# Which way will the scale tip? [duplicate]

This problem has bothered me for quite some time and I can't solve it. I have even tried to make a construction, but it sometimes tips to the left and sometimes tips to the right :).

When we submerge the body in the water the water pushes it up. That is an action. At the same time the body pushes water down. That is a reaction. So the balance should be maintained.

But, since the water level rises the hydrostatic pressure on the bottom is greater so the right side should go down.

There is another similar problem but it's not the same.

• This looks different, the bouyant mass is hanging from the scales not an external hook. Actually this makes all the difference to the answer. – JMLCarter Mar 16 '17 at 23:04
• I have seen the problem you have suggested and it is not the same as JMLCarter already wrote – user3368512 Mar 16 '17 at 23:21
• Are we assuming that the acceleration due to gravity is constant throughout the height difference for the purpose of this question? (I didn't think I had to ask this but maybe it should be cleared up) – JMac Mar 16 '17 at 23:47
• Again, this question is not a duplicate. I dont see how someone can say that. That means he/she have not read that or this question. Only thing that is the same is the title because the problems are similar, but certainly not the same. Cheers – user3368512 Mar 17 '17 at 8:07
• This is not a duplicate. In the referred question some portion of the weight of the steel ball is carried by the string that is not connected to the scale. But in the current question, the string is connected to the scale. – lucas Mar 17 '17 at 8:39

The balance will be maintained because there is no EXTERNAL force applied on left or right side. This is because of the same reason you can't push a car while sitting in it!

The scale will not move.

You don't need to think about buoyancy at all to answer this question. In the first picture, the scale is balanced, because the net force on each side (the weight) is equal. No mass is added to or removed from either side, so the net forces remain the same.

If you consider everything put in the right side of balance scale as one system, only the internal forces are changed which will not affect the balance. No external force is applied and hence the balance will be maintained.

So various forces neglected (including gravity, air resistance, thermal forces, air pressure) - but your question suggest this is an inquiry about the effect of bouyancy.

That being the case, lowering the mass into the water brings into effect its bouyancy. This is a force lifting the mass and acting down on the water. However, as the mass is tethered to the scale the force acting down the string is reduced by the same amount as the bouyancy, and the net effect on the scale is zero.

(Also neglecting rotating frames of reference, light pressure, stress and young's modulus, magnetically induce currents etc)

Let $V$ be the volume of the hanging block, $A$ the area of the water surface, $\rho$ the density of water, and $g$ the acceleration due to gravity.

The extra hydrostatic force felt by the bottom of the container after lowering the block is $$\Delta F = \rho gA\Delta h,$$ where $\Delta h$ is the change in water height ($\rho g \Delta h$ would be the change in hydrostatic pressure). Because water is not compressible, $A\Delta h$ must be the volume of the submerged block. $$\Delta F = \rho g V.$$ Notice that this is exactly equal to the buoyant force on the block, so the two new forces created after the block is submerged negate each other.

## EDIT:

Since this answer is not being very well received I will just say that if we neglect gravitational forces due to the height difference between the mass in the upper and lower pictures; then yes, the scales will be balanced as there is simply no net torque.

However, since the question did not stipulate these assumptions then the answer below is the correct one:

Look at the RHS of the scale and consider Newton's Law of Gravitation $$F=-\frac{G \cdot m\cdot m_E}{d^2}$$ On the RHS consider $m$ to be the mass of block suspended by the string and $m_E$ is the mass of the Earth $\approx 6\times 10^{24}$kg. $G$ is the gravitational constant and $d$ is the distance of the string between the block in the top picture and the one in the bottom plus the distance to the centre of the earth.

Now let $m_{s}$ be the mass of the submerged block in the container of the lower RHS and $d_E$ is the distance to the centre of the earth. $$F=-\frac{G \cdot m_{s}\cdot m_E}{d_E^2}$$

The gravitational force acting on the RHS of lower system is therefore greater by Newton's inverse square law.

Can you now understand why once the block is dropped into the fluid the RHS of the lower image will weigh more and hence tilt clockwise?

• I don't think Newtons Law of Gravity applies here... these forces are completely negligible compared to the force of Earth's gravity. – JMac Mar 16 '17 at 23:13
• I'm saying this entire thing is being driven by acceleration due to gravity. You very rarely (I've never really seen it) use the force of gravity between two bodies when doing questions like this. In fact, draw FBD's for the gravitational force they will exert on each other. There's a gravitational force down on the mass and an equal force up on the water. Both the water and string are attached to the scale, the net force on the scale is still 0 due to the gravitational force between them. Instead we assume one of the masses is Earth then we know acceleration in freefall is $\approx$ g. – JMac Mar 16 '17 at 23:19
• @JMac Yes, but they are at different heights with respect to the earth, so Newton's law of gravity is applicable here. – BLAZE Mar 16 '17 at 23:21
• Use some realistic numbers here. The mass of Earth is $5.972 x 10^{24}$ kg, the radius is $6371$ km. Compare that to any values you would expect to see in this situation and you can recognize that even moving a 1 ton mass 100 meters won't have an appreciable effect on acceleration due to gravity. – JMac Mar 16 '17 at 23:23
• I agree with JMac. Earth's gravity is way greater. However it is very interesting to take into the consideration changed gravity due to the changed distance between two bodies, although it is negligible. – user3368512 Mar 16 '17 at 23:25