Questions: I'm sorry, but you didn't provide the content of the question from the given problems (3.39 and 3.40). Could you please provide the specific text of the question(s) you need extracted?

Solution Steps

• There are four forces acting on the circular object: two 30 lb forces and two 50 lb forces.
• The forces are applied tangentially to the circle at different points.

• The radius \( r \) of the circle is given as 3 ft.
• The moment arm for each force is the perpendicular distance from the line of action of the force to the center of the circle, which is the radius \( r \).

• The moment (or torque) \( M \) is calculated using the formula \( M = F \times r \).
• For the 30 lb forces: \( M_{30} = 30 \, \text{lb} \times 3 \, \text{ft} = 90 \, \text{lb-ft} \).
• For the 50 lb forces: \( M_{50} = 50 \, \text{lb} \times 3 \, \text{ft} = 150 \, \text{lb-ft} \).

• The direction of the moment is determined by the right-hand rule.
• The 30 lb force on the left and the 50 lb force on the right create counterclockwise moments.
• The 50 lb force on the left and the 30 lb force on the right create clockwise moments.

• Sum the counterclockwise moments: \( M_{\text{ccw}} = 90 \, \text{lb-ft} + 150 \, \text{lb-ft} = 240 \, \text{lb-ft} \).
• Sum the clockwise moments: \( M_{\text{cw}} = 150 \, \text{lb-ft} + 90 \, \text{lb-ft} = 240 \, \text{lb-ft} \).

Final Answer