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for the system of parallel forces, determine the magnitude and the position (from point A) of the resultant force

The system in the image shows a beam subjected to three loads: 
1. A point load of 900 lb applied at 4 ft from point A.
2. A uniformly distributed load of 60 lb/ft applied over a span of 18 ft (from point A to point D).
3. A point load of 400 lb applied at 18 ft from point A (at point D).

We are tasked with determining the magnitude and position (from point A) of the resultant force acting on the beam.

### Step 1: Calculate the total load from the distributed force
The distributed load is 60 lb/ft over a length of 18 ft, so the total distributed load is:
\[
\text{Total distributed load} = 60 \times 18 = 1080 \, \text{lb}
\]
This distributed load acts at the centroid of the distributed load area, which is at the midpoint of the distribution. Since the distributed load covers 18 ft, the resultant force acts at:
\[
\text{Distance from A} = \frac{18}{2} = 9 \, \text{ft}
\]

### Step 2: Determine the total resultant force
The total resultant force is the sum of all the forces acting on the beam:
\[
R = 900 \, \text{lb} + 1080 \, \text{lb} + 400 \, \text{lb} = 2380 \, \text{lb}
\]

### Step 3: Calculate the moment about point A
We need to find the position of the resultant force by taking moments about point A. The moment of each force about point A is given by multiplying the force by its distance from point A.

- The 900 lb point load acts at 4 ft from A, so its moment is:
  \[
  \text{Moment from 900 lb} = 900 \times 4 = 3600 \, \text{lb-ft}
  \]
  
- The resultant of the distributed load acts at 9 ft from A, so its moment is:
  \[
  \text{Moment from distributed load} = 1080 \times 9 = 9720 \, \text{lb-ft}
  \]
  
- The 400 lb point load acts at 18 ft from A, so its moment is:
  \[
  \text{Moment from 400 lb} = 400 \times 18 = 7200 \, \text{lb-ft}
  \]

The total moment about point A is the sum of all these moments:
\[
\text{Total moment} = 3600 + 9720 + 7200 = 20520 \, \text{lb-ft}
\]

### Step 4: Calculate the position of the resultant force
The position \( x_R \) of the resultant force from point A is found by dividing the total moment by the total resultant force:
\[
x_R = \frac{\text{Total moment}}{\text{Resultant force}} = \frac{20520}{2380} \approx 8.62 \, \text{ft}
\]

### Conclusion
The magnitude of the resultant force is 2380 lb, and the position of the resultant force is approximately 8.62 ft from point A.

---

Do you want more details or have any questions?

Here are some related questions to explore:

1. How would you find the reactions at points \( R_A \) and \( R_C \)?
2. How does the length of the beam affect the position of the resultant force?
3. What changes in the moment calculation if the distributed load is non-uniform?
4. How would the analysis change if an additional point load were added at mid-span?
5. How can shear and moment diagrams help in understanding beam behavior?

**Tip:** Always check the location and type of loads to simplify moment and force calculations!

For the system of parallel forces, determine the magnitude and the position (from point A) of the resultant force.

Learn how to calculate the magnitude and position of the resultant force for a beam subjected to a 900 lb point load, a 60 lb/ft distributed load over 18 ft, and a 400 lb point load. This problem involves using the principle of moments and superposition of forces.

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