Applied Physics Capstone Project

Welcome to the Capstone Project

In this final lesson, you will perform a system-level analysis of a mechanical structure. You'll synthesize everything you've learned: vector addition, dot products, and cross products.

Welcome to your capstone project. Today, we move beyond simple classroom exercises to perform a full system-level analysis of an industrial sky-crane. We will combine vector decomposition, torque, and work to evaluate the crane's stability and energy efficiency.

• Transition from simple calculations to complex engineering analysis
• Synthesize all vector operations into one cohesive project

The Scenario: Industrial Sky-Crane

We are analyzing a crane with a position vector r (the arm) and a force vector F (the weight of the load).

Imagine you are the lead engineer. The crane's arm, or jib, is represented by vector r. Hanging from it is a steel beam, creating a downward force vector F. To ensure safety, we must calculate the rotational stress and the energy needed to move this load.

• Position vector (r) represents the crane's jib
• Force vector (F) represents the weight of the beam

Phase 1: Vector Decomposition

Resolve the jib vector r and the load vector F into their i and j components. Assume the jib is 12m at 60°.

First, we must define our coordinate system. Let's resolve our vectors into components. For a 12-meter jib at 60 degrees, r becomes 6i plus 10.39j. The 2,000 Newton load acts purely in the negative y direction: 0i minus 2000j.

• Use trigonometry to resolve 2D vectors
• Define the coordinate system with the pivot at (0,0)

Phase 2: Rotational Stability

Calculate the torque (τ) at the pivot point using the cross product:
τ = r × F.

To prevent the crane from tipping, we calculate the torque. Torque is the cross product of r and F. In our 2D plane, this results in a value of negative 12,000 k Newton-meters. The negative sign tells us the crane has a strong clockwise rotation tendency.

• Torque measures the tendency of the crane to rotate
• The cross product results in a vector perpendicular to the plane

Phase 3: Energy and Displacement

Determine the work (W) done by gravity as the crane moves the beam 5m horizontally.

Now, let's look at energy. Suppose the crane moves the beam 5 meters horizontally. Using the dot product, we find that gravity does zero work. Why? Because the downward force of gravity is perfectly perpendicular to the horizontal movement.

• Work is the dot product of force and displacement
• Perpendicular vectors result in zero work

Project Checklist

To successfully complete your report, follow this workflow.

When presenting your findings, follow this four-step workflow. Start with a geometric diagram. Move to the algebraic setup using unit vectors. Show your step-by-step derivations for torque and work. Finally, provide a physical justification—tell us what the numbers mean for the safety of the crane.

• Geometric Diagram (Tip-to-Tail)
• Algebraic Setup (i, j, k)
• Mathematical Derivation
• Physical Justification

Final Challenge: Stability Check

A new load is added. The jib is at r = 8i + 6j and the force is F = 0i - 3000j. Calculate the torque and explain if it's clockwise or counter-clockwise.

Time for your final evaluation. A new load has been rigged. Calculate the torque for the given vectors r and F. Type your calculation and your interpretation of the rotation direction.

• Apply cross product to a new scenario
• Interpret the sign of the result