This problem involves calculating the position of a 100 lb load on a 20 ft uniform pole, such that the man supports twice as much as the boy.

Let’s solve it using the principle of moments (torque), assuming the pole is in equilibrium, which means the sum of forces and moments about any point is zero.

Step 1: Define variables and positions

• Length of the pole: $20 \, \text{ft}$
• Weight of the pole: $80 \, \text{lb}$
• Distance from the end $A$ to the boy: $2 \, \text{ft}$
• Distance from the end $B$ to the man: $5 \, \text{ft}$
• Load weight: $100 \, \text{lb}$

Let:

• $F_b$ be the force exerted by the boy.
• $F_m$ be the force exerted by the man.
• The load is placed at distance $x$ from point $A$.
• The center of mass of the uniform pole is at its midpoint (10 ft from both ends).

Step 2: Forces acting on the pole

The forces acting on the system are:

• The weight of the pole ($80 \, \text{lb}$) acting downward at its center (10 ft from $A$).
• The load ($100 \, \text{lb}$) acting downward at a point $x$ ft from $A$.
• The upward forces from the boy ($F_b$) and the man ($F_m$).

Step 3: Conditions of equilibrium

For equilibrium, the sum of the vertical forces must equal zero, and the sum of the moments about any point must also be zero.

Vertical force equation:

$F_b + F_m = 80 + 100 = 180 \, \text{lb}$

Moment equation:

Let's take moments about point $A$:

• The moment due to the weight of the pole is $80 \times 10 = 800 \, \text{ft-lb}$.
• The moment due to the load is $100 \times x \, \text{ft-lb}$.
• The moment due to the boy's force is $F_b \times 2 \, \text{ft}$.
• The moment due to the man's force is $F_m \times 15 \, \text{ft}$ (since he is 5 ft from the end $B$, which is 15 ft from $A$).

So, the total moment equation about $A$ is: $F_b \times 2 + F_m \times 15 = 80 \times 10 + 100 \times x$ $F_b \times 2 + F_m \times 15 = 800 + 100x$

Step 4: Relation between forces

It is given that the man supports twice as much as the boy: $F_m = 2 F_b$

Step 5: Solve the system of equations

We have two equations:

1. $F_b + F_m = 180$
2. $F_b \times 2 + F_m \times 15 = 800 + 100x$

Substitute $F_m = 2 F_b$ into both equations.

Substitution into the first equation:

$F_b + 2F_b = 180$ $3F_b = 180$ $F_b = 60 \, \text{lb}, \quad F_m = 120 \, \text{lb}$

Substitution into the moment equation:

$60 \times 2 + 120 \times 15 = 800 + 100x$ $120 + 1800 = 800 + 100x$ $1920 = 800 + 100x$ $100x = 1120$ $x = 11.2 \, \text{ft}$

Final Answer:

The load of 100 lb should be placed 11.2 ft from end $A$.

Would you like a more detailed explanation or have any questions?

Related Questions:

1. How does the center of mass of a uniform object affect its balance?
2. What is the principle of moments, and how is it applied in physics?
3. How would the solution change if the boy and the man were at different positions?
4. What happens if the load were heavier than 100 lb?
5. Can this same method be used for non-uniform poles?

Tip: When using moments to solve problems, always select a point to take moments where it simplifies the calculation by eliminating unknown forces.