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Member BD exerts on member ABC a force P directed along line BD. Force P must have a 345-lb horizontal component.

A figure shows a diagonal member B D fixed to an L-shaped member C B A.The diagonal member B D is attached to the L-shaped member A B C at point B. Point C at the top of the L-shaped member and point D of the diagonal member are attached to the horizontal member. Point A is at the bottom left of the L-shaped member. The angle between B D and B C is 35 degrees. Load Q acts vertically down from point A.

Determine the vertical component of the force.

The magnitude of the vertical component of the force is 734.28  Numeric ResponseEdit Unavailable. 734.28 incorrect.lb.

This force is pointing downwards  Correct.



To solve this problem, let's break it down step by step.

### Problem Breakdown:
- **Given:** The horizontal component of the force \( P \) along line BD is 345 lb.
- **Angle between BD and BC:** \( \theta = 35^\circ \).
- **Objective:** Determine the vertical component of the force \( P \).

### Step 1: Analyze the Force Components
Force \( P \) can be broken down into horizontal and vertical components:
\[
P_x = P \cos(\theta)
\]
\[
P_y = P \sin(\theta)
\]

Where:
- \( P_x \) is the horizontal component (345 lb).
- \( P_y \) is the vertical component.

### Step 2: Calculate the Magnitude of Force \( P \)
Given that the horizontal component \( P_x \) is 345 lb:
\[
P_x = P \cos(35^\circ)
\]
So,
\[
P = \frac{P_x}{\cos(35^\circ)} = \frac{345 \text{ lb}}{\cos(35^\circ)}
\]

Calculating \( P \):
\[
P \approx \frac{345 \text{ lb}}{0.8192} \approx 421.25 \text{ lb}
\]

### Step 3: Calculate the Vertical Component \( P_y \)
Now, we find the vertical component:
\[
P_y = P \sin(35^\circ)
\]
\[
P_y \approx 421.25 \text{ lb} \times 0.5736 \approx 241.67 \text{ lb}
\]

### Conclusion:
- The vertical component \( P_y \) should be approximately **241.67 lb**.

### Discrepancy:
You mentioned that the vertical component is **734.28 lb** and marked it incorrect. This discrepancy suggests that perhaps either the angle provided or the understanding of the problem setup might need reassessment, or there could be other forces at play not considered in this setup.

Would you like to revisit the setup or explore any other aspect of this problem?

---

### Relative Questions:
1. How does the angle between BD and BC affect the force components?
2. Can the direction of force P change the vertical component's magnitude?
3. What role does the vertical force Q play in determining P's vertical component?
4. What would happen if the angle \( \theta \) were different?
5. How would the solution change if the horizontal component of P were different?

### Tip:
When resolving forces into components, always ensure that the angles and directions are correctly interpreted relative to the coordinate system.

Member BD exerts on member ABC a force P directed along line BD. Force P must have a 345-lb horizontal component.

A figure shows a diagonal member B D fixed to an L-shaped member C B A.The diagonal member B D is attached to the L-shaped member A B C at point B. Point C at the top of the L-shaped member and point D of the diagonal member are attached to the horizontal member. Point A is at the bottom left of the L-shaped member. The angle between B D and B C is 35 degrees. Load Q acts vertically down from point A.

Determine the vertical component of the force.

The magnitude of the vertical component of the force is 734.28 Numeric ResponseEdit Unavailable. 734.28 incorrect.lb.

This force is pointing downwards Correct.

Learn how to decompose forces using trigonometry by calculating the vertical component of a force P exerted along BD with a 345-lb horizontal force. This problem explores key concepts in vector decomposition, trigonometry, and force components. Suitable for advanced high school students.

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