1/7/25 - 1 Intro to Statics
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1.1 Newton’s Laws of Motion
o First law – an object will remain at rest or in uniform motion in a straight line
unless acted upon by an external force.
o Second law – F = ma
o Third law – for every action, there is an equal and opposite reaction.
1.2 Units
o
Force
Mass
Acc.
SI
Newton (N)
kilogram (kg)
m/s2
English
pound (lbf or #f)
pound (lbm or #m)
ft/s2
o Weight is a force, a is the gravitational constant (9.81 m/s 2 or 1 ft/s2)
1.3 Forces
o Point forces, forces that act at a single point
o Distributed forces, forces that are spread out over a line, area, or volume
o Body forces, distributed forces acting on the volume of a body
o Loads are forces an object must support to perform its function
o Reaction forces, or reactions, are forces that constrain an object in
equilibrium
o Internal forces are forces that hold the parts of an object together
1/9/25 - 2 Forces and Other Vectors
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2.1 Vectors
o Vectors have direction and magnitude (F)
2.2 One-Dimensional Vectors
o Adding multiple vectors finds the resultant vector
o You can also subtract multiple vectors
o Multiplying or dividing a vector changes just the magnitude of the vector
2.3 2D Coordinate Systems and Vectors
o P = (x, y)
o F = <Fx, Fy>
o P = (r; θ)
o Rectangular to polar points
▪ r = √(x2 + y2)
▪ θ = tan-1(y/x)
▪ P = (r; θ)
o Polar to rectangular points
▪ x = rcosθ
▪ y = rsinθ
▪ P = (x, y)
o Rectangular to polar forces
▪ A = √(Ax2 + Ay2)
▪ θ = tan-1(x/y)
▪ A = (A; θ)
o Polar to rectangular forces
▪ Ax = Acosθ
▪ Ay = Asinθ
1/14/25 - 2 Forces and Other Vectors
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2.4 3D Coordinate Systems and Vectors
o P = (x, y, z)
o F = <Fx, Fy, Fz>
o |F| = √(Fx2 + Fy2 + Fz2)
o Polar Coordinates
▪ cosθx = Fx / |F|
▪ cosθy = Fy / |F|
▪ cosθz = Fz / |F|
o Spherical Coordinates
▪ F = (F; θ; ϕ)
▪ F’ = Fsinϕ
▪ Fx = F’cosθ = Fsinϕcosθ
▪ Fy = F’sinθ = Fsinϕsinθ
▪ Fz = Fcosϕ
2.5 Unit Vectors
o i points in +x direction
o j points in +y direction
o k points in +z direction
o F = |F| * ̂F
̂ = F / |F|
o F
2.6 Vector Addition
o Triangle Rule – Place tail of one vector to tip of another vector, then draw
resultant from first vector’s tail to second vector's’ tip (Law of Sines)
o Parallelogram Rule – Place both vector’s tails at origin, then complete
parallelogram with parallel vectors, the resultant is equal to the diagonal
from tails to opposite corner (Law of Cosines)
o FR = Σ(Fxi) + Σ(Fyj) + Σ(Fzk)
1/16/25 - 2 Forces and Other Vectors
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2.7 Dot Products
A * B = AxBx + AyBy + AzBz
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o A * B = |A||B|cosθ
o A * B = A(Bcosθ) = A * ||projAB||
o ||projAB|| = (A * B) / |A|
o projAB = ||projAB|| * A
o BT = B – projAB
2.8 Cross Products
o A * B = |A||B|sinθu, u = -k
o A * B = -B * A
o A * B = (AyBz - AzBy)i - (AxBz - AzBx)j+ (AxBy - AyBx)k
1/21/25 - 3 Equilibrium of Particles
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3.1 Equilibrium
o ΣF = 0
1/23/25 - Moments, Part 1
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M = F * d⟂; units of N-m or ft-lb
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Direction is clockwise or counterclockwise, force acts perpendicular to rotation
Vector is clockwise or counterclockwise, scalar is -z or +z
Varignon’s Theorem states that the moment of a force equals the sum of the
moments of its components, F∥ does not produce a moment
Cross products: M = r x F; -M = F x r
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1/28/25 - Moments, Part 2 (3D)
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M1 = r1 x F1, M2 = r2 x F2, M = M1 + M2
M = (r x F) * u
1/30/25 - Couples and Static Equivalence
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M = F * d⟂
ΣM = Σ(r x F) + ΣMc
2/4/25 - Rigid Body Equilibrium
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Static equilibrium: ΣF = 0, ΣM = 0
Engaged: provides a non-zero reaction
Available: present but zero reaction
2/11/25 - Determinacy and Stability
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Determinate means it’s possible to find unknowns, indeterminate means it’s not
possible
Stable means supports restrain a body from moving at any load, unstable means a
body can move
2/13/25 - Trusses and Method of Joints
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Trusses are made of multiple members and are arranged in triangular sub-units and
attached at ends; the members carry only axial forces
Planar trusses are 2D structures that have only two-force members; the joints are
frictionless pins and typically have three reaction forces
Method of joints solves for unknown forces in members and treats joints as
particles; start with a known force and solved forces can be applied to other joints
2/18/25 - Methods of Sections
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Method of sections solves for unknown forces in select truss members by making
imaginary cuts; cut through no more than 3 unknowns
2/20/25 - Zero-Force Members and Frames and Machines
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If two non-collinear members meet at an unloaded joint, then both are zero-force
members
If three forces (interaction, reaction, or applied forces) meet at a joint and two are
collinear, then the third is a zero-force member
Frames are made of multiple members that support loads (rigid, stationary) and at
least one member is multi-force
Machines are made of multiple members with parts designed move and at least on
member is multi-force, there is an input and output force
3/6/25 - Shear and Moment Diagrammes
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Shear diagram
o dV/dx = w(x)
o ΔV = ∫(w(x)dx)
Moment diagram
o dM/dx = V(x)
o ΔM = ∫(V(x)dx)
3/18/25 - Friction, Slipping, and Tipping
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Dry friction is a force that opposes motion between two solid surfaces in contact;
distributed force modeled as concentrated
Types of friction
o Static: prevents motion
o Kinetic: allows but hinders motion
Modelling friction: parallel (friction) and perpendicular (normal) forces at the
surface; created by force P
Phases of friction: static with no impending motion, static with impending motion,
kinetic
Friction at impending motion
o F = μs N
o tanΦ = (F / N) = μs
Slipping: applied force overcomes friction; need to know μs or Φ
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Tipping: normal force is beyond object; locate N at opposite corner to calculate,
don’t need μs for this
3/20/25 - Wedges and Screws
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Wedges: friction occurs on inclined and flat surfaces; always under compression
Screws: inclined plane in helical shape; input force is moment; used for lifting
objects; almost always right-handed thread pattern
o Applied force opposed impending motion: M = Wr*tan(Φs + α)
o Applied force supports impending motion (self-locking): M’ = Wr*tan(Φs - α)
o Applied force supports impending motion (unwind with load): M’’ = Wr*tan(α
- Φs)
o Applied force supports impending motion (impending motion): M’’’ = 0
3/25/25 - Screws and Flexible Belts
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Contact angle is the angle of chord on contact of rope to pulley; must be in radians
ΣFx = 0
o dN = (dT / μ)
ΣFy = 0
o dN = Tdθ
(T+ / T-) = eμβ
3/27/25 - Center of Gravity/Mass and Centroid Calculations
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Center of gravity is the object’s weight concentrated at one point
o Weighted average: x = ΣxiWi / ΣWi
Centroid pertains to geometric center
o Weighted average: x = ΣxiAi / ΣAi
4/1/25 - Centroids Using Integration
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A = ∫(dA)
o x = ∫(xeldA) / ∫(dA) = Qy / A
o y = ∫(yeldA) / ∫(dA) = Qx / A
o Qx = ∫(yeldA)
o Qy = ∫(xeldA)
4/3/25 - Distributed Loads and Fluid Statics
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Distributed loads: can be represented as concentrated loads at the centroid
Fluid statics: pressure varies with height, fluid depth imparts linear distributed
force, pressure felt in all directions
o P = ρgh (Pa)
o Pa = 1N / m2
4/8/25 - Moment of Inertia
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Area moment of inertia: geometric property of the cross-section of a beam, relates
to its ability to bend
o First moment of area
▪ Qx = ∫(yeldA)
▪ Qy = ∫(xeldA)
o Second moment of area
▪ Ix = ∫(y2dA)
▪ Iy = ∫(x2dA)
Parallel axis theorem: the moment of inertia about any parallel axis can be
calculated from the moment of inertia about the centroidal axis
o I = I + Ad2
4/10/25 - Polar Moment of Inertia, Radius of Gyration, and Product of Inertia
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Polar moment of inertia: moment of inertia about perpendicular axis, applied in
torsional stress
o Jo = ∫(r2dA) = Ix + Iy
Radius of gyration
o kx = √(Ix / A), ky = √(Iy / A), ko = √(Jo / A)
Product of inertia: may be +, -, or 0, will be 0 if symmetrical about x- or y-axis
o Ixy = ∫(xydA)
o Ixy = Ix’y’ + Axy
4/15/25 - Exam 2
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Shear and moment diagramme
o Shear is the derivative of moment
Friction (1 belt/friction and 1 screw)
o Ff = μsN
o tanΦ = (F / N) = μs
o (T+ / T-) = eμβ
o M = Wrtan(Φs + α)
Centroid
Moment of inertia