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