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Co-planar Forces

December 8, 2016
Published By : Pratik Kataria
Categorised in:

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, $b = {ΣM}/{R_x}$ = ${ΣM}/{ΣF_x}$