Lab-03.pdf - Lab 3 VECTOR ADDITION EQUILIBRIUM OF FORCES...

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Lab 3 VECTOR ADDITION & EQUILIBRIUM OF FORCES Procedure 1: Graphical method of finding the resultant of vectors Scale: 1cm = 20 gm A is 230 gm (11.5 cm) force directed 30 relative to the X axis. B is 300 gm ( 20 cm) force directed 70 relative to the horizontal axis. The Magnitude R was found at 500 gm (25 cm) The direction was found at 50 relative to the x axis. Vector Mass(gm) Orientation ϴ (degrees) A 230 30 B 300 70 R 500 50 Procedure 2: Analytical method of finding the resultant of vectors. Procedure 2a Vector Mass(gm) Orientation ϴ (degrees) X-component(gm) Y-component (gm) A 230 30 199.18 115 B 300 70 102.60 281.90 R 498.59 52.75 301.78 396.9 Using the vectors given in part 1 the resultants where found as shown: 𝑅 ? : ? ? = 230 cos(30°) 𝑅 ? : ? ? = 230 sin(30°) = 199.18 𝑔? = 115 𝑔? ? ? = 300 cos(70°) ? ? = 300 sin(70°) + = 102.60 𝑔? = 281.90 𝑔? = 301.78 𝑔? = 396.9 𝑔? Formula used to find Resultant and Angle:
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|𝑅| = √(𝑅?) 2 + (𝑅?) 2 𝜃 = 𝑡𝑎? −1 ( 𝑅? 𝑅? ) |𝑅| = √(301.78 𝑔?) 2 + (396.9 𝑔?) 2 𝜃 = 𝑡𝑎? −1 ( 396.9 301.78 ) = 498.59 𝑔? = 52.75° Procedure 2b Calculate the magnitude and direction of (a) the resultant R and (b) the equilibrant E Vector
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