Question

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?

Answer 1

any 1 pleeeese help

Answer 2

hey jamesj pleese help

Answer 3

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.

Answer 4

Make sense?

Answer 5

wow
i am extremely convinced

Answer 6

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

Answer 7

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

Answer 8

F_g is particularly easy

Answer 9

is f_g is mg?

Answer 10

yes

Answer 11

so u dono the expressions?

Answer 12

I do know. Of course.

But I want you to learn by doing, not watching.

Answer 13

k pleese wait

Answer 14

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 ...

Answer 15

buoyant force=weight of water displaced

Answer 16

i.e F_b=mg

Answer 17

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

Answer 18

ya

Answer 19

so now can u solve

Answer 20

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.

Answer 21

so wat's the use of relative density

Answer 22

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

Answer 23

of cousre it is m/d

Answer 24

d is the density

Answer 25

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?

Answer 26

i cant understand the last equation

Answer 27

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

Answer 28

what don't you understand about the last equation.

Answer 29

no other guy was able to answer my qusetion

Answer 30

i'll send u a frnd request in face book

Answer 31

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

Answer 32

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.

Answer 33

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

Answer 34

wow wow wo o

Answer 35

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.

Answer 36

k by the way pleese tell yor fb username

Answer 37

No, I don't do that, sorry.