a wooden block of mass 8 kg is tied to a string attatched to the bottom of the tank.In equilibrium the block is completely immersed in water. if relative density of wood is 0.8 and g=10m/s^2 then what is the value of tension in the string?

- anonymous

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- chestercat

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- anonymous

any 1 pleeeese help

- anonymous

hey jamesj pleese help

- JamesJ

There are three forces acting on the block of wood
F_g = force of gravity (down)
F_b = force of buoyancy (up)
F_s = force from the string (down)
The block is in equilibrium hence the sum of the three is zero.
Use that to solve for F_s. The magnitude of F_s is the tension in the string.

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## More answers

- JamesJ

Make sense?

- anonymous

wow
i am extremely convinced

- anonymous

but for more convinciation can u pleese solve the problem for me

- JamesJ

No ... I want you to take the next steps. What are expressions for
\[ F_g \]
and
\[ F_b \] ?

- JamesJ

F_g is particularly easy

- anonymous

is f_g is mg?

- JamesJ

yes

- anonymous

so u dono the expressions?

- JamesJ

I do know. Of course.
But I want you to learn by doing, not watching.

- anonymous

k pleese wait

- JamesJ

To find F_b, go back to first principles and definitions
The buoyant force is equal to the weight of water displaced.
hence you need to find the weight of water displaced.
The find the weight of water displaced, you'll need to find the volume of water displaced and multiply it by the density of water.
Then to find the volume of water displaced ...

- anonymous

buoyant force=weight of water displaced

- anonymous

i.e F_b=mg

- JamesJ

Ok, yes, where m here is the mass of the water displaced

- anonymous

ya

- anonymous

so now can u solve

- JamesJ

We use rho, \( \rho \) for density. Hence you have by definition
\[ \rho_{water} = \frac{m_{water}}{V_{water}} \]
where V is volume. Hence
\[ m_{water} = \rho_{water}.V_{water} \]
Now find \( V_{water} \). It must be equal to \( V_{block} \). Hence you have to find \( V_{block} \). Use the same density principle. That's why the density of wood was given to you in the problem.

- anonymous

so wat's the use of relative density

- JamesJ

Tell me first what's the equation of the volume of the block?

- anonymous

of cousre it is m/d

- anonymous

d is the density

- JamesJ

Hence
\[ m_{water} = \rho_{water}.V_{water} \]
\[ = \rho_{water} . V_{block} \]
because the volume of water displaced must equal the volume of the block
\[ = \rho_{water}.\frac{m_{block}}{\rho_{block}} \ \ \ \ \text{ by definition of density } \]
\[ = \frac{\rho_{water}}{\rho_{block}}. m_{block} \]
See what to do now?

- anonymous

i cant understand the last equation

- anonymous

k thanks for yor help
iam really happy with the presence of u

- JamesJ

what don't you understand about the last equation.

- anonymous

no other guy was able to answer my qusetion

- anonymous

i'll send u a frnd request in face book

- anonymous

what is your fb username
and i am nt big enough to misuse it

- JamesJ

By definition
\[ \rho_{block}/\rho_{water} \]
is the relative density of the block, i.e., is equal to 0.8
Therefore
\[ m_{water} = \rho_{water}/\rho_{block}.m_{block} \]
\[ = (\rho_{block}/\rho_{water})^{-1}.m_block \]
\[ = (0.8)^{-1}.m_block \]
\[ = (5/4).m_block \]
\[ = (5/4).(8 \ kg) \]
\[ = 10 \ kg \]
Now I want you to make an effort to finish the problem yourself, as I've now shown you 90% of it.

- JamesJ

I should have written
\[ m_{block} \]
not
\[ m_block \]

- anonymous

wow wow wo o

- JamesJ

I'm going to help someone else with a question. Try and figure it out while I'm gone. I'll keep an eye on what's going on over here.

- anonymous

k by the way pleese tell yor fb username

- JamesJ

No, I don't do that, sorry.

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