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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?

Three Ropes A B And C Are Tied Together In One Single Knot K See Figure If The Tension In Rope A Is 505 N Then What Is The Tension In Rope B class=

Sagot :

LammettHashidn

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

• net horizontal force

[tex]F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0[/tex]

• net vertical force

[tex]F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0[/tex]

where the angle [tex]\theta[/tex] that rope C makes with the horizontal axis satisfies

[tex]\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57[/tex]

Solve the first equation for [tex]F_C[/tex].

[tex]F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\sec(x)=\frac1{\cos(x)}[/tex].)

Substitute this into the second equation and solve for [tex]F_B[/tex].

[tex]F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}[/tex].)

[tex]F_B = \dfrac57 F_A[/tex]

[tex]\boxed{F_B \approx 36.1\,\rm N}[/tex]

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