### Problem DescriptionA uniform plank of length 2.00 meters and mass 35.0 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 675-N person is \(d = 0.500 \) meters from the left end.**Given:**- Length of plank, \( L = 2.00 \) m- Mass of plank, \( m = 35.0 \) kg- Weight of person, \( W_P = 675 \) N- Distance from left end where person is standing, \( d = 0.500 \) m**To Find:**- Magnitude of tension in each rope ( \( \vec{T_1}, \vec{T_2}, \vec{T_3} \) ).### Tension in Ropes- Magnitude of \( \vec{T_1} \): \(\_\_\_ \) N- Magnitude of \( \vec{T_2} \): \(\_\_\_ \) N- Magnitude of \( \vec{T_3} \): \(\_\_\_ \) N### Diagram ExplanationThe figure provided is a schematic representation that includes the following elements:1. **Plank Dimensions**: The length of the plank is shown to be 2.00 meters.2. **Mass Representation**: The weight of the plank and the additional weight of a person standing 0.500 meters from the left end.3. **Ropes**: - \( \vec{T_1} \): A tension force directed at an angle of 40.0° from the plank. - \( \vec{T_2} \): A vertical tension force on the left side. - \( \vec{T_3} \): A horizontal tension force on the right side.The numerical details and arrows in the diagram illustrate the forces involved and their directions which are essential in solving for the tension in each rope.### Calculation Steps1. **Determine the forces acting on the system**: - The weight of the plank \( W \) acts midway since it's uniform: \( W = mg = 35.0 \text{ kg} \times 9.81 \text{ m/s}^2 = 343.35 \text{ N} \

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A uniform plank of length 2.00 m and mass 35.0 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 675-N person is d = 0.500 m from the left end.

A uniform plank of length 2.00 m and mass 35.0 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 675-N person is d = 0.500 m from the left end.

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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Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and after loading.

Question

### Problem Description

A uniform plank of length 2.00 meters and mass 35.0 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 675-N person is \(d = 0.500 \) meters from the left end.

**Given:**
- Length of plank, \( L = 2.00 \) m
- Mass of plank, \( m = 35.0 \) kg
- Weight of person, \( W_P = 675 \) N
- Distance from left end where person is standing, \( d = 0.500 \) m

**To Find:**
- Magnitude of tension in each rope ( \( \vec{T_1}, \vec{T_2}, \vec{T_3} \) ).

### Tension in Ropes

- Magnitude of \( \vec{T_1} \): \(\_\_\_ \) N
- Magnitude of \( \vec{T_2} \): \(\_\_\_ \) N
- Magnitude of \( \vec{T_3} \): \(\_\_\_ \) N

### Diagram Explanation

The figure provided is a schematic representation that includes the following elements:

1. **Plank Dimensions**: The length of the plank is shown to be 2.00 meters.
2. **Mass Representation**: The weight of the plank and the additional weight of a person standing 0.500 meters from the left end.
3. **Ropes**:
- \( \vec{T_1} \): A tension force directed at an angle of 40.0° from the plank.
- \( \vec{T_2} \): A vertical tension force on the left side.
- \( \vec{T_3} \): A horizontal tension force on the right side.

The numerical details and arrows in the diagram illustrate the forces involved and their directions which are essential in solving for the tension in each rope.

### Calculation Steps

1. **Determine the forces acting on the system**:
- The weight of the plank \( W \) acts midway since it's uniform: \( W = mg = 35.0 \text{ kg} \times 9.81 \text{ m/s}^2 = 343.35 \text{ N} \

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