Math Problem Statement

As shown, a pipe is anchored to a wall at point A. (Figure 2) During the pipe's installation, several forces are applied to the pipe at different locations. If F1 = 17.7 lb , F2 = 16.5 lb , F3 = 12.6 lb , F4 = 10.1 lb , d1 = 0.400 ft , d2 = 0.750 ft , and d3 = 0.650 ft , what is MRA , the net moment about point A due to these forces? Assume that moments acting counterclockwise about point A are positive whereas moments acting clockwise are negative. Express your answer numerically in pound-feet to three significant figures.

Solution

To determine the net moment MRAM_{RA} about point A, we need to calculate the moments produced by each of the forces F1F_1, F2F_2, F3F_3, and F4F_4 with respect to point A. The moment produced by a force is given by the equation:

M=F×dM = F \times d

Where:

  • FF is the magnitude of the force,
  • dd is the perpendicular distance from point A to the line of action of the force,
  • The direction of the moment (clockwise or counterclockwise) will determine the sign (positive or negative).

Forces and Distances:

  • F1=17.7lbF_1 = 17.7 \, \text{lb} at d1=0.400ftd_1 = 0.400 \, \text{ft}
  • F2=16.5lbF_2 = 16.5 \, \text{lb} at d2=0.750ftd_2 = 0.750 \, \text{ft}
  • F3=12.6lbF_3 = 12.6 \, \text{lb} at d3=0.650ftd_3 = 0.650 \, \text{ft}
  • F4=10.1lbF_4 = 10.1 \, \text{lb}, no distance specified (assumed to act directly at point A, which would contribute zero moment).

Moment Calculation:

  • For F1F_1 at d1d_1, assume it generates a counterclockwise moment (positive): M1=17.7lb×0.400ft=7.08lb-ftM_1 = 17.7 \, \text{lb} \times 0.400 \, \text{ft} = 7.08 \, \text{lb-ft}

  • For F2F_2 at d2d_2, assume it generates a clockwise moment (negative): M2=16.5lb×0.750ft=12.375lb-ftM_2 = -16.5 \, \text{lb} \times 0.750 \, \text{ft} = -12.375 \, \text{lb-ft}

  • For F3F_3 at d3d_3, assume it generates a counterclockwise moment (positive): M3=12.6lb×0.650ft=8.19lb-ftM_3 = 12.6 \, \text{lb} \times 0.650 \, \text{ft} = 8.19 \, \text{lb-ft}

  • For F4F_4, since no distance is provided, we assume the moment is zero: M4=0M_4 = 0

Net Moment:

The net moment about point A is the sum of the individual moments: MRA=M1+M2+M3+M4M_{RA} = M_1 + M_2 + M_3 + M_4 MRA=7.08+(12.375)+8.19+0=2.895lb-ftM_{RA} = 7.08 + (-12.375) + 8.19 + 0 = 2.895 \, \text{lb-ft}

Thus, the net moment about point A is:

MRA=2.90lb-ftM_{RA} = 2.90 \, \text{lb-ft}

Final Answer:

The net moment about point A is 2.90lb-ft\boxed{2.90 \, \text{lb-ft}}.


Do you want details or have any questions? Here are some follow-up questions you might explore:

  1. What are the conditions under which a moment is considered positive or negative?
  2. How would the net moment change if the forces were applied at different angles instead of perpendicular distances?
  3. What would happen if F4F_4 had a specified distance from point A?
  4. How do the units of force and distance affect the calculation of moments?
  5. How can the principle of moments be applied in real-world engineering scenarios?

Tip: When summing moments, always make sure to pay attention to the direction (clockwise or counterclockwise) to ensure the correct sign is applied.

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Math Problem Analysis

Mathematical Concepts

Statics
Moments
Equilibrium

Formulas

Moment = Force × Distance
Net Moment = Sum of Individual Moments

Theorems

Principle of Moments (Varignon's Theorem)

Suitable Grade Level

College-Level Engineering