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Equilibrium and the Law of Sines

F a. F a. F a. F b. F b. F b. R. R. 80.0 o. 50.0 o. 50.0 o. 90.0 o. 90.0 o. E. Equilibrium and the Law of Sines. An 11.0 kg box is suspended by two wires 80.0 o apart. How much force (tension) does each wire have?. 11.0 kg.

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Equilibrium and the Law of Sines

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  1. Fa Fa Fa Fb Fb Fb R R 80.0o 50.0o 50.0o 90.0o 90.0o E Equilibrium and the Law of Sines An 11.0 kg box is suspended by two wires 80.0o apart. How much force (tension) does each wire have? 11.0 kg Fa and Fb are components that overcome (equalize) the weight of the box, so we can say that Fw is the Equilibrant of Fa and Fb or of its Resultant (or vise-versa): Fa and Fb (or their resultant) are equilibrants of Fw! 80.0o 140.0o 140.0o Rest, therefore SF = 0 R + E = 0 R = -E E = Fw = -108 N Therefore, R = 108 N

  2. Fa Fa Fa Fb Fa 50.0o Fb Fb Fb Fb R R 80.0o 50.0o 50.0o 90.0o 90.0o E 80o 40o 100o 80o 100o 40o 40o 40o 80o

  3. 40o Fb A 100o Fa C 40o 50.0o R B Remember, R = 108 N Fa is side a, Fb is side b, and R is side c Angles are opposite their sides b Law of Sines states that: a/sin A = b/sin B = c/sin C So, to find side “a” use two of the equalities c a/sinA = c/sinC a/sin40o = 108 N/sin100o a = 70 N a b/sinB = c/sinC b/sin40o = 108 N/sin100o b = 70 N

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