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Forces and moments. Resolving forces. Forces and moments. Example 1 Drawing to scale. Weight suspended by two ropes. Draw the perpendicular. Identify the angles between the forces A and B and the perpendicular. Draw the triangle using the angles. A. B. 20 o. A. 55 o. 55 o. 70 o. - PowerPoint PPT Presentation
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FORCES AND MOMENTSResolving forces
Forces and moments
Example 1Drawing to scale
70o35o
20o
55o
20o
55o
105o
Draw the perpendicular
Identify the angles between the forces A and B and the perpendicular
2000 Newtons
Weight suspended by two ropes
Draw the triangle using the angles
2000 N
A B
A
B
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)
20o
55o
105o
a
b
2000 N (c)
a/sin A = b/Sin B = c/sin C
angle A = 20o angle B = 55o (opposites to sides a & b)
Angle C = 105o and side c represents 2000N
Using the sine rule (if you know the angles)
20o
55o
105o
a
b
2000 N (c)
a/sin A = c/sin C therefore a/sin 20o = 2000/sin105o
a = 2000 x sin 20o/sin105o
708.17N
b/sin B = c/sin C therefore b/sin 55o = 2000/sin105o
a = 2000 x sin 55o/sin105o
1696.1N
Using the cosine rule ( if you know one angle and two sides)
F2 = 60N
70o
F1 = 30N
F3
Using the cosine rule ( if you know one angle and two sides)
70o
F2 = 60N (C)
F1 = 30N (B)
F3 (A)
A =110o
A2 = B2 + C2 -2BCcosA
(F3)2 = 302 + 602 – 2x60x30x cos110o
= 75.7N
Vertical and horizontal components of forces
FFv
FH
θ
Sketch the diagram
Fv can be drawn at the other
end of the sketch
Vertical and horizontal components of forces
FFv
FH
θ
Sketch the diagram
Fv
sin θ = Fv/F
F.sin θ = Fv
cos θ = FH/F
F.cos θ = FH
Restoring force of two forces
25o
70o
F1(55N)
F2 (25N)
F3
F3 is the restoring force of F1 and F2
25o
70o
Can be drawn to scale
74.8N
Restoring force of two forces
25o
70o
F1(55N)
F2 (25N)
F3
F3 is the restoring force of F1 and F2
Can be solved by resolving the
horizontal components of F1
and F2
Restoring force of two forces
25o
70o
F1(55N)
F2 (25N)
F3
F3 is the restoring force of F1 and F2
F1v = F1.sin70o
55sin70o
= 51.68N
F1h = F1.cos70o
55cos70o
= 18.81N
Restoring force of two forces
25o
70o
F1(55N)
F2 (25N)
F3F2v = F2.sin25o
25sin25o
= 10.57N
F2h = F2.cos25o
25cos25o
= 22.66N
Restoring force of two forces
25o
70o
F1(55N)
F2 (25N)
F3F3v = F1v + F2v
51.68 +10.57= 62.25N
F3h = F1h +F2h18.81 + 22.66
= 41.47N
Restoring force of two forces
F3
62.25N
41.47N
(F3)2 = 62.252 + 41.472
(F3)2 = 5594.82
F3 = 74.80N
Resultant of two forces
F3
62.25N
41.47N
θ
Tan θ = opposite/adjacent
Tan θ = 62.25/41.47
Tan θ = 1.5
θ = 56.33o
Direction of F3 = 180 + 56.33 = 236.33o
Moments of force
2m
4N
4m
2N
Total Anticlockwise moments = Total Clockwise moments
8Nm 8Nm
Moments of force3m4m
2m
4N
2N 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = 12 Nm
