-4.45 A pinata of mass \(M=8.0 \mathrm{~kg}\) is attached to a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.0 \mathrm{~m},\) and the top of the right pole is a vertical distance \(h=0.50 \mathrm{~m}\) higher than the top of the left pole. The pinata is attached to the rope at a horizontal position halfway between the two poles and at a vertical distance \(s=1.0 \mathrm{~m}\) below the top of the left pole. Find the tension in each part of the rope due to the weight of the pinata.

Given information: - Pinata mass: \(M=8.0\,\text{kg}\) - Horizontal distance between poles: \(D=2.0\,\text{m}\) - Vertical distance between the right pole higher than the left pole: \(h=0.50\,\text{m}\) - Vertical distance of the pinata below the left pole: \(s=1.0\,\text{m}\) We need to find the tension in each part of the rope due to the pinata's weight. Let's denote the tension in the left part of the rope by \(T_1\) and the tension in the right part of the rope by \(T_2\).

Let's denote the angle between the left part of the rope and the horizontal plane by \(\alpha\), and the angle between the right part of the rope and the horizontal plane by \(\beta\). Using the given dimensions, we can calculate the angles with the help of the tangent functions: \(\tan(\alpha) = \frac{s}{\frac{D}{2}}\) and \(\tan(\beta) = \frac{s - h}{\frac{D}{2}}\) We'll also need the pinata's weight, which can be calculated using the equation \(W = M \times g\), where \(g = 9.81\,\text{m/s}^2\) is the acceleration due to gravity.

To find the tension in the rope, we'll analyze the forces acting on the pinata in the horizontal and vertical directions. In the horizontal direction, the forces balance each other out: \(T_1 \cos(\alpha) = T_2 \cos(\beta)\) In the vertical direction, the sum of the vertical components of the tensions must equal the weight of the pinata: \(T_1 \sin(\alpha) + T_2 \sin(\beta) = W\)

We have a system of two equations with two unknowns, \(T_1\) and \(T_2\). To solve the system, we can use the substitution method or the elimination method. We choose the elimination method in this case. Divide the first equation by \(\cos(\alpha)\) and the second equation by \(\sin(\alpha)\) to eliminate \(T_1\): \(\frac{T_1 \cos(\alpha)}{\cos(\alpha)} = \frac{T_2 \cos(\beta)}{\cos(\alpha)} \Rightarrow T_1 = T_2 \frac{\cos(\beta)}{\cos(\alpha)}\) \(\frac{T_1 \sin(\alpha)}{\sin(\alpha)} + \frac{T_2 \sin(\beta)}{\sin(\alpha)} = \frac{W}{\sin(\alpha)} \Rightarrow T_1 + T_2 \frac{\sin(\beta)}{\sin(\alpha)} = \frac{W}{\sin(\alpha)}\) Now, substitute for \(T_1\) in the second equation: \(T_2 \frac{\cos(\beta)}{\cos(\alpha)} + T_2 \frac{\sin(\beta)}{\sin(\alpha)} = \frac{W}{\sin(\alpha)}\) Factor out \(T_2\): \(T_2 \left(\frac{\cos(\beta)}{\cos(\alpha)} + \frac{\sin(\beta)}{\sin(\alpha)}\right) = \frac{W}{\sin(\alpha)}\) Now, solve for \(T_2\) and substitute it back into the equation for \(T_1\) to find the values of both tensions. After finding the tensions, we can provide the final answer: The tension in the left part of the rope (due to the pinata's weight) is \(T_1 \approx (T_2 \cdot \frac{\cos(\beta)}{\cos(\alpha)})\) N. The tension in the right part of the rope (due to the pinata's weight) is \(T_2 \approx (\frac{W}{\sin(\alpha)} \cdot \left(\frac{\cos(\beta)}{\cos(\alpha)} + \frac{\sin(\beta)}{\sin(\alpha)}\right)^{-1})\) N.