Solve for the Resultant (Magnitude, Direction, and Location x) for the Nonconcurrent system below. Note: You do not need a y for the location for this problem. 600 N 200 N 45° 2m 4m _2m 100 N
Solve for the Resultant (Magnitude, Direction, and Location x) for the Nonconcurrent system below. Note: You do not need a y for the location for this problem. 600 N 200 N 45° 2m 4m _2m 100 N
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![### Nonconcurrent Force System Analysis
**Problem Statement:**
Solve for the resultant (magnitude, direction, and location \(x\)) for the nonconcurrent system below.
**Note:** You do not need a \(y\) for the location for this problem.
#### Diagram Overview:
The diagram presents a horizontal beam supported on a wall, subjected to three forces:
1. **Force of 600 N:**
- Directed downward and to the right at a 45° angle.
- Applied at the left end of the beam.
2. **Force of 100 N:**
- Directed vertically downwards.
- Applied 2 meters to the right of the left end of the beam.
3. **Force of 200 N:**
- Directed vertically downwards.
- Applied at the right end of the beam.
**Distance Measurements:**
- The leftmost application point of the 600 N force is the reference point.
- The 100 N force is applied 2 meters from the reference point.
- The 200 N force is applied 2 meters from the right application point of the 100 N force, making it 4 meters from the reference point to the application point of the 100 N force.
#### Detailed Descriptions of Forces and Locations:
1. **600 N Force:**
- Acts downward at a 45° angle.
- Applied at the left end (0 meters).
2. **100 N Force:**
- Acts vertically downward.
- Applied 2 meters from the left end.
3. **200 N Force:**
- Acts vertically downward.
- Applied 8 meters from the left end.
For calculations, decompose the 600 N force into its horizontal and vertical components:
- Horizontal Component (600 N \(\times\) cos(45°)) = 424.26 N to the right.
- Vertical Component (600 N \(\times\) sin(45°)) = 424.26 N downwards.
Next, compute the location \(x\) of the resultant using the principle of moments about the reference point (left end):
\[ \sum M_A = 0 \]
By using these diagrams and distance measurements, along with decomposition of forces, the resultant force's magnitude, direction, and location can be solved.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24227ad7-0cee-4cb3-bf8f-36b2c099999b%2F1fe69e70-06c1-4447-8e82-48464a982dbd%2Fli61i6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Nonconcurrent Force System Analysis
**Problem Statement:**
Solve for the resultant (magnitude, direction, and location \(x\)) for the nonconcurrent system below.
**Note:** You do not need a \(y\) for the location for this problem.
#### Diagram Overview:
The diagram presents a horizontal beam supported on a wall, subjected to three forces:
1. **Force of 600 N:**
- Directed downward and to the right at a 45° angle.
- Applied at the left end of the beam.
2. **Force of 100 N:**
- Directed vertically downwards.
- Applied 2 meters to the right of the left end of the beam.
3. **Force of 200 N:**
- Directed vertically downwards.
- Applied at the right end of the beam.
**Distance Measurements:**
- The leftmost application point of the 600 N force is the reference point.
- The 100 N force is applied 2 meters from the reference point.
- The 200 N force is applied 2 meters from the right application point of the 100 N force, making it 4 meters from the reference point to the application point of the 100 N force.
#### Detailed Descriptions of Forces and Locations:
1. **600 N Force:**
- Acts downward at a 45° angle.
- Applied at the left end (0 meters).
2. **100 N Force:**
- Acts vertically downward.
- Applied 2 meters from the left end.
3. **200 N Force:**
- Acts vertically downward.
- Applied 8 meters from the left end.
For calculations, decompose the 600 N force into its horizontal and vertical components:
- Horizontal Component (600 N \(\times\) cos(45°)) = 424.26 N to the right.
- Vertical Component (600 N \(\times\) sin(45°)) = 424.26 N downwards.
Next, compute the location \(x\) of the resultant using the principle of moments about the reference point (left end):
\[ \sum M_A = 0 \]
By using these diagrams and distance measurements, along with decomposition of forces, the resultant force's magnitude, direction, and location can be solved.
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