Question

A block of mass 2M sits at the bottom of a container and is completely submerged in a liquid. A second block of mass M sits on top of the first block with no amount of it submerged in the liquid. The blocks are made of the same material and the ratio of the density of the blocks to liquid is ?b/?l = 0.80.

a) Draw a FBD of bottom block. Ensure accurate relative size between the force vectors and clearly label each.

b) Starting from first principles, calculate the magnitude of the normal force from the container onto the bottom block and express your result as a simplified expression of known quantities.

c) Calculate the acceleration vector needed to be given to the container to cause the bottom block to begin to lose contact with the container. Write your result in terms of known quantities. State any assumptions necessary.

Answer 1

a)

Here,we have given the mass of the bottom at the bottom completely immersed in the liquid is,

And the mass of the block placed above the bottom block which not submerged in watre,

Here,the wieght of the smaller block at the top is exerted downwards on the bottom block as,

The wieght of the bottom block itself downwards,

The buoyant force on the bottom block upwards,

The normal force (Contact reactional force) exerted by the bottom of the container on block upwards,

b)

Consider the free body diagram of bottom block of mass ,

We have,the wieght of the smaller block at the top is exerted downwards on the bottom block as,

The wieght of the bottom block itself downwards,

The buoyant force on the bottom block upwards,

Here we have the buoyant force by the liquid on the bottom block,

But we have the density of the block,

So,Volume of the bottom block,

So,Buoyant force,

Bu given,

So,Buoyant force on bottom block,(acting upwards)

So,We have the block is in equillubrium,e,net force on bottom block is zero,

ie,

So,The normal force by the bottom of container on the bottom block,

c)

Now let assume that the container is accelerated at a rate of m/s^2 downwards.

ie,When the container is moving downward with ana acceleration of m/s^2 downwads,due to pseudoforce on the blocks it will expierience a upward force opposite to the direction of acceleration on the container.So,There will be change in effective wieght into apparent wieght as follows.

Due to this,a pseudoforce is expierenced on the 2 blocks in the container.

So,The apparent wieght of the bottom block is, downwards

And the apparent wieght of the top block is,downwards.

Also,There is no change on the buoyant force on the bottom block by the liquid.ie,

Also the normal force by container on bottom block,

So,Here the net force on the bottom block,

ie,

Also,for the acceleration of the container enough to loose the contact of bottom block with the container,Normal force on container,(Since there is no net contact force downward on the block)

So,

Or,

Acceleration of the container,