A man exerts a force F on the handle of the stationary wheelbarrow at A. The weight of the wheelbarrow along with its load of dirt is 185 lb with center of gravity at G. For the configuration shown, what force F must the man apply at A to make the net moment about the tire contact point B equal to zero?
A man exerts a force F on the handle of the stationary wheelbarrow at A. The weight of the wheelbarrow along with its load of dirt is 185 lb with center of gravity at G. For the configuration shown, what force F must the man apply at A to make the net moment about the tire contact point B equal to zero?
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### Problem Statement:
A man exerts a force \( F \) on the handle of the stationary wheelbarrow at point A. The weight of the wheelbarrow along with its load of dirt is 185 lb with a center of gravity at point G. For the configuration shown, what force \( F \) must the man apply at point A to make the net moment about the tire contact point \( B \) equal to zero?
### Diagram Description:
The diagram shows a wheelbarrow with:
1. A handle extending from point A.
2. A load of dirt placed in the container with its center of gravity at point G.
3. A weight (\( W \)) acting downward at G.
4. A wheel at point B, which is 8 inches from the container's end.
The following dimensions are given:
- The distance between point A and the ground is 24 inches.
- The length from point A to point G along the handle is 38 inches.
- The distance from point A to point B horizontally is 38 inches.
Two forces are acting:
- The force \( F \) at an angle of \( 60^\circ \) from the horizontal applied at point A.
- The weight \( W \) acting vertically downwards through point G.
### Part 1: Component Analysis
Break the force \( F \) into horizontal and vertical components. Enter positive numbers for each component.
#### Incorrect Attempt:
A prompt indicates an incorrect attempt along with the interpretation for further instructions. Below the instructions, there's another diagram that indicates how to resolve the components of force \( F \).
### Diagram Explanation for Part 1:
In the diagram:
- \( F \) is the applied force at 60 degrees to the horizontal.
- \( F_x \) is the horizontal component of \( F \).
- \( F_y \) is the vertical component of \( F \).
To break \( F \) into components:
\[ F_H = F_x = F \cos(60^\circ) = 0 \]
\[ F_V = F_y = F \sin(60^\circ) = 38.9473 F \]
### Answers:
- \( F_H = 0 \)
- \( F_V = 38.9473 \)
This information relates to understanding vector components and equilibrium conditions in static systems, crucial for solving problems in engineering mechanics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F237cab30-be38-4e01-8717-835ae8175bb5%2F27b2dd93-b80d-4afa-a769-b226a2aedd91%2Fsx182yo_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A man exerts a force \( F \) on the handle of the stationary wheelbarrow at point A. The weight of the wheelbarrow along with its load of dirt is 185 lb with a center of gravity at point G. For the configuration shown, what force \( F \) must the man apply at point A to make the net moment about the tire contact point \( B \) equal to zero?
### Diagram Description:
The diagram shows a wheelbarrow with:
1. A handle extending from point A.
2. A load of dirt placed in the container with its center of gravity at point G.
3. A weight (\( W \)) acting downward at G.
4. A wheel at point B, which is 8 inches from the container's end.
The following dimensions are given:
- The distance between point A and the ground is 24 inches.
- The length from point A to point G along the handle is 38 inches.
- The distance from point A to point B horizontally is 38 inches.
Two forces are acting:
- The force \( F \) at an angle of \( 60^\circ \) from the horizontal applied at point A.
- The weight \( W \) acting vertically downwards through point G.
### Part 1: Component Analysis
Break the force \( F \) into horizontal and vertical components. Enter positive numbers for each component.
#### Incorrect Attempt:
A prompt indicates an incorrect attempt along with the interpretation for further instructions. Below the instructions, there's another diagram that indicates how to resolve the components of force \( F \).
### Diagram Explanation for Part 1:
In the diagram:
- \( F \) is the applied force at 60 degrees to the horizontal.
- \( F_x \) is the horizontal component of \( F \).
- \( F_y \) is the vertical component of \( F \).
To break \( F \) into components:
\[ F_H = F_x = F \cos(60^\circ) = 0 \]
\[ F_V = F_y = F \sin(60^\circ) = 38.9473 F \]
### Answers:
- \( F_H = 0 \)
- \( F_V = 38.9473 \)
This information relates to understanding vector components and equilibrium conditions in static systems, crucial for solving problems in engineering mechanics.
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