Question

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

#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

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 \(\theta\).

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) \(x = d\cos{\theta}\)

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.

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 \(y\).

Key concepts

Tension Force

Tension force is an essential concept in understanding how objects in static equilibrium behave, especially when they're suspended by ropes or strings. In this scenario, the tension force is the force exerted by the rope on the pinata, which keeps it hanging without falling. Unlike other forces that may push or pull directly on an object, tension is a force that is transmitted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. To visualize this, imagine the rope as a stretched elastic band between your hands. It's the same kind of force that is keeping the pinata up.

• For the pinata to remain in equilibrium, the tension in the rope must counteract the weight of the pinata.
• Therefore, tension is directly related to the gravitational force pulling down on the pinata, which is its mass times gravity (Mg).
• Mathematically, tension can be split into components along the axis of interest - usually, horizontal and vertical.

This balance of forces keeps the pinata stationary and provides a practical demonstration of how tension works to balance other forces.

Newton's Second Law

Newton's Second Law is crucial in analyzing how forces interact to create equilibrium, which is the main stage of our pinata problem. Newton's Second Law states that the acceleration of an object depends on the net force acting upon it and its mass, formulated as \(F = ma\). In static equilibrium, however, there is no acceleration.
This means the net forces acting on the object are zero, maintaining a balance that helps us solve for other unknowns, like tension force.

• In the case of the pinata, the net force is zero, implying that the upward tension force equals the downward gravitational force.
• This is expressed mathematically as \(T = Mg\) when the angles are zero or simplified.
• Otherwise, the law's principle is used to relate forces in specific directions using trigonometric components, captured by \(T_x = T\cos{\theta}\) and \(T_y = T\sin{\theta} + Mg\).

Understanding Newton's Second Law's role in this setup allows us to solve for the tension by leveraging the equilibrium condition, where sum of forces equals zero.

Trigonometry

Trigonometry helps us relate the angles and distances in the setup, crucial for solving the physics problem of a hanging pinata. The problem involves finding the angle \(\theta\) that the rope makes with the horizontal, a task particularly suited for trigonometric functions.
These relationships are core to breaking the tension forces into their horizontal \(x\) and vertical \(y\) components.

• Use \(\sin{\theta}\) and \(\cos{\theta}\) to express the vertical and horizontal components of various lengths and forces.
• The sine function computes the ratio of the vertical side to the hypotenuse (\(\sin{\theta} = \frac{y}{L}\)).
• The cosine function computes the ratio of the horizontal side to the hypotenuse (\(\cos{\theta} = \frac{x}{L}\)).

By understanding these basic trigonometric identities, one can calculate unknown values like the direction of the rope or compare lengths critical in determining equilibrium conditions.

Pythagorean Theorem

The Pythagorean Theorem is a foundational principle that assists in solving problems involving right-angled triangles, such as in this mechanical equilibrium scenario. It is depicted as \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.
In the pinata problem, this theorem helps determine the relationships between various lengths on the right triangle formed by the rope and poles.

• Given the right triangle's sides \(x\) and \(y\), and hypotenuse \(d\), you use \(d^2 = x^2 + y^2\) to find unknown distances or check geometric validity.
• An additional relationship using the length of the entire rope, \(L\), is set as \((2-x)^2 + (0.5+y)^2 = L^2\).
• This approach ensures accurate determination of the pinata’s position along the rope and helps solve for \(\theta\), the rope's angle.

The Pythagorean Theorem's application confirms the lengths and angles involved, forming the base for the trigonometric calculations in static equilibrium.