A box (M = 140 kg) is sitting on a 6.0 m light uniform board (with negligible mass). The board is supported by two pivots one at each end. The box is sitting x = 1.6 m from the left side of the board as shown. What is the magnitude of the normal force produced by the pivot at the left end (na) measured in Newtons? M -6.0 m- B
A box (M = 140 kg) is sitting on a 6.0 m light uniform board (with negligible mass). The board is supported by two pivots one at each end. The box is sitting x = 1.6 m from the left side of the board as shown. What is the magnitude of the normal force produced by the pivot at the left end (na) measured in Newtons? M -6.0 m- B
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Chapter1: Units, Trigonometry. And Vectors
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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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![### Problem Statement:
A box (M = 140 kg) is sitting on a 6.0 m light uniform board (with negligible mass). The board is supported by two pivots, one at each end. The box is sitting \( x = 1.6 \, \text{m} \) from the left side of the board as shown. What is the magnitude of the normal force produced by the pivot at the left end (\( n_A \)) measured in Newtons?
### Diagram Explanation:
The provided diagram depicts a horizontal board of length 6.0 meters. Here are the key elements seen in the diagram:
1. **Board**: A light uniform bar with a length of 6.0 meters.
2. **Box**: Represented with mass \( M \) (140 kg) located 1.6 meters from the left end of the board (\( x = 1.6 \, \text{m} \)).
3. **Pivots**: Two pivots—one at each end of the board.
- Pivot at the left end is labeled as **A**.
- Pivot at the right end is labeled as **B**.
4. **Normal Forces**:
- Upward force due to the left pivot (\( n_A \)).
- Upward force due to the right pivot (\( n_B \)).
### Forces and Equilibrium Analysis:
To find the normal force at the left end (\( n_A \)), use the equilibrium conditions:
1. **Sum of Forces (Vertical Equilibrium)**:
\[ n_A + n_B = M \times g \]
- \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)).
2. **Sum of Torques (Rotational Equilibrium)** about the left pivot (A):
\[ n_B \times 6.0 - M \times g \times 1.6 = 0 \]
- Taking torques about point \( A \).
Solving these equations will yield the values of the normal forces at each pivot.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcde4937d-e883-446c-8941-8d2ade4d99b0%2F6aff8e08-e47e-4709-a6b2-b738d74fb92d%2F145jqcf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A box (M = 140 kg) is sitting on a 6.0 m light uniform board (with negligible mass). The board is supported by two pivots, one at each end. The box is sitting \( x = 1.6 \, \text{m} \) from the left side of the board as shown. What is the magnitude of the normal force produced by the pivot at the left end (\( n_A \)) measured in Newtons?
### Diagram Explanation:
The provided diagram depicts a horizontal board of length 6.0 meters. Here are the key elements seen in the diagram:
1. **Board**: A light uniform bar with a length of 6.0 meters.
2. **Box**: Represented with mass \( M \) (140 kg) located 1.6 meters from the left end of the board (\( x = 1.6 \, \text{m} \)).
3. **Pivots**: Two pivots—one at each end of the board.
- Pivot at the left end is labeled as **A**.
- Pivot at the right end is labeled as **B**.
4. **Normal Forces**:
- Upward force due to the left pivot (\( n_A \)).
- Upward force due to the right pivot (\( n_B \)).
### Forces and Equilibrium Analysis:
To find the normal force at the left end (\( n_A \)), use the equilibrium conditions:
1. **Sum of Forces (Vertical Equilibrium)**:
\[ n_A + n_B = M \times g \]
- \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)).
2. **Sum of Torques (Rotational Equilibrium)** about the left pivot (A):
\[ n_B \times 6.0 - M \times g \times 1.6 = 0 \]
- Taking torques about point \( A \).
Solving these equations will yield the values of the normal forces at each pivot.
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