X 0150gn f y 0260gn f x 0150gn f y 0260gn vector

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x =0.150gN F y =0.260gN F x =0.150gN F y =0.260gN Vector addition IV F 1 =(0.100)gN, 1 =30 o F 2 =(0.200)gN, 2 =90 o F 3 =(0.300)gN, 3 =225 o F R =0.131gN r =163 o F R =0.131gN r =163 o F R =0.08gN r =141.5 o Vector addition V F 1 =(0.225)gN, 1 =56 o F 2 =( 0.240)gN, 2 =- 62 o F R =0.238gN r =-5.4 o F R =0.239gN r =-6.0 o F R =0.265gN r =-5.5 o References See Figure 1. Graph section Please see Formulas and Calculation s section Data Gathered directly from Force table Formulas and Calculations The weight of an object is the force of gravity on the object and may be calculated as the mass times the acceleration of gravity. Since the weight is a force or magnitude, its SI units is the Newton (N). For an object in free fall, so that gravity is the only force acting on it, then the expression for weight follows from Newton's second law is: F = m. g (1) F Force in Newton's (N) m Mass in Kilograms (g) g Gravity acceleration (9.8 m/s²) – this factor is a constant and is left as a constant in calculations as “g”. When adding force vectors, the resultant force may be determined by the analytical method of components. For two concurrent forces (F1 and F2) acting on the center ring on the force table, the resultant force, FR = F1+F2. To add these force vectors by analitycal method, the magnitude and direction of the resultant force is determined using the law of cosines and the law of sines for every force vector in order to find their components (X and Y).
F1x = F1cos F2x = F2cos (2) F1y = F1sin F2y = F2sin By the addition of the like components in this case FX and FY, the components of the resultant force FR can be determined. Frx = F1x + F2x Fry = F1y + F2y (3) Pythagorean theorem can be used to find the magnitude of the resultant force of the vector FR, assuming that the X component lies on the X- axis with an angle of 0°, and the Y component lies on the Y- axis forming an angle of 90°, therefore by adding those vectors tip to tail graphically the hypotenuse can be determined by drawing a diagonal line from the origin point (0,0) to the tip of the component Y vector creating a triangle rectangle, hence the magnitude of the resultant force vector (Fr) can be calculated. Fr² = Frx² + Fry², F r = F x 2 + F y 2 (4) Knowing the components X and Y of the resultant force vector, the direction of the vector also can be computed using the law of arc tangent in order to find the angle of the triangle rectangle, but in reality the resultant direction of the arc tangent law is not going to be the equilibrated vector for all the other forces. The equilibrated force will be the opposite of the direction of the vector force resultant. ¿ tan 1 F Ry F Rx (5) Calculations Vector Addition I. forces X component (2) Y component (2) F 1 =.200 gN, 1 =30 o F1x=0.200cos 30 =0.173 gN F1y=0.200sin30 =0.100 gN F 2 = .200 gN, 2 =120 o F2x=0.200cos120 =- 0.100gN F2y=0.200 sin12 0 = 0.173gN Sum of components (3) Fx=0.173+(- 0.100)=0.073gN Fy=0.100+0.173=0.273 gN
Resultant vector, Fr at angle, F r = 0.073 2 + 0.273 2 = 0.283gN ¿ tan 1 0.273 0.073 = 75 Vector Addition II. forces X component (2) Y component (2) F 1 =.200 gN, 1 =20 o F1x=0.200cos 20 =0.188 gN F1y=0.200sin20 =0.068 gN F 2 = .150 gN, 2 =80 o F2x=0.150cos80 =0.026 gN F2y=0.150sin80 = 0.148gN Sum of components (3) Fx=0.188+0.026=0.214 gN Fy=0.068+0.148=0.216 gN Resultant vector, Fr at angle, F r = 0.214 2 + 0.216 2 = 0.304gN ¿ tan 1 0.216 0.214 = 45.3 Vector addition III. forces X component (2) Y component (2) F 1 =.200 gN, 1 =0 o F1x=0.200cos 0 =0.2gN F1y=0.200sin0

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