There is a new internet trend called “tensegrity” – an amalgamation of the words tension and integrity. It is basically a trend of videos showing how objects appear to float above a structure while experiencing tensions that appear to pull parts of the floating object downwards.

In the diagram below, the red vectors show the tensions acting on the “floating” object while the green vector shows the weight of the object.

The main force that makes this possible is the upward tension exerted by the string from which the lowest point of the object is suspended. The other tensions are downward and serve to balance the moment created by the weight of the object.

This is a fun demonstration to teach the principle of moments, and concepts of equilibrium.

These tensegrity structures are very easy to build if you understand the physics behind them. Some tips on building such structures:

Make the two strings exerting the downward tensions easy to adjust by using technic pins to stick them into bricks with holes. You can simply pull to release more string in order to achieve the right balance.

The two strings should be sufficiently far apart to prevent the floating structure from tilting too easily to the side.

The centre of gravity of the floating structure must be in front of the string exerting the upward tension.

The base must be big enough to prevent the whole structure from toppling.

Here’s another tensegrity structure that I built: this time, with a Lego construction theme.

While preparing for a bridging class for those JAE JC1s who did not do pure physics in O-levels, I prepared an app on using a vector triangle to “solve problems for a static point mass under the action of 3 forces for 2-dimensional cases”.

For A-level students, they can be encouraged to use either the sine rule or the cosine rule to solve for magnitudes of forces instead of scale drawing, which is often unreliable.

For students who are not familiar with these rules, here is a simple summary:

Sine Rule

If you are trying to find the length of a side while knowing only two angles and one side, use sine rule:

$$\dfrac{A}{\sin{a}}=\dfrac{B}{\sin{b}}$$

Cosine Rule

If you are trying to find the length of a side while knowing only one angle and two sides, use cosine rule:

A ladder rests on rough ground and leans against a rough wall. Its weight W acts through the centre of gravity G. Forces also act on the ladder at P and Q. These forces are P and Q respectively.

Which vector triangle represents the forces on the ladder?

It’s Day 1 of the full home-based learning month in Singapore! As teachers all over Singapore scramble to understand the use of the myriad EdTech tools, I have finally come to settle on a few:

Google Meet to do video conferencing

Google Classroom for assignment that requires marking

Student Learning Space for students’ self-directed learning, collaborative discussion and formative assessment.

Loom for lecture recording

GeoGebra for visualisation

The following is a video that was created using Loom to explain a question on why tension in a rope on which a weight is balanced increases when the rope straightens.

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:

apply the principle of moments to new situations or to solve related problems.

show an understanding that, when there is no resultant force and no resultant torque, a system is in equilibrium.

use a vector triangle to represent forces in equilibrium.

*derive, from the definitions of pressure and density, the equation ?=??ℎ.

*solve problems using the equation ?=??ℎ.

*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:

As I move the fingers inward, there is friction between the ruler and my finger. This friction depends on the normal contact force as $f=\mu N$.

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, $$N_1\times d_1=N_2 \times d_2$$

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.

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.

Recollection of the slides on moment of a force and torque of a couple.

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.

Mention that

axis of rotation is commonly known as where the pivot is

perpendicular distance is also the “shortest distance”

Activity 3: Conditions for Equilibrium

State the conditions for translational and rotational equilibrium

Show how translation equilibrium is due to resultant force being zero using vector addition

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.

Go through example 7 (2 methods: resolution of vectors and closed vector triangle)

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.

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

Derive from definitions of pressure and density that $p = h\rho g$

Note that this is an O-level concept.

Activity 5: Something to sink about

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.

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.