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Home / Science, Engineering & Maths / Through Engineers' Eyes: Engineering Mechanics by Experiment, Analysis and Design / Analysis: Interpreting the tests

Analysis: Interpreting the tests

Although the two methods look quite different, in both cases the weight vector passes through the pivot so that its moment arm is zero.
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9.8
Now we can interpret the experiments on finding centres of gravity by the suspension method. We’ll start with the triangular shape. Here’s an FBD of the triangular cardboard shape. There is a force at the pin and a force from the weight acting at the centre of gravity. Because it can be assumed to be a two-force object, the centre of gravity must be along this line. It must be along this second line, too. So the centre of gravity must be where these two lines cross. We can check it out. In our test, we made a third measurement. Here’s the result. The three lines almost meet at a point. You would expect a little triangle, depending on how careful you had been.
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There’s a standard result for the location of the CG of a triangle. It finds many applications in engineering. The centre of gravity is one-third of the height up from any side taken as base, like this, and this, and this.
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Here’s the card from the test with the one-third point marks. Again, I think the agreement isn’t bad. But where does this result for a triangle come from? The classic proof involves calculus, but you can also see it by considering the triangle to be made up of a lot of narrow strips. Each strip is roughly a rectangle with its CG in the middle, so the CG must be along this line connecting the narrow strips. For fun, you could show that this method gives the one-third height rule. Now to the balancing method– when the card balances, the centre of gravity must lie along the line of balance. Imagine taking moments about the line of balance.
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For equilibrium, the moment arm for the weight must be zero, so the CG must, indeed, be on the line of balance. We rotated the card and found another line of balance. The estimated CG is where they cross. Here, we found the centre of the square by drawing diagonals. What do you think of the agreement?
172
Because our cardboard is uniform in density, we have also found its centre of volume. The centre of gravity might be the same location as the centre of volume, but it isn’t always– for example, if the object is made up of materials which have different densities. And because out cardboard is uniformly thick and a flat shape, we have also found the centre of area. These geometric properties are called centroids. It’s a term that often appears in engineering. Now let’s find out how to calculate centres of gravity and centroids.

Although the two methods look quite different, in both cases the weight vector passes through the pivot so that its moment arm is zero.

This video explains in detail how this condition can be used to locate the centre of gravity.

Data from the experiment are given in the Downloads section below.

For most of the simple shapes used in the tests the location of the centre of gravity is obvious – it’s in the middle. The odd one out is the triangle, but it’s easily found from standard results. Knowing all this you can see how accurate the experimental results turned out to be.

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This content is taken from
UNSW Sydney online course,

Through Engineers' Eyes: Engineering Mechanics by Experiment, Analysis and Design

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Because the centre of gravity of a triangle often appears in engineering analysis, it’s worth learning as part of developing your engineers’ eyes.

Talking points

  • What did you think of the agreement between the theoretical location of the centre of gravity of the triangle, and the data from tests?

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This content is taken from UNSW Sydney online course
Through Engineers' Eyes: Engineering Mechanics by Experiment, Analysis and Design
View Course
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