## Problem 9A 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 ExplanationThe 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.### SolutionTorque ($ \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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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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Estimation

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Question

## 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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