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**Try this problem**Five forces act on an object: the first, 60 N at 90 degrees; the second 40 N at 0 degrees; the third 80 N at 270 degrees; the forth 40 N at 180 degrees; the fifth 50 N at 60 degrees. What is the magnitude and direction of a sixth force that produces equilibrium of the object?**Math**Explanation 1) 60N 80N 40 N 40N 50N 2) H V 0 60 40 0 0 -80 -40 0 25 43.3. 25 23.3 Totals Estimation Box 1) WHY-> Since I am dealing with forces and forces are vector quantities because they must be defined with both magnitude and direction I must add them vectorally by following the rules of vector addition in order to preserve direction. Explain No variables to explain Math Since I chose horizontal and vertical components I get a 90o angle and I can then use trig to solve my direction, and geometry to find my magnitude 2) Since horizontal and vertical are independent of each other and the sum of the horizontal components will give me the horizontal component of the resultant I can add them. I will use a chart to keep myself organized. 3) Since I chose horizontal and vertical components the angle between them is 90o because they are perpendicular. Therefore, I can use trig to find the rest of the components 3) Sin 60o (50N) = 43.3N Cos 60o (50N) = 25N 4) Since I had a 90o angle and found components of my resultant I can use the Pythagorean theorem to find the magnitude. 4) c2 = a2 + b2 c2 = 252 + 23.32 c = 34.17N a2 + b2 = c2 5) Since I had a 90o angle and found components of my resultant I can use the inverse tangent to find my angle c2 = 252 + 232 C = 34 6) Since I found my magnitude and direction and I need both to find a vector I think I am finished and my estimation is close in magnitude and direction to my answer. I need to take the opposite of the direction but I need to keep the same magnitude the figure out what would make it come to equilibrium. 5) Tan θ = .932 θ = 43o 6) My component to reach equilibrium is 34.17N at an angle of 223o above the horizontal position.