A fish has been hooked and is pulling on the line as shown in the figure below. 25.0⁰ N.m 25.0° 35.0⁰ If the pole makes an angle of 25° with the horizontal, the distance between the point where the line is attached to the pole and the anglers hand is L = 2.00 m, and the fish pulls on the line such that the tension in the line is T = 130 N at an angle 35.0° below the horizontal, determine the magnitude (in N m) of the torque exerted by the fish about an axis perpendicular to the page and passing through the angler's hand.
A fish has been hooked and is pulling on the line as shown in the figure below. 25.0⁰ N.m 25.0° 35.0⁰ If the pole makes an angle of 25° with the horizontal, the distance between the point where the line is attached to the pole and the anglers hand is L = 2.00 m, and the fish pulls on the line such that the tension in the line is T = 130 N at an angle 35.0° below the horizontal, determine the magnitude (in N m) of the torque exerted by the fish about an axis perpendicular to the page and passing through the angler's hand.
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Chapter1: Units, Trigonometry. And Vectors
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In the figure, the fishing pole is positioned at an angle of 25° with the horizontal. The distance \( L = 2.00 \, \text{m} \) is the length from the point where the line is attached to the pole to the angler's hand. The fish exerts a pull on the line, creating a tension \( T = 130 \, \text{N} \) at an angle of 35.0° below the horizontal.
**Objective:**
Determine the magnitude of the torque (in Newton meters) exerted by the fish on the pole. The torque is calculated about an axis perpendicular to the page and passing through the angler's hand.
**Diagram Explanation:**
- The pole is inclined at 25° to the horizontal.
- Tension \( \vec{T} \) in the line acts at a 35.0° angle below the horizontal.
- The horizontal and vertical components of these angles are crucial for calculating the torque.
**Calculation:**
\[
\text{Torque} (\tau) = L \cdot T \cdot \sin(\theta)
\]
Where:
- \( L \) is the length of the pole segment (\( 2.00 \, \text{m} \))
- \( T \) is the line tension (\( 130 \, \text{N} \))
- \( \theta \) is the effective angle between the line of action of the force and the lever arm.
Substitute and calculate the value to find the torque exerted by the fish on the pole.
\(\boxed{\text{N} \cdot \text{m}}\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe7370c0b-05cc-48e6-9b0a-29a272be895a%2Fda501e91-e5f4-40e2-9060-ca82a9188f27%2Fmhdd73_processed.png&w=3840&q=75)
Transcribed Image Text:**Fishing Line Torque Calculation**
A fish has been hooked and is pulling on the line as depicted in the figure below.
In the figure, the fishing pole is positioned at an angle of 25° with the horizontal. The distance \( L = 2.00 \, \text{m} \) is the length from the point where the line is attached to the pole to the angler's hand. The fish exerts a pull on the line, creating a tension \( T = 130 \, \text{N} \) at an angle of 35.0° below the horizontal.
**Objective:**
Determine the magnitude of the torque (in Newton meters) exerted by the fish on the pole. The torque is calculated about an axis perpendicular to the page and passing through the angler's hand.
**Diagram Explanation:**
- The pole is inclined at 25° to the horizontal.
- Tension \( \vec{T} \) in the line acts at a 35.0° angle below the horizontal.
- The horizontal and vertical components of these angles are crucial for calculating the torque.
**Calculation:**
\[
\text{Torque} (\tau) = L \cdot T \cdot \sin(\theta)
\]
Where:
- \( L \) is the length of the pole segment (\( 2.00 \, \text{m} \))
- \( T \) is the line tension (\( 130 \, \text{N} \))
- \( \theta \) is the effective angle between the line of action of the force and the lever arm.
Substitute and calculate the value to find the torque exerted by the fish on the pole.
\(\boxed{\text{N} \cdot \text{m}}\)
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