Vector Addition by Components — Comprehensive Biomechanics Notes
Scalars vs vectors
- Scalars have magnitude only; they may be negative as numbers, but they do not have direction in space. Examples: work, energy, speed (scalar), etc.
- Vectors have both magnitude and direction; they are essential in biomechanics to account for both size and direction of quantities like force, velocity, and torque.
- Negative quantities can appear as scalars (e.g., negative work indicating energy decrease) and as components of vectors (negative x or y components indicating direction relative to the coordinate axes).
- Distinguish magnitude vs scalar: magnitudes are the positive sizes of a quantity; scalars can be negative and are used as part of vector expressions (e.g., work negative meaning energy decrease).
- Example: work is a scalar quantity defined as the change in energy, and negative work implies energy has decreased: W = riangle E with negative W indicating a loss in energy.
Vector representations and conventions
- Two primary vector expression forms:
- Magnitude and direction (angle of projection).
- Components (x, y, z components in a coordinate system).
- Converting between forms is standard; component form is often preferred for addition because it scales cleanly to any number of vectors.
- Angle of projection conventions:
- Angles are typically measured from the positive x-axis.
- Clockwise rotation is treated as negative; counterclockwise as positive.
- When dealing with torque, clockwise torque is usually negative and counterclockwise is positive.
- The coordinate system setup governs the sign of components and resulting angle.
- In 3D, include a z component; in this course the focus is on the x-y plane for vector addition.
Express vectors in components (example setup)
- Given two coplanar forces, express each in x and y components:
- For a vector with magnitude $F$ and projection angle $ heta$ (from +x axis):
- F_x = F \, \cos\theta
- F_y = F \, \sin\theta
- Example components (from the lecture):
- Jetpack force: magnitude $F_1 = 2500$ N at angle $\theta = 65^{\circ}$
- Anvil force: magnitude $F_2 = 1800$ N directed straight downward (negative y)
- Compute components:
- F{1x} = 2500 \cos 65^{\circ}, \quad F{1y} = 2500 \sin 65^{\circ}
- F{2x} = 0, \quad F{2y} = -1800
- In numerical terms (approximate):
- F_{1x} \approx 2500 \times 0.4226 \approx 1{,}057\ \text{N}
- F_{1y} \approx 2500 \times 0.9063 \approx 2{,}266\ \text{N}
- F{2x} = 0, \quad F{2y} = -1800\ \text{N}
Net force in components and magnitude
- Net force components (sum of x’s and y’s):
- F{\text{net},x} = F{1x} + F_{2x} = 1{,}057 + 0 \approx 1{,}057\ \text{N}
- F{\text{net},y} = F{1y} + F_{2y} = 2{,}266 - 1{,}800 \approx 466\ \text{N}
- Net force in component form:
- \mathbf{F}{\text{net}} = \langle F{\text{net},x}, F_{\text{net},y} \rangle \approx \langle 1057, \ 466 \rangle \ \text{N}
Magnitude and angle of the net force
- Magnitude of the net force:
- |\mathbf{F}{\text{net}}| = \sqrt{F{\text{net},x}^2 + F_{\text{net},y}^2}
= \sqrt{1057^2 + 466^2} \approx 1.15 \times 10^3 \text{ N}
- Angle of projection (relative to +x axis):
- \theta = \tan^{-1}\left(\frac{F{\text{net},y}}{F{\text{net},x}}\right) = \tan^{-1}\left(\frac{466}{1057}\right) \approx 23.8^{\circ}
- Note on quadrant: since both components are positive, the vector lies in Quadrant I (positive x and positive y).
Graphical methods for vector addition
- Tip-to-tail method (graphical, quantitative):
- Arrange vectors end-to-end and draw from the start of the first to the end of the last to obtain the resultant.
- Works for any number of vectors; order (the sequence) does not affect the final resultant for vector addition.
- Parallelogram method (for two vectors):
- Place both vectors tail-to-tail and construct the parallelogram; the diagonal gives the resultant.
- Cross-product note (not for addition):
- The order of operands matters in cross products: for torque calculations, use (\mathbf{r} \times \mathbf{F}) rather than (\mathbf{F} \times \mathbf{r}).
- Law of cosines as an alternative approach (less general, sometimes convenient):
- For two vectors a and b with angle gamma between them (the included angle):
- c^2 = a^2 + b^2 - 2ab \cos \gamma
- Here, c is the magnitude of the resultant vector when the two are arranged head-to-tail in a triangle.
- Parallelogram approach variant uses the angle theta between them tail-to-tail and yields:
- c^2 = a^2 + b^2 + 2ab \cos \theta
- Relationship between gamma and theta: (\gamma = 180^{\circ} - \theta).
- Matrix method (alternative quantitative approach):
- Build a matrix of the components and perform matrix operations to sum vectors; equivalent to summing components directly.
- Takeaway: for most biomechanics problems, addition by components (and using the Pythagorean and arctangent relationships) is the most general and scalable approach.
Applications to biomechanics and torque generation
- Muscular contributions: muscles can pull in different directions; the resultant net force on a joint depends on the vector sum of individual muscle forces.
- Example: pectoralis major has fibers with different directions; activation levels combine to produce a net tendon pull and torque.
- Changing joint posture (e.g., bench press incline/decline) changes the plane of motion and the relative contributions of different muscle fiber directions, altering the net torque produced.
- Key idea: break forces into perpendicular components relative to the lever arm; decompose into perpendicular (torque-producing) and parallel (compressive or joint-closing) components.
- Resultant force analysis helps determine which postures maximize desired torque while managing compressive forces on joints.
Coplanar vectors and the knee pulley analogy (biomechanics example)
- Visualizing a knee sagittal view: the quadriceps tendon pulls on the patella, redirecting force across the knee like a pulley.
- Consequences of the redirection: a compressive force pushing the patella into the femur’s articular cartilage, which is relevant for osteoarthritis and joint pain.
- Practical exercise setup (from the lecture): assume the quadriceps can maintain a constant 300 N force through the range of motion.
- Question posed: as knee flexion increases (more deep into flexion), how does the net force from the two red vectors change (remain the same, increase, or decrease)?
- Students are asked to sketch the vector composition for differing knee angles and discuss how the resultant vector changes.
- Real-world relevance: exercise prescription could mitigate compressive forces by selecting joint angles that reduce net anterior-posterior loading on the patellofemoral joint.
Practical notes and common pitfalls
- Degree vs radian mode on calculators:
- In this course, work in degree mode for angles unless radian-mode is specifically required later.
- Using radians by mistake can yield incorrect results when applying trigonometric functions with degree inputs.
- Degree reference and angle signs:
- Always reference angles to the +x axis; clockwise is negative, counterclockwise is positive.
- Quadrants and inverse trig:
- When using inverse tangent, be mindful of the quadrant of the resulting vector; use quadrant checks or the atan2 function if available.
- Sign conventions for torques:
- Clockwise torques are typically negative; counterclockwise torques are positive; be consistent with the chosen coordinate system.
- Summary of the main workflow for two coplanar vectors:
- Express each vector in components: F{ix} = Fi \cos\thetai, \quad F{iy} = Fi \sin\thetai
- Sum components: F{\text{net},x} = \sum F{ix}, \quad F{\text{net},y} = \sum F{iy}
- Compute magnitude and angle: |\mathbf{F}{\text{net}}| = \sqrt{F{\text{net},x}^2 + F{\text{net},y}^2}, \quad \theta = \tan^{-1}\left(\frac{F{\text{net},y}}{F_{\text{net},x}}\right)
- Open exercise reminder (friday topic): analyze how net force changes with knee flexion given a constant quadriceps force and two red force vectors; sketch and discuss.
- Component form of a vector: \mathbf{F} = \langle Fx, Fy \rangle,\quad Fx = F \cos\theta,\quad Fy = F \sin\theta
- Net components: F{\text{net},x} = \sumi F{ix}, \quad F{\text{net},y} = \sumi F{iy}
- Net magnitude: |\mathbf{F}{\text{net}}| = \sqrt{F{\text{net},x}^2 + F_{\text{net},y}^2}
- Net angle: \theta = \tan^{-1}\left(\frac{F{\text{net},y}}{F{\text{net},x}}\right)
- Law of cosines (two vectors, triangle method):c^2 = a^2 + b^2 - 2ab \cos\gamma\,, where gamma is the angle between the two vectors.
- Law of cosines (parallelogram tail-to-tail, theta):c^2 = a^2 + b^2 + 2ab \cos\theta\,, with theta the angle between the vectors when arranged tail-to-tail; note that (\gamma = 180^{\circ} - \theta).
- Cross product note (torque): \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}\,, order matters for the cross product (but not for simple vector addition).
End-of-lecture prompts
- How does changing the knee angle affect the net force on the patellofemoral joint, given a constant quadriceps force? Sketch and explain using vector addition concepts.
- Consider how aligning muscle fiber directions with the plane of motion affects the torque and potential joint loading in exercises like bench press variations.