Question

the three ropes in the figure (figure 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.

part a
what is the magnitude of the tension \\( \vec { t } _ { 3 } \\) in the third rope?
express your answer in newtons.

part b
what is the direction of the tension \\( \vec { t } _ { 3 } \\) in the third rope?
express your answer in degrees.

Explanation:

Step1: Find the angle between the two ropes

Using the Pythagorean theorem for the right - angled triangle formed by the two ropes' lengths (\(a = 0.60m\), \(b=0.80m\)), the angle \(\alpha\) between \(T_1\) and \(T_2\) can be found. \(\tan\alpha=\frac{0.60}{0.80}\), so \(\alpha = 36.87^{\circ}\).

Step2: Resolve the forces in \(x\) and \(y\) directions

Let the angle of \(T_3\) with the negative \(x\) - axis be \(\theta\).
In the \(x\) - direction: \(T_3\cos\theta=T_1\sin\alpha + T_2\cos\alpha\)
In the \(y\) - direction: \(T_3\sin\theta=T_2\sin\alpha - T_1\cos\alpha\)

Substitute \(T_1 = 50N\), \(T_2 = 80N\), \(\alpha = 36.87^{\circ}\)
\(T_1\sin\alpha=50\times\sin(36.87^{\circ}) = 50\times\frac{3}{5}=30N\), \(T_1\cos\alpha=50\times\frac{4}{5} = 40N\)
\(T_2\sin\alpha=80\times\frac{3}{5}=48N\), \(T_2\cos\alpha=80\times\frac{4}{5}=64N\)

In the \(x\) - direction: \(T_3\cos\theta=30 + 64=94N\)
In the \(y\) - direction: \(T_3\sin\theta=48-40 = 8N\)

Step3: Calculate the angle \(\theta\)

\(\tan\theta=\frac{T_3\sin\theta}{T_3\cos\theta}=\frac{8}{94}\)
\(\theta=\arctan(\frac{8}{94})\approx4.85^{\circ}\)

Answer:

\(4.85^{\circ}\)