College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Calculating the Total Net Torque on a Rod
#### Problem Statement
What is the total net torque on the rod?
#### Diagram Explanation
- The diagram represents a rod with three applied forces.
- The rod has a total length of 6 meters.
- Forces acting on different points of the rod and at different angles are illustrated.
##### Forces and Distances:
1. **Force 250 N:**
- Applied at a point 3 meters from the left end of the rod.
- Acts at an angle of 45 degrees downward to the rod horizontal axis.
2. **Force 65 N:**
- Applied at the right end of the rod.
- Acts at an angle of 30 degrees upwards to the rod horizontal axis.
3. **Force 180 N:**
- Applied horizontally to the right at the end of the rod.
#### Calculation of Torque
The torque (\( \tau \)) produced by a force is calculated using the formula:
\[ \tau = r \times F \times \sin(\theta) \]
where:
- \( r \) is the distance from the pivot point to the point where the force is applied,
- \( F \) is the magnitude of the force,
- \( \theta \) is the angle between the force vector and the lever arm.
##### For Force 250 N:
\[ \tau_{250} = 3 \text{ m} \times 250 \text{ N} \times \sin(45^\circ) \]
\[ \tau_{250} = 3 \times 250 \times \frac{\sqrt{2}}{2} \]
\[ \tau_{250} = 3 \times 250 \times 0.707 \]
\[ \tau_{250} = 530.25 \text{ Nm (clockwise)} \]
##### For Force 65 N:
\[ \tau_{65} = 6 \text{ m} \times 65 \text{ N} \times \sin(30^\circ) \]
\[ \tau_{65} = 6 \times 65 \times 0.5 \]
\[ \tau_{65} = 195 \text{ Nm (counterclockwise)} \]
##### For Force 180 N:
(Note: This force is horizontal. As it acts parallel to the rod, it does not contribute to the torque.)
\[ \tau_{180}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ec4e4dd-380a-451c-aeb3-9cf66efaad71%2F19b04911-7e6b-4d6d-aaee-d97147f4aab1%2Fx87pzo5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating the Total Net Torque on a Rod
#### Problem Statement
What is the total net torque on the rod?
#### Diagram Explanation
- The diagram represents a rod with three applied forces.
- The rod has a total length of 6 meters.
- Forces acting on different points of the rod and at different angles are illustrated.
##### Forces and Distances:
1. **Force 250 N:**
- Applied at a point 3 meters from the left end of the rod.
- Acts at an angle of 45 degrees downward to the rod horizontal axis.
2. **Force 65 N:**
- Applied at the right end of the rod.
- Acts at an angle of 30 degrees upwards to the rod horizontal axis.
3. **Force 180 N:**
- Applied horizontally to the right at the end of the rod.
#### Calculation of Torque
The torque (\( \tau \)) produced by a force is calculated using the formula:
\[ \tau = r \times F \times \sin(\theta) \]
where:
- \( r \) is the distance from the pivot point to the point where the force is applied,
- \( F \) is the magnitude of the force,
- \( \theta \) is the angle between the force vector and the lever arm.
##### For Force 250 N:
\[ \tau_{250} = 3 \text{ m} \times 250 \text{ N} \times \sin(45^\circ) \]
\[ \tau_{250} = 3 \times 250 \times \frac{\sqrt{2}}{2} \]
\[ \tau_{250} = 3 \times 250 \times 0.707 \]
\[ \tau_{250} = 530.25 \text{ Nm (clockwise)} \]
##### For Force 65 N:
\[ \tau_{65} = 6 \text{ m} \times 65 \text{ N} \times \sin(30^\circ) \]
\[ \tau_{65} = 6 \times 65 \times 0.5 \]
\[ \tau_{65} = 195 \text{ Nm (counterclockwise)} \]
##### For Force 180 N:
(Note: This force is horizontal. As it acts parallel to the rod, it does not contribute to the torque.)
\[ \tau_{180}
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