Of course! Let's break down the problem step-by-step and find the location where the 20 kg weight should be attached to satisfy the given conditions.
### Step-by-Step Solution:
1. Define the given values:
- Length of the pole = 7 meters
- Weight of the pole = 10 kg
- Position of the boy from one end = 2 meters
- Position of the man from the other end = 3 meters
- Weight to be attached = 20 kg
2. Convert the man's position to a consistent reference:
- Position of the man from the same end as the boy:
Since the pole is 7 meters long and the man is 3 meters from the opposite end, the man's position from the reference end (same as the boy's) is:
[tex]\( 7 - 3 = 4 \)[/tex] meters
3. Understand the torque equilibrium condition:
- For the pole to remain in equilibrium, the sum of the torques about any point must be zero. We will take moments about the position where the boy is standing.
4. Locate the center of mass of the pole:
- Since the pole is uniform, its center of mass is at its midpoint. For a 7-meter pole, the center of mass is at:
[tex]\( \frac{7}{2} = 3.5 \)[/tex] meters from either end.
5. Calculate the torque due to the pole's weight:
- The distance from the boy's position (2 meters from one end) to the pole's center of mass (3.5 meters from either end) is:
[tex]\( 3.5 - 2 = 1.5 \)[/tex] meters
- The torque due to the pole around the boy's position is given by:
[tex]\( 10 \text{ kg} \times 1.5 \text{ meters} \)[/tex]
6. Define the torque due to the 20 kg weight:
- Let [tex]\( x \)[/tex] be the distance from the boy's position where the 20 kg weight is attached.
7. Torque equilibrium condition:
- The man's force is supposed to be three times the boy's force.
- Thus, for equilibrium:
[tex]\[
(\text{Torque due to pole}) + (\text{Torque due to 20 kg weight}) = 3 \times (\text{Weight supported by boy}) \times (\text{Distance between boy and man})
\][/tex]
- The distance between the boy and the man is:
[tex]\( 4 - 2 = 2 \)[/tex] meters
8. Solve for [tex]\( x \)[/tex]:
- Using the known values in the torque equilibrium equation, we set up:
[tex]\[
(10 \times 1.5) + (20 \times x) = 3 \times (10/2 + 20) \times 2
\][/tex]
9. Equation solutions:
- Simplifying this, we get:
[tex]\[
15 + 20x = 3 \times (5 + 20) \times 2
\][/tex]
[tex]\[
15 + 20x = 3 \times 25 \times 2
\][/tex]
[tex]\[
15 + 20x = 150
\][/tex]
[tex]\[
20x = 135
\][/tex]
[tex]\[
x = \frac{135}{20} = 6.75 \text{ meters}
\][/tex]
### Final Result:
The 20 kg weight must be attached at a position of 6.75 meters from the end where the boy is standing. This ensures that the man supports three times as much weight as the boy.
