Question

The three ropes in $FIGUREEX6.1$are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with the tensions shown in the figure. What are the magnitude and direction of the tension $\overrightarrow{T_3}$in the third rope?

Short answer

The magnitude of the tension $\overrightarrow{T_3}$is $94.35N$and in the third rope it makes$58°$with the positive x-axis.

Step-by-step solution

Tension :

The pulling force transferred axially by a string, cable, chain, or similar item, or by each end of a rod, truss member, or similar three-dimensional object is known as tension.

Direction of the tension T3→ :

Three ropes connect a ring together, two of which are fixed to the walls at right angles as illustrated in the diagram.

The free body diagram of the ring is shown :

Using Newton's second law for equilibrium condition,

$\sum _{}{F}_x=0$

Resolving the tension components in the three ropes in the x-direction,

$-T_1+T_3\cos \theta =0$

Tension in the first rope =$T_1$

Tension in the third rope=$T_2$

Angle made by the third rope with the positive x-axis=$\theta$

Substitute $50N$for $T_1$.

role="math" localid="1647429381825" $$\begin{gathered} 50N+T_3\cos \theta =0 \\ {T}_3\cos \theta =50N.......\left(1\right) \end{gathered}$$

Using Newton's second law for equilibrium condition,

$\sum _{}{F}_y=0$

Resolving the components of tension in the three ropes in y-direction,

role="math" localid="1647429112409" $T_2-T_3\sin \theta =0$where $T_2$is tension in the second rope.

Substitute $80N$for $T_2$,

role="math" localid="1647429359075" $$\begin{gathered} 80N-T_3\sin \theta =0 \\ {T}_3\sin \theta =80N......\left(2\right) \end{gathered}$$

Divide equation $\left(2\right)$by $\left(1\right)$.

$$\begin{gathered} \frac{T_3\sin \theta }{T_3\cos \theta }=\frac{80N}{50N} \\ \tan \theta =1.6 \\ \theta ={\tan }^{{}^{-1}}(1.6) \\ \theta =58° \end{gathered}$$

Hence, the third rope makes$58°$with the positive x-axis.

Magnitude of the tension T3→ :

Three ropes connect a ring together, two of which are fixed to the walls at right angles as illustrated in the diagram.

Substitute $58°$for $\theta$in equation $\left(1\right)$,

$$\begin{gathered} T_3\cos 58°=50N \\ {T}_3=\frac{50N}{\cos 58°} \\ =94.35N \end{gathered}$$

Hence, $94.35N$is magnitude of the tension $\overrightarrow{T_3}$.