Co-planar Forces
December 8, 2016
Categorised in: 1st Year Mechanics
Magnitude of Resultant:
$R = √{P^2 + Q^2 + 2PQ cosθ}$
Direction of Resultant:
$tan α = {Q sinθ}/{P + Q cos θ}$
Resultant of two or more forces:
$R = √{(ΣF_x)^2 + (ΣF_y)^2} $
Direction of two or more forces:
$tan θ = {ΣF_y}/{ΣF_x}$
$ΣF_x$ = Algebric sum of all x components (or x component of resultant)
$ΣF_y$ = Algebric sum of all y components (or y component of resultant)
θ = Angle of ‘R’ with x-axis
When an object is inclined on a plane at angle θ
Component along plane -> mg sinθ
Component perpendicular to plane -> mg cosθ
Table to determine sign for sine and cosine:
$\table Quadrant →,I,II,III,IV;Ratio ↓,,,,;sinθ,+,+,-,-;cosθ,+,-,-,-$
Moment:
$M_o = F x d$
F-> Force
d -> perpendicular distance
$M_o$ = 2 Area of triangle
Here, Base of triangle = force
Height of triangle = perpendicular distance
Sign Convention:
Clockwise moment: Negative
Anti-clockwise moment: Positive
Varignon’s Theorem (Law of Moments):
Perpendicular distance $x = {ΣM}/{R}$
M -> Moments
R -> Resultant
Horizontal and vertical distance of resultant
Horizontal, $a = {ΣM}/{R_y}$ = ${ΣM}/{ΣF_y}$
Vertical, $b = {ΣM}/{R_x}$ = ${ΣM}/{ΣF_x}$
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