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Chapter-V - Copy.pptx
- 2. Introduction Center of Gravity Center of Lines, Areas, and Volume Center of composite Bodies Determining the centroid by integration
- 3. Centroid: Centroid is defined as the point of the geometric center of an object, where the density is wholly distributed over the body. Fig.1. Centroid of triangle The point where the cutout is balanced perfectly is the center of the object.
- 4. The center of gravity is defined as the exact place in a body around which the instants due to gravity are regarded as zero. It is the point at which the entire body is perfectly balanced in relation to gravity. It is the point where the gravitational force or weight of the body acts in any orientation of the body.
- 5. The center of gravity is abbreviated as C.G or simply G. The gravitational field always affects the center of gravity because when the gravitational field’s value varies, the center of gravity’s value changes. Fig. 2. Point of center of gravity
- 6. A body is composed of an infinite number of particles of differential size. When the body is located within a gravitational field each of these particles have a weight 𝑑𝑊. These weights form a parallel force system. The resultant of this system is the total weight of the body, which passes through a single point called the center of gravity, 𝐺 Fig. 3. classification of force
- 7. To determine the location of the center of gravity of any body, the principle of moments is applied to the parallel system of gravitational forces. The moment of the resultant gravitational force 𝑊 about any axis equals the sum of the moments about the same axis of the gravitational forces 𝑑𝑊 acting on all particles of infinitesimal elements of the body. If the principle of moment is applied about the y-axis: 𝑥𝑊 = 𝑥𝑤 , 𝑡ℎ𝑒𝑛 𝑥 = 𝑥𝑤 𝑊 , 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑧 𝑎𝑥𝑒𝑠 𝑎𝑙𝑠𝑜, 𝑦 = 𝑦𝑤 𝑊 , 𝑧 = 𝑧𝑤 𝑊
- 8. The center of mass is defined as the position at which the entire body is directed. The mass distribution is considered uniform around the center of the mass. Since the center of mass is independent of the gravitational field the body remains unaffected by change in the gravitational field’s force. In a simple rigid bodies with uniform density, the center of mass is located at the center or centroid. In order to study the dynamic response or accelerated motion of a body knowing the location of center of mass of the body is important.
- 9. With the substitution of 𝑊 = 𝑚𝑔 and 𝑑𝑊 = 𝑔𝑑 the expressions for the coordinates of the center of gravity become: 𝑥 = 𝑥𝑑𝑚 𝑚 𝑦 = 𝑦𝑑𝑚 𝑚 𝑧 = 𝑧𝑑𝑚 𝑚 their respective position vectors: 𝑟 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 𝒓 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 𝒓 = 𝑟𝑚 𝑚
- 10. The major distinction between center of gravity and the center of mass is that the center of gravity is the position at which the entire body weight is balanced. While the center of mass is the position at which the entire mass of the body is directed.
- 11. Difference between center of mass and center of gravity. Center of Mass Center of Gravity Is the point where mass distribution is uniform in all directions. Is the point where weight is evenly distributed in all direction. Is based on the mass of the body Is based on the weight of the body It is the center where the entire bodily mass is concentrated It is the point at which the body’s entire weight is suspended Mass of the body is distributed uniformly throughout the body Weight of the body is distributed uniformly throughout the body. When the body moves from left to right mass operating to the left side is equals to mass acting to on the right side. When the body travels on an axis from left to right weight on the left side is equals with weight on the right side. Change in gravitational field has no effect on it Change in gravitational field has effect on it. When spinning around a point it produces angular momentum When spinning around an axis the net torque is zero due to gravitational force
- 12. For a slender rod or wire of length L, cross-sectional area A, and density 𝜌, the body is approximated as a line segment, and 𝑑𝑚 = 𝜌𝐴𝑑𝐿. If 𝜌 and 𝐴 are constant over the length of the rod, the coordinates of the center of mass becomes the coordinates of the centroid 𝐶 of the line segment. Thus, 𝒙 = 𝒙𝒅𝑳 𝑳 𝒚 = 𝒚𝒅𝑳 𝑳 𝒛 = 𝒛𝒅𝑳 𝑳
- 13. If a line segment (or rod) lies within the x–y plane and described by a thin curve 𝑦 = 𝑓 (𝑥), as shown in the figure below, then its centroid is determined from: 𝒙 = 𝒙𝒅𝑳 𝑳 𝒚 = 𝒚𝒅𝑳 𝑳
- 14. The length of the differential element is given by the Pythagorean theorem: 𝒅𝑳 = 𝒅𝒙 𝟐 + 𝒅𝒚 𝟐 This can be written as: 𝒅𝑳 = 𝒅𝒙 𝒅𝒙 𝟐 𝒅𝒙 𝟐 + 𝒅𝒚 𝒅𝒙 𝟐 𝒅𝒙 𝟐 = 𝟏 + 𝒅𝒚 𝒅𝒙 𝟐 𝒅𝒙 In other way: 𝒅𝑳 = 𝒅𝒙 𝒅𝒚 𝟐 𝒅𝒚 𝟐 + 𝒅𝒚 𝒅𝒚 𝟐 𝒅𝒚 𝟐 = 𝒅𝒙 𝒅𝒚 𝟐 + 𝟏 𝒅𝐲
- 15. When a body of density𝜌 has a small but constant thickness 𝑡, the body can model as a surface area 𝐴. The mass of an element becomes 𝑑𝑚 = 𝜌𝑡𝑑𝐴. If 𝜌 and 𝑡 are constant over the entire area, the coordinates of the center of mass of the body becomes the coordinates of the centroid C of the surface area. Thus, 𝑥 = 𝑥𝑑𝐴 𝐴 𝑦 = 𝑦𝑑𝐴 𝐴 𝑧 = 𝑧𝑑𝐴 𝐴 The numerators in the equations are the first moments of area.
- 16. For a general body of volume 𝑉 and density 𝜌, the element has a mass 𝑑𝑚 = 𝜌𝑑𝑉. The density 𝜌 cancels if it is constant over the entire volume. The coordinates of the center of mass also become the coordinates of the centroid C of the body. Thus, 𝒙 = 𝒙𝒅𝑽 𝑽 𝒚 = 𝒚𝒅𝑽 𝑽 𝒛 = 𝒛𝒅𝑽 𝑽
- 17. Principles used to choose the differential element and setting up the integrals to simplify difficulties in the determination of centroid: Order of Element: a first-order differential element should be selected rather a higher-order element which cover the entire figure by a single integration. Example: A 1st order horizontal strip of area 𝑑𝐴 = 𝑙𝑑𝑦 requires only one integration with respect to y to cover the entire figure. The second-order element 𝑑𝑥𝑑𝑥 requires two integrations, with respect to x & y, to cover the entire figure.
- 18. A 1st order element in the form of a circular slice of volume 𝑑𝑉 = 𝜋𝑟2 𝑑𝑦 requires only one integration, & is preferable rather to choose a 3rd order element 𝑑𝑉 = 𝑑𝑥𝑑𝑦𝑑𝑧, which require three integrations..
- 19. Continuity: Choose an element which can be integrated in one continuous operation to cover the entire figure. Example: To determine the centroid the horizontal strip is preferable to the vertical strip, which, requires two separate integrals because of the discontinuity in the expression for the height of the strip at 𝑥 = 𝑥1.
- 20. Discarding Higher-Order Terms: Higher order terms should be dropped compared with lower-order terms. Example: In the figure above the vertical strip of area under the curve is given by the first-order term 𝑑𝐴 = 𝑦𝑑𝑥, and the second-order triangular area 1 2 𝑑𝑥𝑑𝑦 is discarded. In the limit, of course, there is no error.
- 21. Choice of Coordinates: Choose the coordinate system which best matches the boundaries of the figure. Example: The boundaries of the area in the 1st figure is easily described in rectangular coordinates, whereas the boundaries of the circular sector in the 2nd figure is best suited to polar coordinates.
- 22. Centroidal Coordinate of differentials Element: When a 1st or 2nd order differential element is chosen, it is essential to use the coordinate of the centroid of the element for the moment arm in expressing the moment of the differential element. Example: 𝑥 = 𝑥𝑐 𝑑𝐴 𝐴 𝑦 = 𝑦𝑐 𝑑𝐴 𝐴 𝑥 = 𝑥𝑐 𝑑𝑉 𝑉 𝑧 = 𝑧𝑐 𝑑𝑉 𝑉
- 23. #1. Locate the centroid of a circular arc as shown in the figure. Solution: The x-axis is chosen as axis of symmetry. The co-ordinate system is polar coordinates. Take the differential element length of the arc is 𝐝𝐋 = 𝒓𝒅𝜽
- 24. The total length of the arc is: 𝐿 = 2𝛼𝑟 Thus, 𝐿𝑥 = 𝑥𝑐 𝑑𝐿 , ⇒ 2𝛼𝑟𝑥 = −𝛼 𝛼 𝑟𝑐𝑜𝑠𝜃 𝑟𝑑𝜃 2𝛼𝑟𝑥 = 𝑟2 −𝛼 𝛼 𝑐𝑜𝑠𝜃𝑑𝜃 𝑥2𝛼𝑟 = 𝑟2 𝑠𝑖𝑛𝜃 𝛼 −𝛼 = 𝑟2 𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛 −𝛼 ⇒ 𝑥 = 𝑟𝑠𝑖𝑛𝛼 𝛼 For 2𝛼 = 𝜋, 𝑥 = 2𝑟 𝜋
- 25. #2. Determine the distance from the base of a triangle of altitude ℎ to the centroid of its area. Solution: The x-axis is taken to coincide with the base 𝑏 of the triangle. The co-ordinate system is cartesian coordinates The differential stripe area is taken as 𝑑𝐴 = 𝑥𝑑𝑦
- 26. From similarity of triangles: ℎ 𝑏 = (ℎ − 𝑦) 𝑥 ⇒ 𝑥 = 𝑏(ℎ − 𝑦) ℎ Thus, 𝐴𝑦 = 𝑦𝑐𝑑𝐴 , ⇒ 𝑏ℎ 2 𝑦 = 0 ℎ 𝑦𝑥𝑑𝑦 𝑏ℎ 2 𝑦 = 0 ℎ 𝑦 𝑏 ℎ (ℎ − 𝑦)𝑑𝑦
- 27. Integration with respect to 𝑦: ⇒ 𝒃𝒉 𝟐 𝒚 = 𝟎 𝒉 (𝒚𝒃 − 𝒃𝒚𝟐 𝒉 )𝒅𝒚 , ⇒ 𝒃𝒉 𝟐 𝒚 = 𝒃𝒚𝟐 𝟐 − 𝒃𝒚𝟑 𝟑𝒉 𝟎 𝒉 ⇒ 𝒃𝒉 𝟐 𝒚 = 𝒃𝒉𝟐 𝟐 − 𝒃𝒉𝟑 𝟑𝒉 ⇒ 𝒃𝒉 𝟐 𝒚 = 𝒃𝒉𝟐 𝟔 ⇒ 𝒚 = 𝒉 𝟑
- 28. #3. Locate the centroid of the area of a circular sector with respect to its vertex. Solution: Choose the x-axis as the axis of symmetry. The co-ordinate system is polar coordinates Two methods of solution: I. Circular ring is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 2𝑟𝑜𝛼𝑑𝑟𝑜
- 29. The centroid of the differential area is: 𝐱𝐜 = 𝐫𝐨𝐬𝐢𝐧𝛂 𝛂 Thus, 𝐀𝐱 = 𝐱𝐜𝐝𝐀 , ⇒ 𝟐𝛂 𝟐𝛑 𝛑𝐫𝟐 𝐱 = 𝟎 𝐫 𝐫𝐨𝐬𝐢𝐧𝛂 𝛂 𝟐𝐫𝐨𝛂𝐝𝐫𝐨 𝐫𝟐 𝛂𝐱 = 𝟐𝐬𝐢𝐧𝛂 𝟎 𝐫 𝐫𝐨 𝟐 𝐝𝐫𝐨 𝐫𝟐 𝛂𝐱 = 𝟐 𝟑 𝐫𝐨 𝟑 𝐬𝐢𝐧𝛂 𝒓 𝟎 𝐱 = 𝟐 𝟑 𝐫𝐬𝐢𝐧𝛂
- 30. II. Swinging triangle about the vertex is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 𝑟 2 𝑟𝑑𝜃 The centroid of the differential area is: 𝑥𝑐 = 2 3 𝑟𝑐𝑜𝑠𝜃 𝑨𝒙 = 𝒙𝒄𝒅𝑨 , ⇒ 𝟐𝜶 𝟐𝝅 𝝅𝒓𝟐 𝒙 = 𝟎 𝒓 2 3 𝑟𝑐𝑜𝑠𝜃 𝑟 2 𝑟𝑑𝜃 𝒓𝟐 𝜶𝒙 = 1 3 𝒓𝟑 −𝜶 𝜶 𝑐𝑜𝑠𝜃𝒅𝜽 𝒓𝟐 𝜶𝒙 = 𝟏 𝟑 𝒓𝟑 𝑠𝑖𝑛𝜃 𝜶 −𝜶 𝒙 = 2 3 𝑟 𝑠𝑖𝑛𝛼 𝜶 For a semicircular area, 𝟐𝜶 = 𝝅, 𝒙 = 4𝑟 3 𝝅
- 32. #4. Locate the centroid of the area under the curve 𝑥 = 𝑘𝑦3 from 𝑥 = 0 to 𝑥 = 𝑎 Solution: Two methods of solution: I. A vertical differential element is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 𝑦𝑑𝑥
- 33. The x-coordinate of the centroid is: 𝑨𝒙 = 𝒙𝒄𝒅𝑨 , ⇒ 𝒙 𝟎 𝒂 𝒚𝒅𝒙 = 𝟎 𝒂 𝑥𝑦𝑑𝑥 𝒚 = 𝒙 𝒌 𝟏 𝟑 , 𝒌 = 𝒂 𝒃𝟑 , ⇒ 𝒚 = 𝒃 𝒙 𝒂 𝟏 𝟑 𝒙 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝒅𝒙 = 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝒙𝒅𝒙 𝒙 𝟑 𝟒 𝒃 𝒂𝟏 𝟑 𝒙𝟒 𝟑 𝒂 𝟎 = 𝟑 𝟕 𝒃 𝒂𝟏 𝟑 𝒙𝟕 𝟑 𝒂 𝟎 𝒙 = 𝟒 𝟕 𝒂
- 34. The y-coordinate of the centroid is: 𝑨𝒚 = 𝒚𝒄𝒅𝑨 , ⇒ 𝒚 𝟎 𝒂 𝒚𝒅𝒙 = 𝟎 𝒂 𝒚 𝟐 𝑦𝑑𝑥 𝒚 = 𝒙 𝒌 𝟏 𝟑 , 𝒌 = 𝒂 𝒃𝟑 , ⇒ 𝒚 = 𝒃 𝒙 𝒂 𝟏 𝟑 𝒚 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝒅𝒙 = 𝟏 𝟐 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝟐 𝒅𝒙 𝒚 𝟑 𝟒 𝒃 𝒂𝟏 𝟑 𝒙𝟒 𝟑 𝒂 𝟎 = 𝟑 𝟏𝟎 𝒃𝟐 𝒂𝟐 𝟑 𝒙𝟓 𝟑 𝒂 𝟎 𝒚 = 𝟐 𝟓 𝒃
- 35. II. When a horizontal differential element is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 𝑎 − 𝑥 𝑑𝑦 : 𝐱𝐜 = 𝟏 𝟐 𝐱 + 𝐚 𝐱 = 𝐤𝐲𝟑 , 𝐤 = 𝐚 𝐛𝟑 , ⇒ 𝐱 = 𝐚 𝐲 𝐛 𝟑 𝐀𝐱 = 𝐱𝐜𝐝𝐀 ⇒ 𝐱 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝟏 𝟐 𝐚 + 𝐱 𝐚 − 𝐱 d𝐲 𝐱 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝟏 𝟐 𝐚𝟐 − 𝒙𝟐 d𝐲 𝒂𝒚 𝒃 𝟎 − 𝟏 𝟒 𝐚 𝐛𝟑 𝐲𝟒 𝐛 𝟎 𝐱 = 𝟏 𝟐 𝒂𝟐 𝒚 𝒃 𝟎 − 𝐚𝟐 𝟕𝐛𝟔 𝒚𝟕 𝐛 𝟎 𝟑 𝟒 𝐚𝐛 𝐱 = 𝟑𝐚𝟐 𝐛 𝟕 , 𝐱 = 4 7 a
- 36. To determine the centroid of y-co-ordinate: Take the differential stripe area 𝑑𝐴 = 𝑎 − 𝑥 𝑑𝑦 : 𝒚𝐜 = 𝒚 𝐱 = 𝐤𝐲𝟑 , 𝐤 = 𝐚 𝐛𝟑 , ⇒ 𝐱 = 𝐚 𝐲 𝐛 𝟑 𝐀𝒚 = 𝒚𝐜𝐝𝐀 , ⇒ 𝒚 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝒚 𝐚 − 𝐱 d𝐲 𝒚 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝐚𝐲 − 𝒙𝒚 d𝐲 𝒂𝒚 𝒃 𝟎 − 𝟏 𝟒 𝐚 𝐛𝟑 𝐲𝟒 𝐛 𝟎 𝐱 = 𝟏 𝟐 𝒂𝒚𝟐 𝒃 𝟎 − 𝒂 𝟓𝐛𝟑 𝒚𝟓 𝐛 𝟎 𝟑 𝟒 𝐚𝐛 𝐱 = 𝟑𝒂𝒃𝟐 𝟏𝟎 , 𝐱 = 𝟐 𝟓 𝒃
- 37. #6 Determine the y-coordinate of the centroid of the shaded area. Check your result for the special case 𝑎 = 0. Solution: The y-axis is chosen as axis of symmetry. ∴ 𝑥 = 0 The co-ordinate system is cartesian coordinates. Take the differential element area of the triangle is 𝐝𝐀 = 𝟐𝒙𝒅𝒚.
- 38. From similarity of triangle: 2𝑥 𝑦 = 2ℎ 𝑡𝑎𝑛30 ℎ , 2ℎ𝑥 = 2ℎ𝑦 𝑡𝑎𝑛30 2𝑥 = 2𝑦 𝑡𝑎𝑛30
- 39. From the given equation 𝑦 is evaluated as: 2𝑥 = 2𝑦 𝑡𝑎𝑛30 ∴ 𝑑𝐴 = 2𝑦𝑡𝑎𝑛30 𝑑𝑦 𝑦𝑑𝐴 = 𝑦𝑐 𝑑𝐴 , ⇒ 𝑦 2𝑦𝑡𝑎𝑛30 𝑑𝑦 = 𝑦 2𝑦𝑡𝑎𝑛30 𝑑𝑦 = 𝑦2𝑡𝑎𝑛30 𝑎 ℎ 𝑦𝑑𝑦 = 2𝑡𝑎𝑛30 𝑎 ℎ 𝑦2 𝑑𝑦 𝑡𝑎𝑛30𝑦 𝑦2 ℎ 𝑎 = 2 3 𝑡𝑎𝑛30𝑦 𝑦3 ℎ 𝑎 ℎ2 − 𝑎2 𝑡𝑎𝑛30𝑦 = 2 3 ℎ3 − 𝑎3 𝑡𝑎𝑛30 𝑦 = 2 3 ℎ3 − 𝑎3 ℎ2 − 𝑎2 , 𝑤ℎ𝑒𝑛 𝑎 = 0, 𝑦 = 2 3 ℎ
- 40. #1. Determine the distance 𝑥, 𝑦 to the center of gravity of the homogeneous rod.
- 41. #2. Locate the centroid 𝑥 of the parabolic area.
- 42. A composite body consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular, semicircular, etc. The body can be sectioned or divided into its composite parts. The weight and location of the center of gravity of each of these parts are known. This avoids the need for integration to determine the center of gravity for the entire body. Composite body 𝑥 = ∑𝑚𝑥𝑐 ∑𝑚 𝑦 = ∑𝑚𝑦𝑐 ∑𝑚 𝑧 = ∑𝑚𝑧𝑐 ∑𝑚 𝑚1 + 𝑚2 + 𝑚3 𝑥 = 𝑚1𝑥1 + 𝑚2𝑥2 + 𝑚3𝑥3
- 43. In practice the boundaries of an area or volume might not be expressible in terms of simple geometrical shapes as shapes which can be represented mathematically. For such cases method of approximation is used to determine centroid of a body. 𝑚1 + 𝑚2 + 𝑚3 𝑥 = 𝑚1𝑥1 + 𝑚2𝑥2 + 𝑚3𝑥3
- 44. Example: #1. Centroid C of the irregular area Example: #2. The centroid C of an irregular volume 𝑥 = ∑𝐴𝑥𝑐 ∑𝐴 𝑥 = ∑𝐴𝑥𝑐 ∑𝐴 𝑥 = ∑ 𝐴∆𝑥 𝑥𝑐 ∑ 𝐴∆𝑥 𝑥 = ∑𝑉𝑥𝑐 ∑𝑉
- 45. To determine location of the center of gravity or the centroid of a composite geometrical object represented by a line, area, or volume: Using a sketch, divide the body or object into a finite number of composite parts of simpler shapes. Composite body with a hole is considered as body without the hole and the hole as an additional composite part having negative weight or size. Establish the coordinate axes on the sketch and determine the coordinates 𝑥𝑐, 𝑦𝑐 & 𝑧𝑐 of the center of gravity or centroid of each part. Determine 𝑥, 𝑦 & 𝑧 by applying the center of gravity equations. If an object is symmetrical about an axis, the centroid of the object lies on this axis.
- 46. #3. Locate the centroid of the shaded area. Solution: The object is divided into four parts.
- 47. Tabulated values. Thus, Part 𝐴 𝑖𝑛2 𝑥𝑐 𝑖𝑛 𝑦𝑐 𝑓𝑡 𝐴𝑥𝑐 𝑖𝑛3 𝐴𝑦𝑐 𝑖𝑛2 1 120 6 5 720 600 2 30 14 3.33 420 100 3 −14.14 6 1.27 −84.84 −18 4 −8 12 4 −96 −32 ∑ 127.86 2.15 ∑ 𝟗𝟓𝟔. 𝟏𝟔 𝟔𝟓𝟎 𝑥 = ∑𝐴𝑥𝑐 ∑𝐴 , ⇒ 𝑥 = 956.16 127.86 , ∴ 𝑥 = 7.478𝑖𝑛 𝑦 = ∑𝐴𝑧𝑐 ∑𝐴 , ⇒ 𝑦 = 650 127.86 , ∴ 𝑦 = 5.083𝑖𝑛
- 48. #7. Determine the x- and y- coordinates of the centroid of the shaded area. Solution: The co-ordinate system is cartesian coordinates. Take the differential element area of the triangle is 𝐝𝐀 = 𝒚𝟐 − 𝒚𝟏 𝒅𝒙.
- 49. Centroid of the shaded area can be evaluated as: 𝐀𝒙 = 𝒙𝐜𝐝𝑨 , ⇒ 𝒙 𝑦2 − 𝑦1 𝑑𝑥 = 𝒙 𝑦2 − 𝑦1 𝑑𝑥 From the given equation: 𝑦1 = 𝑎𝑥 1 2 , 𝑦2 = 𝑥3 𝑎2
- 50. From the given equation 𝑥 is evaluated as: 𝒙 0 𝑎 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 0 𝑎 𝑥 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 𝑥 2 3 𝑎1 2 𝑥3 2 − 1 4 𝑥4 𝑎2 𝑎 0 = 2 5 𝑎1 2 𝑥5 2 − 1 5 𝑥5 𝑎2 𝑎 0 5 12 𝑎2 𝑥 = 1 5 𝑎3 𝒙 = 𝟏𝟐 𝟐𝟓 𝒂 … … 𝑨𝒏𝒔.
- 51. From the given equation 𝑦 is evaluated as: 𝐀𝒚 = 𝒚𝒄𝐝𝑨 , ⇒ 𝒚 𝑦2 − 𝑦1 𝑑𝑥 = 𝒚 𝑦2 − 𝑦1 𝑑𝑥, 𝒚𝒄 = 𝒚 = 𝟏 𝟐 𝒚𝟏 + 𝒚𝟐 𝒚 0 𝑎 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 1 2 0 𝑎 𝑎𝑥 1 2 + 𝑥3 𝑎2 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 𝒚 0 𝑎 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 1 2 0 𝑎 𝑎𝑥 − 𝑥6 𝑎4 𝑑𝑥 = 𝑦 2 3 𝑎1 2 𝑥3 2 − 1 4 𝑥4 𝑎2 𝑎 0 = 1 2 𝑎𝑥2 − 1 7 𝑥7 𝑎4 𝑎 0 5 12 𝑎2 𝑦 = 3 7 𝑎3 , ⇒ 𝒚 = 𝟑𝟔 𝟑𝟓 𝒂 … … 𝑨𝒏𝒔.