Fluids Help

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  • #1
DDS
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A light spring of constant k = 153 N/m rests vertically on the bottom of a large beaker of water (as seen in the picture below).

An m=4.63 kg block of wood (density = 675 kg/m3) is connected to the spring and the mass-spring system is allowed to come to static equilibrium (b).
What is the elongation, ΔL, of the spring?

I have tried this but it doesnt seem to give me the right answer:

L= (Volume of water - mass of block/ K) * 9.81

can anyone please give me some detailed help?
 

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  • #2
FredGarvin
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Are you familiar with Archimede's Principle? Use that to determine the weight of the block in the water and then use Hooke's Law to determine the deflection in the spring.
 
  • #3
DDS
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I am not. What is Archimede's Principle?
 
  • #4
DDS
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i have found his principle but how would i determine the amount of volume displaced?
i am using

w=D*V*g
 
  • #5
FredGarvin
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The volume displaced is the volume of the block.

Specifically, you are looking for the buoyant force. It will be equal to the weight of the water displaced by the block.
 
  • #6
DDS
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what am i doing wrong here:

v=m/d
v=4.63/675
v=6.8959e-3

Fb=W=D*V*g
Fb=W=675*6.8959e-3*9.81
Fb=W=45.4203

now using Hookes Law:

F=-kx
-(45.4203/153)=x
x=-0.297 m

which is wrong any suggestions where i went wrong?
 
  • #7
FredGarvin
Science Advisor
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Your calculation for the buoyant force is incorrect. Think of it this way:

The block has a volume, that you already calculated, of V= 6.86x10^-3 m^3. How much does that same volume of water weigh? That is going to be:
[tex]F_b = (\rho_{water} )(V_{block})(g)[/tex]

[tex]F_b = (1000 \frac{kg}{m^3})(6.86x10^-3 m^3)(9.81 \frac{m}{s^2})[/tex]

[tex]F_b = 67.3 N [/tex]

Now go back and use Hooke's Law to recalculate the deflection. I get .44 m for delta L.
 
  • #8
DDS
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hmm thats odd wen i put the asnwer in , it says its incorect

ive tired both positive and negative versions as well

any ideas where we went wrong?
 
  • #9
DDS
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ive gone over the calculations and they all seem correct...could it be a matter of units?
 
  • #10
DDS
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can anyone tell me where i went wrong??
 
  • #11
OlderDan
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DDS said:
can anyone tell me where i went wrong??
I think you failed to include the weight of the block as a force acting downward. The net force on the block is the spring force plus weight (downward) plus buoyant force (upward).
 
  • #12
DDS
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So

w=mg
=4.63*9.81
=45.4

45.4+67.3 =112.7203

F=-(kx)
-(112/153)=x
x=0.736 m ??

is this correct because i only have one attempt left
 
  • #13
OlderDan
Science Advisor
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DDS said:
So

x=0.736 m ??

is this correct because i only have one attempt left
No. Draw the forces acting on the block. Be careful with the directions.
 
  • #14
DDS
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okay the negative direction of that is it correct... because the fources acting on the block are:

Fb=up
Fsp=down
Fg=down

however i feel that the negative direction of that answer is incorrect
 
  • #15
DDS
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can anyone please verify
 
  • #16
DDS
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a verification or further suggestion would help because i have afeeling its wrong
 
  • #17
FredGarvin
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DDS said:
hmm thats odd wen i put the asnwer in , it says its incorect

ive tired both positive and negative versions as well

any ideas where we went wrong?
Sorry. I fat-fingered my answer. I get a deflection of .14m, not .44m.

In my FBD, I have 3 forces acting: the buoyant force (Fb) acting in the positive direction, the weight of the block (W) acting in the negative direction and the spring force (Fs) acting in the negative direction.

[tex]F_B - W - F_s = 0[/tex]

[tex]67.3 N - (4.63)(9.81) - F_s = 0[/tex]

[tex]67.3 - 45.4 - F_s = 0[/tex]

[tex]F_s = 21.9 N[/tex]

[tex]F_s = k \Delta L[/tex]

[tex]21.9 N = 153 \frac{N}{m} \Delta L[/tex]

[tex]\Delta L = .14 m [/tex]

Does that answer jive? Hopefully I haven't missed anything...
 
  • #18
DDS
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Thats what i had and was waiting for someone to verify , thanks so much
 

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