Physics: Buoyant forces. Can you help me?

Question

Okay, I have a bunch of problems like this in my homework; if you can show me how to work out one, I should be able to work out the others like it.

A 10.0 kg block of metal measuring 12.0 cm x 10.0 cm x 10.0 centimeters is suspended from a scale and immersed in water. The 12.0 cm surface is vertical, and the top of the block is 5.0 cm below the surface of the water.


Using P(0)=1.013x10^5 N/m^2, what are the forces acting on the top and bottom of the block?

What is the reading of the spring scale?

Show that the buoyant force equals the difference between the forces at the top and bottom of the block.


Help is much appreciated, but please don't just give me an answer (I can get that from the back of the book!)

Thanks!

Answer 1

Ok, first we need to remember the Arquimedes principle :

The weight of the block must be equal to the push of the water exerted to the block, then :

The forces acting on the top of the block are the force by the pressure of the air, and the tension of the cord ( by the sprind scale)

Pressure = P(0)=1.013x10^5 N/m^2

Force = 1.013*10^5*Area

Area = (10 / 100)^2

Force = 1.013*10^5*(10 / 100)^2

Force = 1013 Newtons

Now, we have the tension of the cord, the weight f the block, of mass 10 kg, and the push from the water :

Push of the wate : 10^3*gravity*(volume in water)

The block is 5 cm below the water :

Volume in water = 5*10*10*10^-6

Push of the water : 10^3*9.8*5*10*10*10^-6

Push = 4.9 Newtons

Force by the pressure of air = 1013 Newtons

Weight of the block = 98 Newtons

Tension = T

T + 4.9 = 1013 + 98

T = 1106.1 NEWTONS

T = lecture on the scale

Hope that helps

Answer 2

Here is what is happening. You do the math.
A body is buoyed up by a force equal to the weight of fluid that it displaces. So the weight of the 500 Cu Cm of displaced water will be subtracted from the weight of the block to give the scale reading. The scale reading is the upward force applied at the top of the block. The buoyant force upward in pushing upward ar the bottom, the weight of the block is pushing downward.. Since the block is not moving the two forces upward equal the one force downward.
It is in equalibrium.

Something else is happening which is equivalent. The pressure on the projected area of the bottom is pushing upward with a force. F=PA That pressure depends on the depth. In this case it is simple because the shape is simple and the sides are vertical.
What is the pressure at the 5 cm depth ? That pressure times the 100 sq cm area will also give that buoyant force. If you know the fluid, you can make a pressure versus depth graph. In English units, one pound of pressure (psi) occurs for every 2.31 feet of depth of fresh water. (Salt water is heavier so a ship going from the ocean through the Panama canal will sit deeper in the fresh water canal than in either ocean.)

Answer 3

including extra water won't enhance the buoyant stress. it will enhance for a 2d to take the forged as much as the point proportional to its preliminary point yet after that, it is going to alter into comparable because it substitute into interior the beginning up. Buoyant stress = Density of liquid x quantity of reliable x g = ?Vg It relies upon on density of liquid and quantity of the forged displaced.