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Chapter 4: Problem 46

A pinata of mass \(M=12\) kg hangs on a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.0 \mathrm{~m}\), the top of the right pole is a vertical distance \(h=0.50 \mathrm{~m}\) higher than the top of the left pole, and the total length of the rope between the poles is \(L=3.0 \mathrm{~m}\). The pinata is attached to a ring, with the rope passing through the center of the ring. The ring is frictionless, so that it can slide freely

Short Answer

Expert verified
#subject# Physics #question# Determine the tension in the rope and the position of the pinata along the rope. #answer# To determine the tension in the rope and the position of the pinata along the rope, we need to follow these steps: 1. Determine the angle between the rope and the horizontal using the given values and the Pythagorean theorem. 2. Calculate the horizontal distance (x) and vertical distance (y) from the left pole to the location where the pinata hangs using trigonometric functions. 3. Calculate the tension (T) in the rope using Newton's second law of motion and known values for the forces acting on the pinata. 4. Calculate the position of the pinata along the rope using the calculated values of x, y, and the angle. By following these steps, we can find both the tension in the rope and the position of the pinata along the rope.

Step by step solution

01

Determine the angle between the rope and the horizontal

We will first use the given values \(D\), \(h\), and \(L\) to find the angle \(\theta\) between the rope and the horizontal. To do this, we will need to use the Pythagorean theorem, relating the length of all the sides of a right triangle formed by the rope. Let \(x\) be the horizontal distance from the left pole to the location where the pinata hangs, \(y\) be the vertical distance from the top of the left pole to the location where the pinata hangs, and \(d\) be the distance between the top of the left pole and the pinata. Now we can form a right triangle with side lengths \(x\), \(y\), and \(d\). Since we know that \(L^2 = (D-x)^2 + (h+y)^2\), we can plug in known values to eventually solve for \(\theta\).
02

Calculate \(x\) and \(y\)

Based on the triangle formed by the rope, we can create the following equations using the Pythagorean theorem: (1) \(d^2 = x^2 + y^2\) (2) \((3-x)^2 + (0.5+y)^2 = L^2\) Plug in the known values for \(D\), \(h\), and \(L\): (2) \((2-x)^2 + (0.5+y)^2 = 9\) Now we can use the trigonometric function sine to solve for the values of \(x\) and \(y\): (3) \(\sin{\theta} = \frac{y}{d}\) (4) \(\cos{\theta} = \frac{x}{d}\) (5) \(y = d\sin{\theta}\) (6) \(x = d\cos{\theta}\)
03

Calculate the tension in the rope

Now that we have equations for \(x\), \(y\), and \(d\), we can use Newton's second law of motion to calculate the tension in the rope, \(T\). The forces acting on the pinata are the tension force and the gravitational force, \(Mg\). Since the pinata is in equilibrium, the tension force must balance the gravitational force. In the horizontal direction, we have: \(T_x = T\cos{\theta}\) In the vertical direction, we have: \(T_y = T\sin{\theta} + Mg\) Now, we can divide the vertical equation by the horizontal equation: \(\frac{T_y}{T_x} = \frac{T\sin{\theta} + Mg}{T\cos{\theta}}\) Now, we can solve for \(T\): \(T\cos{\theta}(\frac{T_y}{T_x}) - T\sin{\theta} = Mg\) \(T = \frac{Mg}{\cos{\theta}(\frac{T_y}{T_x}) - \sin{\theta}}\) Now, we can plug in the known values for \(M\), \(g\), \(\theta\), \(T_x\), and \(T_y\) to calculate the tension force in the rope.
04

Calculate the position of the pinata along the rope

Using equations (5) and (6), we can calculate the position of the pinata along the rope: \(x = d\cos{\theta}\) \(y = d\sin{\theta}\) Finally, we can plug in the known values for \(d\) and \(\theta\) to find the position of the pinata in terms of \(x\) and \(y\).

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