Question

a 4.10 kg box is suspended from a system of massless wires attached to a ceiling. the first wire makes an angle of \\( \theta _ { 1 } = 51.6 ^ { \circ } \\) with the ceiling, whereas the second wire makes an angle of \\( \theta _ { 2 } = 37.9 ^ { \circ } \\) with the ceiling, as shown in the figure. determine the magnitudes of the tensions in the first and second wires, \\( t _ { 1 } \\) and \\( t _ { 2 } \\), respectively.

Explanation:

Step1: Analyze the vertical forces

The weight of the box \(W = mg\), where \(m = 4.10\space kg\) and \(g=9.8\space m/s^{2}\). So \(W=4.10\times9.8 = 40.18\space N\).
Since the system is in equilibrium, the sum of the vertical components of \(T_{1}\) and \(T_{2}\) equals the weight of the box.
The vertical component of \(T_{1}\) is \(T_{1}\sin\theta_{1}\) and the vertical component of \(T_{2}\) is \(T_{2}\sin\theta_{2}\). So \(T_{1}\sin\theta_{1}+T_{2}\sin\theta_{2}=mg\).

Step2: Analyze the horizontal forces

The sum of the horizontal forces is zero. The horizontal component of \(T_{1}\) is \(T_{1}\cos\theta_{1}\) and the horizontal component of \(T_{2}\) is \(T_{2}\cos\theta_{2}\). So \(T_{1}\cos\theta_{1}=T_{2}\cos\theta_{2}\), which gives \(T_{1}=T_{2}\frac{\cos\theta_{2}}{\cos\theta_{1}}\).
Substitute \(\theta_{1} = 51.6^{\circ}\) and \(\theta_{2}=37.9^{\circ}\), \(\cos\theta_{1}=\cos(51.6^{\circ})\approx0.621\), \(\cos\theta_{2}=\cos(37.9^{\circ})\approx0.789\). Then \(T_{1}=T_{2}\frac{0.789}{0.621}\approx1.27T_{2}\).

Step3: Substitute \(T_{1}\) into the vertical - force equation

Substitute \(T_{1} = 1.27T_{2}\) into \(T_{1}\sin\theta_{1}+T_{2}\sin\theta_{2}=mg\).
\(\sin\theta_{1}=\sin(51.6^{\circ})\approx0.784\), \(\sin\theta_{2}=\sin(37.9^{\circ})\approx0.614\).
\(1.27T_{2}\times0.784+T_{2}\times0.614 = 40.18\).
\((1.27\times0.784 + 0.614)T_{2}=40.18\).
\((0.996 + 0.614)T_{2}=40.18\).
\(1.61T_{2}=40.18\).
Solve for \(T_{2}\): \(T_{2}=\frac{40.18}{1.61}\approx24.96\space N\).

Step4: Calculate \(T_{1}\)

Since \(T_{1}=1.27T_{2}\), then \(T_{1}=1.27\times24.96\approx31.7\space N\).

Answer:

\(T_{1}\approx31.7\space N\), \(T_{2}\approx25.0\space N\)