Tensegrity Explained

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 (shown below) 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. The centre of gravity of the “floating” structure lies just in front of the supporting string. The two smaller downward vectors at the back due to the strings balance the moment due to the weight, and give the structure stability sideways.

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

The next image labels the forces acting on the upper structure. Notice that the centre of gravity lies somewhere in empty space due to its shape.

Only the forces acting on the upper half of the structure are drawn in this image to illustrate why it is able to remain in equilibrium

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

  1. Make the two strings exerting the downward tensions are 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.
  2. The two strings should be sufficiently far apart to prevent the floating structure from tilting too easily to the side.
  3. The centre of gravity of the floating structure must be in front of the string exerting the upward tension.
  4. The base must be wide enough to provide some stability so that the whole structure does not topple.

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

Apart from using Lego, I have also 3D-printed a tensegrity structure that only requires rubber bands to hold up. In this case, the centre of gravity of the upper structure is somewhere more central with respect to the base structure. Hence, 3 rubber bands of almost equal tension will be used to provide the balance. The STL file for the 3D model can be downloaded from Thingiverse.com.

The main challenge in assembling a tensegrity structure is the adjustment of the tensions such that the upper structure is balanced. One way to simplify that, for beginners, is to use one that is supported by rubber bands as the rubber bands can adjust their lengths according to the tensions required.

3D-printed tensegrity table balanced by rubber bands

Another tip is to use some blu-tack instead of tying the knots dead such as in the photo below. This is a 3D printed structure, also from Thingiverse.

3D-printed tensegrity table held up and balanced by nylon string

(This post was first published on 18 April 2020 and is revised on 24 August 2022.)

Unequal masses attached to rod in free fall

Came across a question recently that many students answered incorrectly.

Close to the surface of the Earth the gravitational field strength is uniform. A pair of unequal masses are joined by a light, rigid horizontal bar and suspended by a string from their centre of gravity as shown. The mass M of the ball on the left is larger than the mass m of the ball on the right.

The supporting string is now cut and the system begins to fall. Air resistance is negligible.

Which statement is correct?

AThe bar will remain horizontal as it falls.
BThe bar will rotate clockwise as it falls.
CThe bar will rotate anti-clockwise as it falls.
DThe bar will first rotate clockwise and then rotate anticlockwise as it falls.

Without air resistance

This question supposes that air resistance is negligible and so the only forces initially acting on the object is weight. The answer that many students gave incorrectly as B because they assume that the larger weight acting on the larger mass will bring about a larger acceleration.

Since the object begins in equilibrium, and the acceleration of both objects is just gravitational acceleration, the bar will remain horizontal.

With air resistance

This then invites a question: What if there is air resistance?

To consider the vertical acceleration on both balls, we need to consider the net force $F_{net}$, which is the vector sum of weight $W$ and air resistance $F_R$, ignoring the tension exerted by the rod at the initial stage of the fall.

$$F_{net} = W – F_R = V \rho_{ball}g – \dfrac{1}{2} \rho_{air}v^2C_DA$$

The volume V of a sphere is proportional to $r^3$ and its cross-sectional area A is proportional to $r^2$,

A larger radius will imply a larger increase in V than A, and hence, a large $W$ than $F_R$. This will then allow the larger mass to experience a larger acceleration than the smaller mass in the initial stage.

Internal Resistance and Terminal Potential Difference


This applet demonstrates how terminal potential difference (as measured by the voltmeter across the terminals of the battery) changes depending on :

  1. internal resistance r
  2. external resistance R
  3. emf E
  4. when a switch is turned on and off
<iframe scrolling="no" title="Internal Resistance and Terminal Potential Difference" src="https://www.geogebra.org/material/iframe/id/puvfjxk5/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>

Man in Elevator

I just took the elevator in my apartment building with the PhyPhox mobile app and recorded the acceleration in the z-direction as the lift went down and up. This was done in the middle of the night to reduce the chances of my neighbours getting into the elevator along the way and disrupting this experiment, and more importantly, thinking I was crazy. The YouTube video below is the result of this impromptu experiment and I intend to use it in class tomorrow.

I used to do this experiment with a weighing scale, and a datalogger, but with smartphone apps being able to demonstrate the same phenomenon, it was worth a try.

To complement the activity, I will be using this simulation as well. Best viewed in original format: https://ejss.s3.ap-southeast-1.amazonaws.com/elevator_Simulation.xhtml, this simulation done in 2016 was used to connect the changes in acceleration and velocity to the changes in normal contact force as an elevator makes its way up or down a building.

Sky-Diving and Terminal Velocity


This is a wonderful applet created by Abdul Latiff, another Physics teacher from Singapore, on how air resistance varies during a sky-dive with a parachute. It clearly demonstrates how two different values of terminal velocity can be achieved during the dive.

Incidentally, there is a video on Youtube that complements the applet very well. I have changed the default values of the terminal velocities to match those of the video below for consistency.

Also relevant is the following javascript simulation that I made in 2016 which can show the changes in displacement, velocity and acceleration throughout the drop.