# Fluids Help

1. Jun 10, 2005

### DDS

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. Jun 10, 2005

### FredGarvin

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. Jun 10, 2005

### DDS

I am not. What is Archimede's Principle?

4. Jun 10, 2005

### DDS

i have found his principle but how would i determine the amount of volume displaced?
i am using

w=D*V*g

5. Jun 10, 2005

### FredGarvin

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. Jun 10, 2005

### DDS

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. Jun 11, 2005

### FredGarvin

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:
$$F_b = (\rho_{water} )(V_{block})(g)$$

$$F_b = (1000 \frac{kg}{m^3})(6.86x10^-3 m^3)(9.81 \frac{m}{s^2})$$

$$F_b = 67.3 N$$

Now go back and use Hooke's Law to recalculate the deflection. I get .44 m for delta L.

8. Jun 11, 2005

### DDS

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. Jun 11, 2005

### DDS

ive gone over the calculations and they all seem correct...could it be a matter of units?

10. Jun 11, 2005

### DDS

can anyone tell me where i went wrong??

11. Jun 11, 2005

### OlderDan

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. Jun 11, 2005

### DDS

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. Jun 11, 2005

### OlderDan

No. Draw the forces acting on the block. Be careful with the directions.

14. Jun 11, 2005

### DDS

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. Jun 11, 2005

### DDS

16. Jun 12, 2005

### DDS

a verification or further suggestion would help because i have afeeling its wrong

17. Jun 12, 2005

### FredGarvin

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.

$$F_B - W - F_s = 0$$

$$67.3 N - (4.63)(9.81) - F_s = 0$$

$$67.3 - 45.4 - F_s = 0$$

$$F_s = 21.9 N$$

$$F_s = k \Delta L$$

$$21.9 N = 153 \frac{N}{m} \Delta L$$

$$\Delta L = .14 m$$

Does that answer jive? Hopefully I haven't missed anything...

18. Jun 12, 2005

### DDS

Thats what i had and was waiting for someone to verify , thanks so much