In ordinary sense the center of gravity is perceived as a point to which it is possible to put equally effective all forces operating on a body. The simplest example is a teeter in the form of a usual board. Without any calculations any child will pick up a board support so that to counterbalance (or perhaps and to move) on a swing the heavy man. In case of difficult bodies and sections not to do without exact calculations and the corresponding formulas. Even if bulky expressions turn out, the main thing is not to be frightened them, and to remember that initially it is about almost elementary task.
Instruction
1. Consider the simplest lever (see rice 1) which is in position of balance. Arrange a point of support on a horizontal axis with an abscissa x ₁₂ and place at edges material points of mass of m ₁ and m ₂. Consider their coordinates on an axis 0th known and equal x ₁ and x ₂. The lever is in position of balance if the moments of forces of weight P ₁=m₁g and P =mg are equal. The moment is equal to the work of force on her shoulder which can be found as length of the perpendicular lowered from a point of application of force on a vertical x = x ₁₂. Therefore, according to figure 1, m₁g ℓ₁ = m₂g ℓ₂, ℓ₁ = x ₁₂-x ₁, ℓ₂ = x -x ₁₂. Then m ₁ (x ₁₂-x ₁) =m ₂ (x -x ₁₂). Solve this equation and receive x ₁₂ = (mx +mx ₂) / (m +m ₂).
2. ₁₂ apply the same reasonings and calculations to clarification of ordinate of the center of gravity of y, as well as on a step 1. Still follow the illustration given on figure 1 where m₁gh ₁ = m₂gh ₂, h =y -y ₁, h =y -y ₁₂. Then m ₁ (y -y ₁) =m ₂ (y -y ₁₂). Result - at ₁₂ = (mu +mu ₂) / (m +m ₂). Further consider that instead of a system from two points there is one point of M ₁₂ (x12, u12) lump (m +m ₂).
3. From two points add one more weight to a system (m ₃) with coordinates (x ₃, at ₃). At calculation it is still necessary to consider that you deal with two points where the second of them has weight (m +m ₂) and coordinates (x12, u12). Repeating already for these two points all actions of steps 1 and 2, you will come to coordinates of the center of gravity of a system of three points x ₁₂₃ = (mx +mx +mx ₃) / (m +m +m ₃), at ₁₂₃ = (mu +mu ₂+m₃y ₃) / (m +m +m ₃). Further add the fourth, fifth and so on points. After repeated repetition of the same procedure make sure that for system n of points of coordinate of the center of gravity are calculated on a formula (see fig. 2). Note for yourself the fact that in the course of work the acceleration of gravity of g was reduced. Therefore coordinates of the center of masses and weight coincide.
4. Imagine that in the considered section some area D, which area density ρ=1 is located. From above and from below the figure is limited to schedules of curves at =φ (x) and at =ψ (x), x є [and, b]. Break area D x=x₍i-1 verticals ₎, x=x₍i ₎ (i=1.2, …, n) into thin strips, such that they can be considered approximately rectangles with xi bases (see fig. 3). At the same time consider the midpoint of xi put coinciding with an abscissa of the center of masses ξi= (1/2) [xi+x(i-1)]. Consider height of a rectangle approximately equal [φ (ξi)-ψ (ξi)]. Then ordinate of the center of mass of the elementary area ηi= (1/2) [φ (ξi)+ψ (ξi)].
5. Owing to uniform distribution of density consider that the center of mass of a strip will coincide with its geometrical center. The corresponding elementary weight ∆mi=ρ [φ (ξi)-ψ (ξi)] xi= [φ (ξi)-ψ (ξi)] is concentrated by xi in a point (ξi, ηi). There came the moment of the return transition from the weight presented in a discrete form to continuous. According to formulas of calculation of coordinates (see fig. 2) the center of gravity the integrated sums illustrated in figure 4a are formed. Upon limit transition at ∆xi→0 (ξi→xi) from the sums to certain integrals, receive the final answer (fig. 4b). In the answer weight is absent. Equality of S=M should be understood only as quantitative. Dimensions are excellent from each other here.