A body of mass $m$ and volume $V$ is immersed into the sea. The body moves under the action of two forces: gravity and buoyancy (Archimedes).

Gravity force has a magnitude $mg$, where $g$ is a constant acceleration due to gravity. Buoyancy force has a magnitude $\rho V g$, where $\rho$ is the water density, and acts in the direction opposite to the gravity force.

Part 1.

Assume that the water density is a constant, $\rho V >m$ always holds, and at the initial time moment the body is released from rest at the depth of $h>0$ metres below the sea surface. Assume that the origin is at the sea surface and the positive direction of motion is towards the centre of the Earth.

(a) Complete the diagram by showing all forces acting on the body, and the initial displacement of the body.

(b) Write down Newton's second law of motion. Solve this equation, i.e. write down the expressions for the velocity and the displacement as functions of time.

(c) Show that the body will move towards the sea surface, and when it reaches the surface it has the velocity $v= -\sqrt{\frac{2gh(\rho V-m)}{m}}$.

1a) I can do

1b) I know that Newton's second law is $mx=F$ with $F$ being the forces. I have written the formula as $mx=\rho Vg - mg$ but I am not sure if this is correct. Then I've divided both sides by $m$ to get $x$ on its own and integrated with respect to t (time) to get $x$ as a function of time. Giving me $x=-gt+\frac{\rho Vgt}{m} + c$

1c) I'm not sure where the $h$ has come from, or how I could input it into my previous equation


closed as off-topic by Abhimanyu Pallavi Sudhir, jinawee, Brandon Enright, John Rennie, Qmechanic Nov 27 '13 at 17:28

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  • $\begingroup$ In regards to (b): If you integrate over $t$, you don't have $x(t)$, you have $\int x\,dt$. $\endgroup$ – Kyle Kanos Nov 27 '13 at 13:55
  • 2
    $\begingroup$ I am not sure if it is clear to you that your $x$ would be the acceleration (and not the displacement of the body, which $x$ usually is used for). $\endgroup$ – fibonatic Nov 27 '13 at 14:05

1b) fibontic correctly pointed out that your expression for Newton's 2nd law is not correct. It should be $$ma=F_\text{net}.$$ You have an $x$ instead of an $a,$ which is causing one of your problems in part b.

By writing $ma=\rho Vg-mg,$ you should be aware that you've already implicitly imposed a coordinate system where up is positive. This is probably very convenient for this problem, but it is good to be aware of such things, as it will play a role in parts b and c.

If you want to use this to solve for the equations of motion, first solve your Newton's 2nd law equation for the acceleration $a$ and note that it is constant; it does not depend on time. For the case of constant acceleration, you probably had some kinematic expressions previously in your physics course, such as $\Delta v = a\Delta t,$ to name just one. These are the go-to kinematic equations for this case since you have constant acceleration. The more difficult part of this will be being careful about initial position; use your coordinate system and choose where the vertical position is zero.

1c) One way to see that your expression was incorrect is unit analysis: determine the dimension of each term (e.g., $x$ and $gt$) and note that they aren't all the same. It is not legal to add things with different units.

To show that the object will move up when released from rest, you can show that the force upward is greater than the force downward (since this would cause the object to accelerate upward). You were given enough information in one of the inequalities to address this.

The $h$ has come from using a convenient kinematic equation and solving for the speed of the particle after it has moved a distance $h$ upward; make sure you were careful in setting up your coordinate system to get the correct answer.


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