Static Equilibrium and Vector Component Analysis Notes

Physics Vector Analysis: Static Equilibrium and Component Method

• Scenario Context: A crew is cutting down a large spruce tree to be transported and displayed at Rockefeller Plaza in New York City.
• Physical Setup: Three ropes are attached to the tree trunk, each pulled by crew members exerting specific forces at designated angles.
• Physical State: Prior to being felled, the tree remains in static equilibrium, meaning the vector sum of all forces acting on the tree is zero ($\sum \mathbf{F} = 0$).
• Methodology: The Component Method is used to resolve each applied force vector into rectangular components ($x$ and $y$) to determine the net applied force and the reaction force exerted by the tree.

Applied Force Vector Diagram

• Force Vectors Exerted by Ropes:
  • Vector 1 ($\mathbf{F}_1$): Magnitude of $180\,\text{lbs}$ at an angle of $160^\circ$
  • Vector 2 ($\mathbf{F}_2$): Magnitude of $200\,\text{lbs}$ at an angle of $75^\circ$
  • Vector 3 ($\mathbf{F}_3$): Magnitude of $250\,\text{lbs}$ at an angle of $52.5^\circ$

Component Method Analysis

• Mathematical Formulas for Components:

  • Horizontal Component ($x$-axis): $F_x = F \cos(\theta)$
  • Vertical Component ($y$-axis): $F_y = F \sin(\theta)$

• Force Component Table:

  • Force 1 ($180\,\text{lbs}$ at $160^\circ$):
  • $F_{1x} = 180 \cos(160^\circ) = -169.1\,\text{lbs}$
  • $F_{1y} = 180 \sin(160^\circ) = 61.6\,\text{lbs}$
  • Force 2 ($200\,\text{lbs}$ at $75^\circ$):
  • $F_{2x} = 200 \cos(75^\circ) = 51.8\,\text{lbs}$
  • $F_{2y} = 200 \sin(75^\circ) = 193.0\,\text{lbs}$
  • Force 3 ($250\,\text{lbs}$ at $52.5^\circ$):
  • $F_{3x} = 250 \cos(52.5^\circ) = 152.1\,\text{lbs}$
  • $F_{3y} = 250 \sin(52.5^\circ) = 198.3\,\text{lbs}$

• Summation of Components:

  • Sum of horizontal components ($\sum F_x$):
  • $\sum F_x = -169.1\,\text{lbs} + 51.8\,\text{lbs} + 152.1\,\text{lbs} = 34.8\,\text{lbs}$
  • Sum of vertical components ($\sum F_y$):
  • $\sum F_y = 61.6\,\text{lbs} + 193.0\,\text{lbs} + 198.3\,\text{lbs} = 453.1\,\text{lbs}$

Question 1: Reaction Force Exerted by the Tree

• Problem Statement: While in static equilibrium, what force must the tree exert when pulled by the three forces? What is the direction of this "tree" force?

• Static Equilibrium Condition:

  • An object in static equilibrium experiences zero net force:
  • $\sum \mathbf{F} = \mathbf{F}_{\text{applied}} + \mathbf{F}_{\text{tree}} = 0$
  • Therefore, the force exerted by the tree ($\mathbf{F}_{\text{tree}}$) is equal in magnitude and opposite in direction to the net applied force ($\mathbf{F}_{\text{applied}}$).

• Magnitude Calculation of Net Applied Force:

  • Formula: $F = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$
  • Substitution: $F = \sqrt{(34.8)^2 + (453.1)^2}$
  • Calculated Result: $F = 454.5\,\text{lbs}$

• Direction Angle Calculation of Net Applied Force:

  • Formula: $\theta = \tan^{-1}\left(\frac{\sum F_y}{\sum F_x}\right)$
  • Substitution: $\theta = \tan^{-1}\left(\frac{453.1}{34.8}\right)$
  • Angle: $\theta = 82.9^\circ$

• Tree Reaction Vector ($\mathbf{F}_{\text{tree}}$):

  • Magnitude: $454.5\,\text{lbs}$
  • Direction: Opposite to the resultant pull angle ($82.9^\circ + 180^\circ = 262.9^\circ$)
  • Final Solution: $F_{\text{tree}} = 454.5\,\text{lbs}\text{ @ }262.9^\circ$

Question 2: Direction of Tree Fall

• Problem Statement: What direction will the tree fall?
• Physical Reasoning:
  • When the trunk is cut and the restraining force of the tree is removed, the tree accelerates and falls in the direction of the net applied force vector exerted by the ropes.
• Final Answer:
  • The tree will fall in the direction of the net applied force: $82.9^\circ$ (or approximately $83^\circ$).