9. A honey extractor is a centrifuge that spins honey out of comb. Less expensive models are hand-cranked. If a 200 N force is exerted on the crank handle as shown (the center of the circle is the rotation axis), then how much torque is applied to the crank? 200 N 15 cm 1150
9. A honey extractor is a centrifuge that spins honey out of comb. Less expensive models are hand-cranked. If a 200 N force is exerted on the crank handle as shown (the center of the circle is the rotation axis), then how much torque is applied to the crank? 200 N 15 cm 1150
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![## Problem 9
A honey extractor is a centrifuge that spins honey out of comb. Less expensive models are hand-cranked. If a 200 N force is exerted on the crank handle as shown (the center of the circle is the rotation axis), then how much torque is applied to the crank?
### Diagram Explanation
The diagram provided shows a circular crank handle with the following details:
- The length of the crank handle (radius) is 15 cm.
- A force of 200 N is applied at an angle of 115° from the handle.
### Solution
Torque (\( \tau \)) is calculated using the formula:
\[ \tau = r \cdot F \cdot \sin(\theta) \]
Where:
- \( r \) is the radius of the crank (15 cm or 0.15 m)
- \( F \) is the force applied (200 N)
- \( \theta \) is the angle between the force and the lever arm (115°)
Let's plug the values into the equation:
\[ \tau = 0.15 \, m \cdot 200 \, N \cdot \sin(115°) \]
Calculating \( \sin(115°) \):
\[ \sin(115°) \approx 0.906307787 \]
Now, calculate the torque:
\[ \tau \approx 0.15 \, m \cdot 200 \, N \cdot 0.906307787 \]
\[ \tau \approx 27.1892 \, Nm \]
Therefore, the torque applied to the crank is approximately \( 27.19 \, Nm \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1bcd67a-5279-440a-b724-241f4d0f844f%2F6f5e499a-ff69-4720-a841-360dfbe4a09f%2Fvbf58oe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem 9
A honey extractor is a centrifuge that spins honey out of comb. Less expensive models are hand-cranked. If a 200 N force is exerted on the crank handle as shown (the center of the circle is the rotation axis), then how much torque is applied to the crank?
### Diagram Explanation
The diagram provided shows a circular crank handle with the following details:
- The length of the crank handle (radius) is 15 cm.
- A force of 200 N is applied at an angle of 115° from the handle.
### Solution
Torque (\( \tau \)) is calculated using the formula:
\[ \tau = r \cdot F \cdot \sin(\theta) \]
Where:
- \( r \) is the radius of the crank (15 cm or 0.15 m)
- \( F \) is the force applied (200 N)
- \( \theta \) is the angle between the force and the lever arm (115°)
Let's plug the values into the equation:
\[ \tau = 0.15 \, m \cdot 200 \, N \cdot \sin(115°) \]
Calculating \( \sin(115°) \):
\[ \sin(115°) \approx 0.906307787 \]
Now, calculate the torque:
\[ \tau \approx 0.15 \, m \cdot 200 \, N \cdot 0.906307787 \]
\[ \tau \approx 27.1892 \, Nm \]
Therefore, the torque applied to the crank is approximately \( 27.19 \, Nm \).
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