Question

Three ropes A, B and C are tied together in one single knot K. (See figure.)
If the tension in rope A is 50.5 N, then what is the tension in rope B?

Answer 1

Assuming point K is held in equilibrium, by Newton's second law we have

• net horizontal force

$F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0$

• net vertical force

$F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0$

where the angle $\theta$ that rope C makes with the horizontal axis satisfies

$\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57$

Solve the first equation for $F_C$.

$F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)$

(Recall that $\sec(x)=\frac1{\cos(x)}$.)

Substitute this into the second equation and solve for $F_B$.

$F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)$

$F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)$

$F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)$

(Recall that $\tan(x)=\frac{\sin(x)}{\cos(x)}$.)

$F_B = \dfrac57 F_A$

$\boxed{F_B \approx 36.1\,\rm N}$