Buoyancy force when a ramp is under water

[Kolika28]

Homework Statement

A solid block, M of mass 1kg, is sliding down a ramp which makes an angle θ=15∘ with the horizontal and has a kinetic coefficient of friction of 0.06. Now assume that you put the ramp and block into a bucket of water so that everything is immersed in water (the block is fully covered). The block is sitting at a height h on the ramp. You can assume the density of water is the same everywhere. Draw the new force diagram and write an equation for the initial acceleration of the block down the ramp. You can ignore friction from the water. The block is made of concrete which has a weight of 2400kg/m3. Assume the water has a weight of 1000kg/m3.

Relevant Equations

$T_b$=weight of water displaced by the block.

So I have made force diagram

And I think that I should find the acceleration by using these equations:
$\sum Fx=w\sin(15)-f_k-T_{x-buoyancy}$
$\sum F_y=N+T_{y-buouancy}-w$
I know that the volume of water displaced must be $V=\frac{1}{2400}m^3$ and the mass of the water is then $m=\frac{1000}{2400}kg$
So $Tb=\frac{1000}{2400}kg*g$

But the thing that confusses me is that if the block is supposed to slide down the ramp, ∑Fy=0. But then I would have to change the weight of the block, so the new weight would be greater to compensate for the buoyancy force in the y-dircetion. That does not give sense. Does anyone have a clue about these kinds of problems?

 

[hutchphd]

The mass of the water is 1000kg. Make sure each equation you write is dimensionally correct when you write it. Always.

 

[Kolika28]

The mass of the water is 1000kg. Make sure each equation you write is dimensionally correct when you write it. Always.

Hmm, I'm sorry, but I don't understand. Why does the water displaced have the mass of 1000 kg? I thought like this

$p_{concrete}=m_{block}/V_{block}$
$V_{block}=1/2400 m^3=V_{water}$
$p_{water}=m_{water}/V_{water}$
$m_{water}=1000/2400 kg$

 

[hutchphd]

The problem is badly stated. ... kg is a unit of mass and kg/m₃ is a mass density. We are very sloppy in everyday usage. For the ratio it will be the same of course

 

[hutchphd]

That being said look at the units for every equation you write down and you will find your mistake.

 

[Kolika28]

Yeah, I also think it was weird that they used the word weight, because I also thought kg/m^3 is density. So I just assumed the person who wrote the problem mixed up the words. But I struggle with finding the acceleration, do you know how?

 

[hutchphd]

Hmm, I'm sorry, but I don't understand. Why does the water displaced have the mass of 1000 kg? I thought like this

$p_{concrete}=m_{block}/V_{block}$
$V_{block}=1/2400 m^3=V_{water}$
$p_{water}=m_{water}/V_{water}$
$m_{water}=1000/2400 kg$

Look at the units. Some equations are very wrong. Do it always.

 

[Kolika28]

Look at the units. Some equations are very wrong. Do it always.

I have been looking at the units but I still don't see it
$2400 kg/m^3=\frac{1kg}{V_{block}}$
$V_{block}=\frac{1}{2400}m^3$
$p_{water}=1000 kg/m^3=\frac{mass}{\frac{1}{2400}m^3}$
$mass=p_{water}*V_{water}=1000kg/m^3* \frac{1}{2400}m^3= \frac{1000}{2400}kg$

 

[hutchphd]

I have been looking at the units but I still don't see it
$2400 kg/m^3=\frac{1kg}{V_{block}}$
$V_{block}=\frac{1}{2400}m^3$
$p_{water}=1000 kg/m^3=\frac{mass}{\frac{1}{2400}m^3}$
$mass=p_{water}*V_{water}=1000kg/m^3* \frac{1}{2400}m^3= \frac{1000}{2400}kg$

OK. Now I see what you are doing.
So how much is the buoyant force and what is the new "effective" weight of the block?

 

[Kolika28]

OK. Now I see what you are doing.
So how much is the buoyant force and what is the new "effective" weight of the block?

The buoyant force is $F_b=\frac{1000}{2400}kg*g$
The effective force is $m=1kg-$mass of water$=(1-\frac{1000}{2400})kg$
But what is the importance of the effective weight in this problem? Are these equations wrong by the way

And I think that I should find the acceleration by using these equations:
∑Fx=wsin⁡(15)−fk−Tx−buoyancy
∑Fy=N+Ty−buouancy−w

 

[Delta2]

Your second equation needs $w\cos 15$ instead of just $w$.

You can solve the problem with those two equations, OR, consider the effective weight of the block (which is weight-buoyancy force) and proceed to solve the problem like there is no water and buoyancy force at all , but instead of the real weight you will use the effective weight.

 

[hutchphd]

This is why I could kill the author of the question...the solution to this (and many similar but more complicated) problems is to replace g by an "effective" g which is what you have calculated. The mass does not change but effectively g (and hence the weight) does. Everything else is the same and with that replacement mutatis mutandis the problem is solved.

 

[Kolika28]

Your second equation needs $w\cos 15$ instead of just $w$.

You can solve the problem with those two equations, OR, consider the effective weight of the block (which is weight-buoyancy force) and proceed to solve the problem like there is no water and buoyancy force at all , but instead of the real weight you will use the effective weight.

Hmm, I see. But if I were to use the two equations, I would have to make sure that $\sum F_y=mg*cos(15)+T_b sin(75)-mgcos(15)=0$ ?
And that would make the m=408.3 kg

 

[Delta2]

I don't understand why you replace N with $mg\cos(15)$. N is the normal force from the ramp right? Just solve for N from this equation and then calculate the friction force and go to the equation for the x-components(the first equation).

 

[Kolika28]

I don't understand why you replace N with $mg\cos(15)$. N is the normal force from the ramp right? Just solve for N from this equation and then calculate the friction force and go to the equation for the x-components(the first equation).

Yeah, I see that now. Thank you both for the help by the way!

 

[jbriggs444]

Homework Statement:: A solid block, M of mass 1kg, is sliding down a ramp which makes an angle θ=15∘ with the horizontal and has a kinetic coefficient of friction of 0.06. Now assume that you put the ramp and block into a bucket of water so that everything is immersed in water (the block is fully covered). The block is sitting at a height h on the ramp. You can assume the density of water is the same everywhere. Draw the new force diagram and write an equation for the initial acceleration of the block down the ramp. You can ignore friction from the water.

The quarrel I have with the problem is highlighted above.

Friction is not required to make water flow. Pressure makes water flow too. We know that the water is flowing. The fact that we are using the concept of buoyancy at all requires it. If the object were accelerating into the water and the water were not accelerating in behind it then buoyancy would cease to function.

Even a fluid with zero viscosity resists the motion of an object that is accelerating into it -- because the fluid must flow in response and flow involves energy. We have already accounted for 100% of the available energy, putting it all into the kinetic energy of the sliding block. That leaves no energy available for the flowing fluid. And that contradicts conservation of energy.

To do the problem more correctly, we'd need to add a term for the effective hydrodynamic mass displaced by the block. This effective mass depends on things such as the geometry of the container. Information which is not given in the problem statement.