The weight of a rectangular block of low-density material is...

The weight of a rectangular block of low-density material is $15.0 \mathrm{~N}$. With a thin string, the center of the horizontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When $25.0 \%$ of the block's volume is submerged, the tension in the string is $10.0 \mathrm{~N}$. (a) Find the buoyant force on the block. (b) Oil of density $800 \mathrm{~kg} / \mathrm{m}^{3}$ is now steadily added to the beaker, forming a layer above the water and surrounding the block. The oil exerts forces on each of the four sidewalls of the block that the oil touches. What are the directions of these forces? (c) What happens to the string tension as the oil is added? Explain how the oil has this effect on the string tension. (d) The string breaks when its tension reaches $60.0 \mathrm{~N}$. At this moment, $25.0 \%$ of the block's volume is still below the water line. What additional fraction of the block's volume is below the top surface of the oil?

The weight of a rectangular block of low-density material is $15.0 \mathrm{~N}$. With a thin string, the center of the horizontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When $25.0 \%$ of the block's volume is submerged, the tension in the string is $10.0 \mathrm{~N}$.
(a) Find the buoyant force on the block. (b) Oil of density $800 \mathrm{~kg} / \mathrm{m}^{3}$ is now steadily added to the beaker, forming a layer above the water and surrounding the block. The oil exerts forces on each of the four sidewalls of the block that the oil touches. What are the directions of these forces?
(c) What happens to the string tension as the oil is added? Explain how the oil has this effect on the string tension.
(d) The string breaks when its tension reaches $60.0 \mathrm{~N}$. At this moment, $25.0 \%$ of the block's volume is still below the water line. What additional fraction of the block's volume is
below the top surface of the oil?

Key Concepts

Pressure Distribution on Submerged Surfaces

The pressure in a fluid acts perpendicular to the surfaces in contact with it. When an object is in contact with or surrounded by a fluid, the forces due to pressure on different sides of the object contribute to the net buoyant force. This concept explains how lateral forces on side surfaces often cancel out and do not contribute directly to net vertical forces.

Archimedes' Principle

This principle states that any object wholly or partially submerged in a fluid experiences an upward force equal in magnitude to the weight of the fluid displaced. It is fundamental in determining the buoyant force acting on submerged or partially submerged bodies and is used to analyze the floating or sinking behavior of the object.

Hydrostatic Pressure

Hydrostatic pressure refers to the pressure exerted by a fluid at rest, which increases with depth due to the weight of the fluid above. This concept is essential for understanding how forces are distributed over the surfaces of objects immersed in fluids and is critical when evaluating the pressure forces from different layers such as water and oil.

Force Equilibrium in Fluids

In problems involving buoyancy and suspending forces, force equilibrium is the condition where the sum of the upward and downward forces, including buoyant forces, gravitational weight, and any additional tensions or applied forces, equals zero. This balance is crucial for determining unknown forces such as the tension in a supporting string.

Effects of Multiple Immiscible Fluids

When more than one fluid is present, such as water and oil, each fluid exerts its own buoyant force and pressure distribution based on its density. Understanding the behavior of layered fluids is essential to analyze how additional forces arise at interfaces and how they affect overall force balances on submerged objects, including adjustments in support tensions.

Transcript

00:01 Hi there, so for this problem, we are told that the weight of a rectangular block is given, and that is 15 newtoms.

00:11 So let's call this t.

00:21 And then the tension in the string is also given.

00:30 Oh, sorry, this is the weight.

00:34 So let's call this fg.

00:35 And the tension is also given, and that is 10 newtoms.

00:39 So with that information for part a of this problem, we are asked to, find the buoyant force on the block.

00:46 Remember that 25 % of the block is submerged.

00:49 So with that set, and to obtain the buoyant force with some, all of the forces obtained on this block, so those forces up to zero, because the system is an equilibrium.

01:03 So we have the buoyant force that is upward, this minus the tension, this minus the weight, and then we set this equal to zero.

01:13 Now, if we solve for this, for the buoyant force that will be the tension plus the weight.

01:19 So that will be 15 newtoms plus 10 newtoms.

01:24 So that will be 25 newtoms.

01:29 So that's a solution for part a of this problem.

01:33 Now for part b, we are asked about oil of density of 800 kilograms per cubic meter is now steadily added to the beaker forming a layer above the water and surrounded the block.

01:51 The oil asserts a force each of the four sides walls of the block that the oil toches.

01:59 So the question is, what are the directions of these forces? and the answer for that is that the oil poshs horizontally inward.

02:19 So that's a solution.

02:20 Horizomptively inward on each side.

02:24 Of block on each side of this block.

02:29 Now, for part c of this problem, we are asked about what happens to the string tension as the oil is added.

02:47 Now, when this happens, the string tension increases.

03:02 The water under the block pushes up on the block more strongly than before, because the water is under higher pressure due to the weight of the oil above.

03:14 Okay, so that's a solution for part c of this problem.

03:21 Now, for part e, we are asked about, okay, we are asked about in part d is the string rates when the, it's tension reaches a value, so the tension reaches a value of 60.

03:53 Sorry, everything deletes in here, but okay, so we are part d.

03:58 So, yes, the tension is 16 newtoms, and it is 25 % of the low volume is still below the water line.

04:11 So what additional fraction of the blobs volume is below the top surface of the oil.

04:17 So with that set, we know the pressure of the oil's weight on the water, that pressure is equal to the density, of oil times the acceleration due to gravity times the hide.

04:34 What the hide is the height of the oil.

04:37 This pressure is transmitted to the bottom of the block...

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