essential piece of information with which to calculate the buoyant force acting on the platform?
 
  • #21
678
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You still are not getting this.

According to the OP, the tank has a base area of 0.5 m2.

Don't you see how this is an essential piece of information with which to calculate the buoyant force acting on the platform?
sorry , it should be 0.5 x h x 1000 x 9.81 ....
how to get the value of h so that maximum buoyant force can be obtained?
 
  • #22
SteamKing
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sorry , it should be 0.5 x h x 1000 x 9.81 ....
how to get the value of h so that maximum buoyant force can be obtained?
Well, now that we've got the force calculation into proper form, let's review the problem statement:

Homework Statement


A platform is supported by a rectangular plastic tank below.

When there's no external load act on it , it oscillates on the water surface with amplitude 0.05m.

When it's in equilibrium, the water reaches 0.07m of the height of tank submerged in water.
P.S. : the tank has a height of 0.35m, and base area of 0.5 (m^2).

Assume it is undamped oscillation. What is the maximum load it can withstand so that the platform still can float on water?

The total depth of the tank is 0.35 m, and the tank has a base area of 0.5 m2.

When it is floating at equilibrium (with no oscillations, presumably), the draft of the tank is 0.07 m. From this information, you should be able to calculate the weight of the platform in the unloaded condition.

When the platform is loaded and oscillating, you don't want to submerge the tank, so what must the maximum equilibrium draft be, given that the amplitude of oscillation is 0.05 m and the total depth is as given above?

Once you know this, you should be able to work out how much load the platform can support.

If it helps, draw a picture.
 
  • #23
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Well, now that we've got the force calculation into proper form, let's review the problem statement:



The total depth of the tank is 0.35 m, and the tank has a base area of 0.5 m2.

When it is floating at equilibrium (with no oscillations, presumably), the draft of the tank is 0.07 m. From this information, you should be able to calculate the weight of the platform in the unloaded condition.

When the platform is loaded and oscillating, you don't want to submerge the tank, so what must the maximum equilibrium draft be, given that the amplitude of oscillation is 0.05 m and the total depth is as given above?

Once you know this, you should be able to work out how much load the platform can support.

If it helps, draw a picture.
if there's no oscillation , the maximum height of tank submerged is 0.35-0.07 = 0.28m , at h = 0.28 , the platform support the max load

when there's oscillation , the maximum height of water is 0.35 -0.07-0.05 = 0.23m , am i right ?
 
  • #24
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if there's no oscillation , the maximum height of tank submerged is 0.35-0.07 = 0.28m , at h = 0.28 , the platform support the max load

when there's oscillation , the maximum height of water is 0.35 -0.07-0.05 = 0.23m , am i right ?
 

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  • #25
SteamKing
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if there's no oscillation , the maximum height of tank submerged is 0.35-0.07 = 0.28m , at h = 0.28 , the platform support the max load

when there's oscillation , the maximum height of water is 0.35 -0.07-0.05 = 0.23m , am i right ?
When the tank is at maximum load, the draft at no load does not come into play, since the tank is now loaded.

The oscillation is going to affect only the maximum draft the tank can have when loaded.

In order to find the maximum load which the platform can support and still remain floating, you want to subtract the minimum draft from the maximum draft. That's what 0.23 m represents in your calculation above, the difference in the minimum and maximum drafts of the tank.

Now, what is the load which the platform can carry?
 
  • #26
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When the tank is at maximum load, the draft at no load does not come into play, since the tank is now loaded.

The oscillation is going to affect only the maximum draft the tank can have when loaded.

In order to find the maximum load which the platform can support and still remain floating, you want to subtract the minimum draft from the maximum draft. That's what 0.23 m represents in your calculation above, the difference in the minimum and maximum drafts of the tank.

Now, what is the load which the platform can carry?
what is the meaning of draft here ? what doi you mean by max and min draft ?
 
  • #27
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When the tank is at maximum load, the draft at no load does not come into play, since the tank is now loaded.

The oscillation is going to affect only the maximum draft the tank can have when loaded.

In order to find the maximum load which the platform can support and still remain floating, you want to subtract the minimum draft from the maximum draft. That's what 0.23 m represents in your calculation above, the difference in the minimum and maximum drafts of the tank.

Now, what is the load which the platform can carry?
max weight = (0.23)(0.5)(9.81)(1000) N ?
 
  • #28
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what is the meaning of draft here ? what doi you mean by max and min draft ?
draft is the distance at which the waterline is located above the bottom of the tank, i.e., how deep the tank sinks into the water when it is floating.
 
  • #29
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max weight = (0.23)(0.5)(9.81)(1000) N ?
Turn the crank and how many newtons is that?

BTW, per the discussion in Post #25, this is the maximum weight the tank can carry and remain floating.
 
  • #30
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Turn the crank and how many newtons is that?

BTW, per the discussion in Post #25, this is the maximum weight the tank can carry and remain floating.
so , the max weight = (0.23)(0.5)(9.81)(1000) N ? is correct ?
 
  • #31
haruspex
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so , the max weight = (0.23)(0.5)(9.81)(1000) N ? is correct ?
If we assume the amplitude of the oscillation is still 0.05m, yes. But the question does not make that clear. Why should the amplitude stay the same? With the extra load, the energy flux is greater for the same amplitude. If we were to add some mass, gently, when the tank is at the equilibrium position the amplitude would decrease. Add the right extra mass at the lowest position and it would stop oscillating entirely.
Low marks to the question setter.
 
  • #32
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If we assume the amplitude of the oscillation is still 0.05m, yes. But the question does not make that clear. Why should the amplitude stay the same? With the extra load, the energy flux is greater for the same amplitude. If we were to add some mass, gently, when the tank is at the equilibrium position the amplitude would decrease. Add the right extra mass at the lowest position and it would stop oscillating entirely.
Low marks to the question setter.
do you mean we can only find the maximum mass hat the tank can support with assumption of no oscillation ?
 
  • #33
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If we assume the amplitude of the oscillation is still 0.05m, yes. But the question does not make that clear. Why should the amplitude stay the same? With the extra load, the energy flux is greater for the same amplitude. If we were to add some mass, gently, when the tank is at the equilibrium position the amplitude would decrease. Add the right extra mass at the lowest position and it would stop oscillating entirely.
Low marks to the question setter.
can i find the maximum buoyant force by not considering the amplitude of the oscilllation of the platform when there is no extra load on it ? if so , the maximum buoyant force = (0.28)(0.5)(9810)N ???
 
  • #34
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If we assume the amplitude of the oscillation is still 0.05m, yes. But the question does not make that clear. Why should the amplitude stay the same? With the extra load, the energy flux is greater for the same amplitude. If we were to add some mass, gently, when the tank is at the equilibrium position the amplitude would decrease. Add the right extra mass at the lowest position and it would stop oscillating entirely.
Low marks to the question setter.
assuming the oscillation is 0.05m when the platform is empty and fully loaded , is the maximum buoyant force = 0.23)(0.5)(9.81)(1000) N ?
 
  • #35
haruspex
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assuming the oscillation is 0.05m when the platform is empty and fully loaded , is the maximum buoyant force = 0.23)(0.5)(9.81)(1000) N ?
Yes, I confirmed that in post #31. I think you will have to assume that.
 

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