Equilibrium and the Law of Sines
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Equilibrium and the Law of Sines 1.0 kg box is suspended by two s 80.0 o apart. How much force sion) does each wire have? 11.0 kg 80.0 o 50.0 o 50.0 o 80.0 o 140.0 o 140.0 o 90.0 o 90.0 o F a F b F a and F b are components that overcome (equalize) the weight of the box, so we can say that F w is the Equilibrant of F a and F b or of its Resultant (or vise- versa): F a and F b (or their resultant) are equilibrants of F w ! F a F b R F a F b R E Rest, therefore F = 0 R + E = 0 R = -E E = F w = -108 N Therefore, R = 108 N
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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. - PowerPoint PPT Presentation
Transcript of Equilibrium and the Law of Sines
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
80.0o
50.0o 50.0o
80.0o
140.0o 140.0o
90.0o90.0o
Fa Fb
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!
Fa
Fb
R Fa
Fb
R
E
Rest, therefore F = 0R + E = 0R = -E E = Fw = -108 NTherefore, R = 108 N
FaFb
R
E
Fa
Fb
R
80.0o
50.0o 50.0o
90.0o90.0o
50.0o
Fa
Fb Fb
80o
80o
100o
100o
Fa Fb80o
40o 40o
40o
40o
Fa
Fb
R
50.0o
100o
40o
40o
a
b
c
Fa is side a, Fb is side b, and R is side c
Angles are opposite their sides
B
C
A
Law of Sines states that:a/sin A = b/sin B = c/sin C
So, to find side “a” use two of the equalities
a/sinA = c/sinC
a/sin40o = 108 N/sin100o
a = 70 N
Remember, R = 108 N
b/sinB = c/sinC
b/sin40o = 108 N/sin100o
b = 70 N
