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Transcript 
FORCES AND
MOMENTS
Resolving forces
Forces and moments
Example 1
Drawing to scale
Weight
suspended by two
Draw the perpendicular
ropes
Identify the angles between the forces A and B and the
perpendicular
A
B
55o
Draw the triangle using the angles
20o
35o
55o
70o
A
105o
2000 N
B
20o
2000
Newtons
The length of the
sides of the
triangle represent
the magnitude of
the forces NOT
the length of rope
Using the sine rule (if you know the
angles)
a/sin A = b/Sin B = c/sin C
angle A = 20o angle B = 55o (opposites
to sides a & b)
55
a
o
Angle C = 105o and side c
represents 2000N
105o
2000 N
(c)
b
20o
Using the sine rule (if you know the
angles)
a/sin A = c/sin C therefore
a/sin 20o = 2000/sin105o
a = 2000 x sin 20o/sin105o
55
o
708.17N
b/sin B = c/sin C therefore
b/sin 55o = 2000/sin105o
a
105o
2000 N
(c)
b
a = 2000 x sin 55o/sin105o
1696.1N
20o
Using the cosine rule ( if you know one
angle and two sides)
F2 = 60N
F3
70o
F1 = 30N
Using the cosine rule ( if you know one
angle and two sides)
A2 = B2 + C2 -2BCcosA
(F3)2 = 302 + 602 – 2x60x30x cos110o
F2 = 60N (C)
= 75.7N
F3 (A)
A =110o
F1 = 30N (B)
70o
Vertical and horizontal components of
forces
Sketch the diagram
Fv
can be drawn at the other
end of the sketch
Fv
F
θ
FH
Vertical and horizontal components of
forces
Sketch the diagram
sin θ = Fv/F
Fv
F
θ
FH
Fv
F.sin θ = Fv
cos θ = FH/F
F.cos θ = FH
Restoring force of two forces
F3 is the restoring force of F1 and F2
F1(55N)
F3
25o
Can be drawn
to scale
74.8N
70o F2 (25N)
25o
70o
Restoring force of two forces
F3 is the restoring force of F1
and F2
F1(55N)
F3
70o F2 (25N)
25o
Can be solved by
resolving the
horizontal
components of F1
and F2
Restoring force of two forces
F3 is the restoring force of F1 and F2
F1(55N)
F3
F1v = F1.sin70o
55sin70o
= 51.68N
70o F2 (25N)
25o
F1h = F1.cos70o
55cos70o
= 18.81N
Restoring force of two forces
F1(55N)
F3
F2v = F2.sin25o
25sin25o
= 10.57N
70o F2 (25N)
25o
F2h = F2.cos25o
25cos25o
= 22.66N
Restoring force of two forces
F1(55N)
F3
F3v = F1v + F2v
51.68 +10.57
= 62.25N
70o F2 (25N)
25o
F3h = F1h +F2h
18.81 + 22.66
= 41.47N
Restoring force of two forces
F3
(F3)2 = 62.252 + 41.472
62.25N
(F3)2 = 5594.82
F3 = 74.80N
41.47N
Resultant of two forces
Tan θ =
opposite/adjacent
F3
62.25N
θ
Tan θ = 62.25/41.47
Tan θ = 1.5
41.47N
θ = 56.33o
Direction of F3 = 180 + 56.33 = 236.33o
Moments of force
4m
2m
2N
4N
Total Anticlockwise moments
moments
8Nm
= Total Clockwise
8Nm
Moments of force
3m
4m
2m
2N
2N
4N
Total Anticlockwise moments
8Nm + 4Nm = 12Nm
= Total Clockwise moments
=
12 Nm
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