Problem 5 - College Physics

Question

\(\bullet\) An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope (Figure 5.38 ). The rope will break if the tension in it exceeds \(2.50 \times 10^{4} \mathrm{N} .\) Our hero's mass is 90.0 \(\mathrm{kg}\) . (a) If the angle \(\theta\) is \(10.0^{\circ}\) , find the tension in the rope. Start with a free-body diagram of the archaeologist. (b) What is the small- est value the angle \(\theta\) can have if the rope is not to break?

Step-by-Step Solution

Answer

(a) Tension is about 2530 N. (b) Minimum angle θ is approximately 10.1°.

Step 1: Draw the Free-Body Diagram

First, sketch the scenario. Draw the archaeologist suspended in the middle of the rope, with the tension forces in the rope pulling up along the rope at an angle \( \theta \) from the horizontal on each side. Also, show the gravitational force acting downward, equal to the archaeologist's weight: \( W = mg \).

Step 2: Analyze Forces Vertically

Consider the vertical components of the tension forces, which must balance the weight of the archaeologist. The vertical component of the tension on each side is given by \( T \sin(\theta) \). Since there are two sides, the total vertical tension is \( 2T \sin(\theta) \). Set this equal to the weight: \[ 2T \sin(\theta) = mg \] where \( m = 90.0 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \).

Step 3: Solve for Tension in the Rope

Rearrange the equation from Step 2 to solve for the tension \( T \): \[ T = \frac{mg}{2\sin(\theta)} \] Substitute the values: \[ T = \frac{(90.0)(9.8)}{2\sin(10.0^{\circ})} \] Calculate \( T \) to find the tension in the rope.

Step 4: Calculate the Critical Angle

To find the smallest angle \( \theta \) that will not cause the rope to break, set the tension \( T \) equal to the maximum tension the rope can handle, \( 2.50 \times 10^{4} \text{ N} \), and solve for \( \theta \):\[ 2.50 \times 10^{4} = \frac{(90.0)(9.8)}{2\sin(\theta)} \]Rearrange to solve for \( \theta \):\[ \sin(\theta) = \frac{(90.0)(9.8)}{2 \times 2.50 \times 10^{4}} \]Calculate \( \theta \) to find the smallest angle that prevents the rope from breaking.

Key Concepts

Free-Body Diagrams

A Free-Body Diagram (FBD) is a crucial starting point in solving physics problems that involve forces. Imagine our brave archaeologist hanging at the center of a rope stretched between two cliffs. To better understand how forces interact, we draw a diagram focusing solely on the archaeologist, representing the forces acting upon him without any surrounding details or the rope's specifics.
In our scenario, there are three key forces at play:

• The gravitational force acting vertically downward, representing the weight of the archaeologist which can be calculated using the product of mass (\( m \)), and gravitational acceleration (\( g \)). This is expressed as \( W = mg \).
• The tension in the rope on the left pointing up and towards the left, at an angle \( \theta \) from the horizontal.
• The tension in the rope on the right pointing up and towards the right, at the same angle \( \theta \).

Drawing these vectors accurately in the FBD helps in visualizing how forces balance out or result in motion. The angles and directions are essential to understanding how these forces come into play, especially when breaking them into components for analysis.

Tension in Rope

Understanding the concept of tension is pivotal when analyzing ropes and strings in physics. Tension is essentially the force that is transmitted through a rope when it is pulled tight by forces acting from opposite ends.
In the case of our archaeologist, the rope tension must counteract his weight when he pauses midway. This tension involves a vertical component that supports the weight and a horizontal component that keeps the rope taut.
We analyze the vertical component of the tension, as this is responsible for supporting the weight of the archaeologist. If the tensions on both sides are termed \( T \), their vertical components are \( T \sin(\theta) \) each. Therefore, the combined contribution from both sides is \( 2T \sin(\theta) \), which must equal the archaeologist's weight \((mg)\) to maintain equilibrium. Solving \( T \) for specified parameters gives insight into whether the rope can sustain the archaeologist's weight without breaking.

Forces in Equilibrium

Forces in equilibrium occur when all the forces acting on an object result in no net force, meaning the object remains at rest or continues to move at a constant velocity. In our scenario, the archaeologist resting at the midpoint of the rope is a classic example of equilibrium. Here, the forces should balance perfectly to ensure he remains stationary.

The forces include:

• The downward force representing gravity, given as \( W = mg \).
• The upward tension forces on each side of the rope, addressed in physics through their vertical components, \( 2T \sin(\theta) \).

Equilibrium is achieved when these forces satisfy the equation \( mg = 2T \sin(\theta) \), indicating that the summation of upward forces equals the downward gravitational force. It's essential to evaluate each component accurately, ensuring no external forces alter the balance, as any imbalance could result in motion or, worse, the breaking of the rope.

Trigonometric Analysis

Trigonometric concepts are indispensable when dealing with forces at angles. In many physics problems, resolving forces into their components along the horizontal and vertical axes is necessary for accurate analysis. This is where trigonometry shines.

For our problem, the angle \( \theta \) plays a significant role in determining how much of the tension is used to counteract the archaeologist’s weight. The relationship between the angle and the tension components is defined as:

• The vertical component: \( T \sin(\theta) \), responsible for supporting the weight.
• The horizontal component: \( T \cos(\theta) \), which maintains tension along the rope.

The trigonometric function \( \sin \) helps in calculating the vertical proportion while \( \cos \) would help in other contexts like horizontal analysis.
For a breaking point analysis, solving to find \( \theta \) when tension \( T \) is maximum requires evaluating \( \sin(\theta) = \frac{mg}{2T} \), utilizing these trigonometric relationships to ensure the calculations effectively prevent the rope from surpassing its tensile limits.