A uniform plank of length 2.00 m and mass 28.5 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 720–N person is d = 0.500 m from the left end.
magnitude of T1
    	  
magnitude of T2
    	  
magnitude of T3
    	 

A person stands on a horizontal 2.00 m long plank a distance d from the left end. The plank is supported by three ropes.
A rope at the right end of the plank goes up and right, making a 40.0° angle with the horizontal and applying a force vector T1 on the plank outward along the rope.
A rope at the left end of the plank goes vertically up, applying a force vector T2 on the plank outward along the rope.
A rope at the left end of the plank goes horizontally left, applying a force vector T3 on the plank outward along the rope.


To find the tension in each rope, we need to use the principles of static equilibrium. This means that the sum of all forces and the sum of all torques (moments) acting on the plank must be zero.

### 1. Sum of Forces in the Vertical Direction
The forces acting in the vertical direction include the weight of the plank, the weight of the person, and the vertical components of the tensions T1 and T2. Let's denote the weights as \( W_{plank} \) and \( W_{person} \).

\[ W_{plank} = mg = 28.5 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 279.3 \text{ N} \]
\[ W_{person} = 720 \text{ N} \]

The vertical component of \( T1 \) is \( T1 \sin(40^\circ) \), and the tension \( T2 \) acts vertically upwards. Therefore, the sum of the vertical forces is:

\[ T2 + T1 \sin(40^\circ) = W_{plank} + W_{person} \]
\[ T2 + T1 \sin(40^\circ) = 279.3 + 720 \]
\[ T2 + T1 \sin(40^\circ) = 999.3 \text{ N} \quad \text{(Equation 1)} \]

### 2. Sum of Forces in the Horizontal Direction
The horizontal components of the forces include \( T1 \cos(40^\circ) \) and \( T3 \):

\[ T3 = T1 \cos(40^\circ) \quad \text{(Equation 2)} \]

### 3. Sum of Torques (Moments)
We choose the pivot point at the left end of the plank to sum torques. The torques due to the tensions \( T1 \) and \( T2 \), the weight of the plank, and the weight of the person are calculated with their respective lever arms.

For \( T1 \) at the right end of the plank (2.00 m):
\[ \text{Torque of } T1 = T1 \sin(40^\circ) \cdot 2.00 \]

For \( T2 \) at the left end of the plank (0 m, so no torque):
\[ \text{Torque of } T2 = 0 \]

For the weight of the plank at its center (1.00 m):
\[ \text{Torque of } W_{plank} = 279.3 \text{ N} \cdot 1.00 \]

For the weight of the person at 0.500 m:
\[ \text{Torque of } W_{person} = 720 \text{ N} \cdot 0.500 \]

Setting the sum of torques around the left end to zero:

\[ T1 \sin(40^\circ) \cdot 2.00 - 279.3 \cdot 1.00 - 720 \cdot 0.500 = 0 \]

\[ 2.00 T1 \sin(40^\circ) = 279.3 + 360 \]
\[ 2.00 T1 \sin(40^\circ) = 639.3 \]

\[ T1 \sin(40^\circ) = 319.65 \]
\[ T1 = \frac{319.65}{\sin(40^\circ)} \]
\[ T1 \approx 497.47 \text{ N} \]

### Solving for \( T2 \) and \( T3 \)

From Equation 1:
\[ T2 + 497.47 \sin(40^\circ) = 999.3 \]
\[ T2 + 319.65 = 999.3 \]
\[ T2 = 999.3 - 319.65 \]
\[ T2 \approx 679.65 \text{ N} \]

From Equation 2:
\[ T3 = 497.47 \cos(40^\circ) \]
\[ T3 \approx 381.1 \text{ N} \]

### Summary
- The magnitude of \( T1 \) is approximately \( 497.47 \text{ N} \).
- The magnitude of \( T2 \) is approximately \( 679.65 \text{ N} \).
- The magnitude of \( T3 \) is approximately \( 381.1 \text{ N} \).

Would you like further details or have any questions?

### Additional Questions:
1. What would be the effect on the tensions if the person stood at the midpoint of the plank?
2. How would the tension values change if the mass of the plank were doubled?
3. What if the angle of \( T1 \) changed to 45 degrees?
4. How would the system change if there were only two supporting ropes?
5. Can we generalize this problem for any position \( d \) of the person?
6. What happens to the tensions if the plank length is increased to 3.00 m?
7. How do we handle the problem if the plank is not uniform?
8. How does the choice of the pivot point affect the torque calculations?

### Tip
Always draw a free body diagram to visualize the forces and their directions before starting the calculations for static equilibrium problems.

Learn how to calculate tensions in three ropes supporting a uniform plank with a person standing on it. Includes detailed calculations using static equilibrium principles and vector components. Suitable for advanced high school physics students.

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