`x_1 = y^(1//3)` (this is variable in this problem) `x_2 = 2` (this is fixed in this problem) Now, for the y coordinate, we need to find: Next, using the formula for the x-coordinate of the centroid we have: `bar(y)="total moments"/"total area"` `=1/Aint_a^b (^2-^2)/2 dx` Example 4įind the centroid of the area bounded by y = x 3, x = 2 and the x-axis. Method 2: We can also keep everything in terms of x by extending the "Alternate Method" given above: Method 1: We take moments about the y-axis and so we'll need to re-express the expressions x 2 and x 1 as functions of y. `bar(x)="total moments"/"total area"` `=1/Aint_a^b x\ (y_2-y_1)\ dx`įor the y coordinate, we have 2 different ways we can go about it. The "typical" rectangle indicated has width Δ x and height y 2 − y 1, so the total moments in the x-direction over the total area is given by: This is true since for our thin strip (width `dx`), the centroid will be half the distance from the top to the bottom of the strip.Īnother advantage of this second formula is there is no need to re-express the function in terms of y. (I've used a different curve for the `bary` case for simplification.)Īlternate method: Depending on the function, it may be easier to use the following alternative formula for the y-coordinate, which is derived from considering moments in the x-direction (Note the " dx" in the integral, and the upper and lower limits are along the x-axis for this alternate method). Of course, there may be rectangular portions we need to consider separately. Also note the lower and upper limits of the integral are `c` and `d`, which are on the `y`-axis. Notice this time the integration is with respect to `y`, and the distance of the "typical" rectangle from the `x`-axis is `y` units. `bar(x)="total moments"/"total area"` `=1/Aint_a^b x\ f(x)\ dx`Īnd, considering the moments in the y-direction about the x-axis and re-expressing the function in terms of y, we have: If we do this for infinitesimally small strips, we get the `x`-coordinates of the centroid using the total moments in the x-direction, given by: Generalizing from the above rectangular areas case, we multiply these 3 values (`x`, `f(x)` and `Deltax`, which will give us the area of each thin rectangle times its distance from the `x`-axis), then add them. The "typical" rectangle indicated is `x` units from the `y`-axis, and it has width `Δx` (which becomes `dx` when we integrate) and height y = f( x). To find the centroid, we use the same basic idea that we were using for the straight-sided case above. Taking the simple case first, we aim to find the centroid for the area defined by a function f( x), and the vertical lines x = a and x = b as indicated in the following figure. This idea is used more extensively in the next section. `bar(y)=("total moments in"\ y"-direction")/"total area"` `bar(x)=("total moments in"\ x"-direction")/"total area"` We would use this process to solve the tilt slab construction problem mentioned at the beginning of this section. Taking moments with respect to the y-axis, we have: Right rectangle: `"Area" = 2 × 4 = 8\ "sq unit"`. Left rectangle: `"Area" = 3 × 2 = 6\ "sq unit"`. We divide the area into 2 rectangles and assume the mass of each rectangle is concentrated at the center. We wish to replace these masses with one single mass to give an equivalent moment. We have 3 masses of 10 kg, 5 kg and 7 kg at 2 m, 2 m and 1 m distance from O as shown. We now aim to find the centre of mass of the system and this will lead to a more general result. (Clockwise is regarded as positive in this work.) In this case, there will be a total moment about O of: Moment = mass × distance from a point Example 1 Clearly, the greater the mass (and the greater the distance from the point), the greater will be the tendency to rotate. The moment of a mass is a measure of its tendency to rotate about a point. In this section we'll see how to find the centroid of an area with straight sides, then we'll extend the concept to areas with curved sides where we'll use integration. Tilt-slab construction (aka tilt-wall or tilt-up) How do we find the center of mass for such an uneven shape? We don't want the wall to crack as we raise it, so we need to know the center of mass of the wall. In tilt-slab construction, we have a concrete wall (with doors and windows cut out) which we need to raise into position.
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