VERY difficult question on center of masses?

Question

A thin uniform wire has length 30cm is bent to form a framework in the shape of a right angled triangle PQR where PQ=5cm, QR =12cm and RP =13cm.
Find the distance of the center of mass of the framework from:
a)PQ
b)QR
The framework is freely suspended from the point P and hangs in equilimbrim.
c)find the acute angle that PQ makes with the vertical

Couild you please show your working out so that i can understand how you got your answer, thanks

Answer 1

Let us draw the triangle into a coordinate system by such a way that P ≡ (0, 5), Q ≡ (0, 0) and R ≡ (12,0).

Let us use the mass of 1 cm wire as a mass unit.

The total mass of the triangle is 30, the masses of its sides are 5, 12 and 13.

Centers of masses of the sides of the triangle are (0, 2.5), (6, 0) and (6, 2.5).

The center of mass C of the framework may be calculated as the average of the mass centers of the sides of the triangle weighted by their masses:

C ≡ {[(5*0 + 12*6 + + 13*6)/30], [(5*2.5 + 12*0 + 13*2.5)/30]}

C ≡ (5, 1.5)

a)

The distance of the center of mass of the framework from PQ is 5.

b)

The distance of the center of mass of the framework from QR is 1.5.

c)

The tangens of the acute angle (
Therefore the angle is arctng(5/3.5) = 55.01 degrees.

BTW, the center of mass of the framework is not identical with the centroid of the triangle. That is because the mass is distributed around the perimeter of the triangle and not in the area of the triangle. The coordinates of the centroid are (4, 1.67) and it is different point than the center of mass of the triangular framework.

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Answer 2

Actually, it doesn't have to be that hard. After all, if we're not bothering first to prove that the center of mass in a triangle made of a single, smoothly distributed material is the triangle's centroid, then why bother deriving the centroid itself?

Use instead the fact that the centroid exists at exactly one-third (1/3) the height and one-third the base of a right triangle. So if QR is the height and PQ the base, then the center of mass will be 1/3 * 5 from QR (= 5/3) and 1/3 * 12 from PQ (= 4).

As for c), we just need to use trigonometry: consider the sine, say, of angle RPQ. It is 12/13. So if we take the arcsine (sin^-1) of 12/13, we find the angle to be 67.38014 degrees. I'd bet the instructor isn't looking for significant digits, but rather just one decimal place so 67.4 degrees.


Added:

In response to oregfiu's error: when the mass is totally in a uniform band at the very edges of a shape with the interior of said shape utterly empty of material, as in the item described in this problem, it is effectively the same as in an identically shaped item made of a uniform sheet of the material.

And the centroid is still the center of mass. Only if the item were somehow non-uniform would the centroid no longer be the center of mass.

The earth is an example of this. From the outside, we perceive the gravitational force to behave as if all the earth's mass were concentrated essentially at the center of the earth (with some minor discrepancies I shall not go into) exactly as we would expect from a uniform distribution of mass in a perfect geometric object. I might point out that a sphere's center is its centroid. However, if we were somehow at the very center of the earth, the earth's gravity would seem (in its action upon us, at the earth's center) to be concentrated in a band of arbitrary thickness forming the surface of the sphere. Of course, we could detect this is not actually so if we saw its interaction with something between the center of the earth (where we are) and the surface, at, say, 2,500 miles deep.

But basically, the point here is that he is wrong about the center of mass not being at the centroid in the described item. So one still needs to find the centroid to correctly answer the problem's question.

Answer 3

You just need to realize that the center of mass lies on the centroid of the triangle (intersection of the 3 medians).

Let Q = (0,0), P = (5,0), R = (0,12)

Find the equation for 2 medians and find the intersection.

Q = (0,0), mPR = (5/2,6)
y = 12x/5

P = (5,0), mQR = (0,6)
y = -6x/5+6

12x/5 = -6x/5+6
18x/5 = 6
x = 5/3
y = 4

Center of Mass = (5/3,4)
Note: You can verify that the third median does pass through this point.

a. PQ is a horizontal segment. Distance from center of mass to PQ is 4.

b. QR is a vertical segment. Distance from center of mass to QR is the 5/3.

c. tan(Θ) = QR/PQ = 12/5
Θ = 67.38 degrees