04 Forces

Lesson Plan for Online Lecture on Forces

I am using this post as a way to document my brief plans for tomorrow’s Google Meet lecture with the LOA students as well as to park the links to the resources and tools that I intend to use for easy retrieval.

Instruction Objectives:

  1. apply the principle of moments to new situations or to solve related problems.
  2. show an understanding that, when there is no resultant force and no resultant torque, a system is in equilibrium.
  3. use a vector triangle to represent forces in equilibrium.
  4. *derive, from the definitions of pressure and density, the equation ?=??ℎ.
  5. *solve problems using the equation ?=??ℎ.
  6. *show an understanding of the origin of the force of upthrust acting on a body in a fluid.

Activity 1: Find CG of ruler demonstration

Having shown them the demonstration last week, I will explain the reason why one can find the CG this way:

  1. As I move the fingers inward, there is friction between the ruler and my finger. This friction depends on the normal contact force as $f=\mu N$.
  2. Drawing the free-body diagram of the ruler, there are two normal contact forces acting on the ruler by my fingers. The sum of these two upward forces must be equal to the weight of the ruler. These forces vary depending on their distance from the CG. Taking moments about the centre of gravity, $$N_1\times d_1=N_2 \times d_2$$
  3. The finger that is nearer to the CG will always have a larger normal contact force and hence, more friction. Hence, the ruler will tend to stop sliding along that finger and allow the other finger to slide nearer. When that other finger becomes closer to the CG, the ruler also stops sliding along it and tends to then slide along the first finger.
  4. This keeps repeating until both fingers reach somewhere near the CG.

Activity 2: Moments of a Force at an Angle to the line between Pivot and Point of Action.

  1. Recollection of the slides on moment of a force and torque of a couple.
  2. Give them a MCQ question to apply their learning using Nearpod’s Quiz function https://np1.nearpod.com/presentation.php?id=47032717
  3. Ask students to sketching on Nearpod’s “Draw It” slides the “perpendicular distance between axis of rotation and line of action of force” and “perpendicular distance between the lines of action of the couple” for Example 5 and 6 of the lecture notes respectively.
  4. Mention that
    1. axis of rotation is commonly known as where the pivot is
    2. perpendicular distance is also the “shortest distance”

Activity 3: Conditions for Equilibrium

  1. State the conditions for translational and rotational equilibrium
  2. Show how translation equilibrium is due to resultant force being zero using vector addition
  3. Show how rotational equilibrium is due to resultant moment about any axis being zero by equating sum of clockwise moments to sum of anticlockwise moments.
  4. Go through example 7 (2 methods: resolution of vectors and closed vector triangle)
  5. Useful tip: 3 non-parallel coplanar forces acting on a rigid body that is in equilibrium must act through the same point. Use 2006P1Q6 as example.
  6. Go through example 8. For 8(b), there are two methods: using concept that the 3 forces pass through the same point or closed triangle.

For next lecture (pressure and upthrust):

Activity 4: Hydrostatic Pressure

  1. Derive from definitions of pressure and density that $p = h\rho g$
  2. Note that this is an O-level concept.

Activity 5: Something to sink about

Get students to explain how the ketchup packet sinks and floats.

Students are likely to come up with answers related to relative density. As them to draw a free body diagram of the ketchup packet. However, we will use the concept of the forces acting on the ketchup packet such as weight and upthrust to explain later.

Activity 6: Origin of Upthrust

I designed this GeoGebra app to demonstrate that forces due to pressure at different depths are different. For a infinitesimal (extremely small) object, the forces are equal in magnitude even though they are of different directions, which is why we say pressure acts equally in all directions at a point. However, when the volume of the object increases, you can clearly see the different in magnitudes above and below the object. This gives rise to a net force that acts upwards – known as upthrust.

Hydrostatic Pressure and Upthrust

This app is used to demonstrate how a spherical object with a finite volume immersed in a fluid experiences an upthrust due to the differences in pressure around it.

Given that the centre of mass remains in the same position within the fluid, as the radius increases, the pressure due to the fluid above the object decreases while the pressure below increases. This is because hydrostatic pressure at a point is proportional to the height of the fluid above it.

It can also be used to show that when the volume becomes infinitesimal, the pressure acting in all directions is equal.

The following codes can be used to embed this into SLS.

<iframe scrolling="no" title="Hydrostatic Pressure and Upthrust" src="https://www.geogebra.org/material/iframe/id/xxeyzkqq/width/640/height/480/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false" width="640px" height="480px" style="border:0px;"> </iframe>

How Does Siphoning Work?

A siphon operates through the combined effects of gravity and air pressure, which work together to move liquid from a higher elevation to a lower one. Gravity is the primary force driving the flow, as it pulls the liquid from the higher container down through the siphon tube to the lower container. The liquid’s potential energy, due to its elevated position, is converted into kinetic energy as it flows downward.

Air pressure plays a crucial supporting role by maintaining the continuous flow of liquid. Atmospheric pressure on the liquid’s surface in the higher container pushes the liquid into the siphon tube. This pressure counteracts gravity’s pull that might otherwise cause the liquid to fall back into the higher container. As the liquid moves downwards, it creates a partial vacuum in the upper part of the tube, allowing atmospheric pressure to push more liquid into the tube, sustaining the flow.

Thus, a siphon can continue to operate as long as the outlet is lower than the liquid surface in the source container, the tube remains filled with liquid, and atmospheric pressure supports the flow.

Water in an Inverted Cup

This demonstration can be modified for use as a magic trick.

Materials:

  1. Glass of water
  2. Piece of cardboard that is larger than the mouth of the glass.

Procedure:

  1. Fill the glass up with water.
  2. Place the piece of cardboard over the mouth of the glass.
  3. Holding the cardboard against the mouth of the glass, invert the glass.
  4. Release the hand slowly.

Explanation

Water can remain in an inverted glass with the piece of cardboard underneath because atmospheric pressure is acting upward on the cardboard, holding it up together with the water. There is little air pressure within the g;ass, so the downward force acting on the cardboard is mainly the weight of the water, which is to the order of several newtons whereas atmospheric pressure exert an upward force of several thousand newtons.

Modification:

  1. Drill a small hole in a plastic cup, near the base.
  2. Seal the hole with your thumb and fill the cup with water.
  3. Place the cardboard over the mouth of the cup.
  4. Invert the cup together with the cardboard, while keeping your thumb over the hole.
  5. Using a magic word as the cue, shift your thumb slightly to allow a little air into the cup. This will cause the cardboard and water to fall. As the air pressure within the cup is equal to that of the atmosphere.

Hanging Forks

This simple demonstration can be done anywhere at home using the following items:

  1. an empty glass
  2. a toothpick
  3. two forks
The video below demonstrates how to do it. When the forks are balanced on the mouth of the glass with the toothpick, the centre of gravity of the forks-and-toothpick system will adjust itself so that it lies vertically below the pivoting point. This is possible because the forks form a V-shape within which the centre of gravity can exist.