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In the diagram we have two bodies, with density and volume $\delta$ and $V$, and a fluid $\ell$.

$$\delta_1=0,7 \rm\frac{gr}{cm^3}$$ $$\delta_2=2 \rm\frac{gr}{cm^3}$$ $$V_1=100 \rm {cm^3}$$ $$V_2=10 \rm {cm^3}$$ $$\delta_\ell = 1,2 \rm\frac{gr}{cm^3}$$

The system is static.

enter image description here

I need to find $(1)$ The tension of the rope $(2)$ the buoyancy $1$ receives due to the liquid.

I solved it, but I'm getting strange results:

Let $W$ be the weighs, $B$ be the buoyancy and $T$ the tension. Then, since the system is static:

$$B_1+B_2-W_1-W_2=0$$

(the tensions cancel out)

But $W_1= \delta_1 \cdot g \cdot V_1$, $W_2= \delta_2 \cdot g \cdot V_2$ and $B_2 = \delta_\ell \cdot g \cdot V_2$ (since the body is fully submerged):

$$\eqalign{ 10\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}\left( {70{\text{gr}} + 20{\text{gr}} - 12{\text{gr}}} \right) = B_1 \cr 10\frac{\rm m}{{{\rm s^2}}}\left( {0,078\rm{kg} } \right) = B{_1} \cr 0,78{\text{N}} = B{_1} \cr} $$

It seems strange that the buoyancy is greater than the weigh of the body ($0,7 \rm N$), is that possible? Also, I can't figure out how to obtain the tension.

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1 Answer 1

up vote 2 down vote accepted

Remember that buoyancy is just a force. There's no reason it necessarily has to bear any particular relationship to weight.

Try drawing a free-body diagram for each of the two objects individually, and writing the corresponding force equation, and you should see what's going on.

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But is it legitimate to "separate" the system? If the rope is cut, $1$ will probably float upwards, and $2$ will sink, I suppose. –  Peter Tamaroff Jun 29 '12 at 4:09
1  
Yes, it is. When you have a system that consists of multiple individual objects, you can always draw a free body diagram (and write the force equation) for any of the objects individually, or for the system as a whole, or indeed for any possible grouping of the objects. All these equations are going to be satisfied. Drawing a diagram for part of the system doesn't mean cutting the rope. –  David Z Jun 29 '12 at 4:11
    
I didn't mean I couldn't draw the FBD. I think I get what you mean - I can consider $T_2+B_2-W_2=0$ for body $1$. Ideal conditions will ensure $T_1=-T_2$. –  Peter Tamaroff Jun 29 '12 at 4:14
    
Yeah... though how does the weight of object 2 $W_2$ act on body 1? ;-) –  David Z Jun 29 '12 at 4:15
    
I don't follow that last hint =/. I've determined $T=0,08N$, though. –  Peter Tamaroff Jun 29 '12 at 4:19

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