moments & couples
TRANSCRIPT
ENGINEERING SCIENCE
Chapter 4:Moments and Couples
4.1 Moment of a Force about an Axis
The Moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis. This is different from the tendency for a body to move, or translate, in the direction of the force. In order for a moment to develop, the force must act upon the body in such a manner that the body would begin to twist. This occurs every time a force is applied so that it does not pass through the centroid of the body. A moment is due to a force not having an equal and opposite force directly along it's line of action.
4.1 Moment of a Force about an Axis
Moment of a force formula: M = Fd
Unit: Nm
4.1 Moment of a Force about an Axis
The principle of moment: For a body in equilibrium the sum of the clockwise moments is
equal to the sum of the anticlockwise moments. Σ clockwise moments = Σ anticlockwise moments
F2d2 = F1d1
4.2 Levers
A lever is simply a rod or bar capable of turning about a fixed axis called the fulcrum, which may be a spindle or a knife edge. The lever may be straight or curved or cranked, and the forces acting upon it may be parallel or otherwise.
4.2 Levers
EFDF
D EC
W 5kg
400 mm 180 mm
N
smkg
gmF EE
05.49
)/81.9)(5( 2
A uniform lever is pivoted at its mid point C. A body having a mass of 5 kg is suspended at a point E, 180 mm to the right of C. Calculate the mass to be suspended at a point D, 400 mm to the left of C to maintain the lever in balance.
kgN
g
Fm
mgF
W
N.m.
m).N)(.(F
m).N)(.(m).(F
dFdF
MM
D
D
D
D
EEDD
cc
25.281.9
0725.22
:for mass So,
07252240
1800549
180054940
Example
A force of 25 N is applied to a spanner at an effective length of 140 mm from the centre of a nut. Calculate: the moment of the force applied to the nut, If the nut have a same magnitude of moment as above to be open,
determine the force required if we used spanner with effective length 100mm.
140mm
25N
Solution
A force of 25 N is applied to a spanner at an effective length of 140 mm from the centre of a nut. Calculate:a. the moment of the force applied to the nut,b. If the nut have a same magnitude of moment as above to be open,
determine the force required if we used spanner with effective length 100mm.
140mm
25N
a) Moment = Fd = (25N)(0.14m)
= 3.5Nm
b) Since the moment to open the nut was 3.5Nm and spanner length is 100mm;Moment = Fd3.5Nm = F(0.1m) F = 3.5/0.1 F = 35N
So, the force used for 100mm spanner is bigger than 140mm spanner.
Exercise
The figure show the brake pedal assembly for a car. Determine the force that need to apply by driver to push the brake pedal if the force from the Push Rod FR=1.2kN perpendicular to Brake Pedal
FR
Fp
50mm
200mm
Solution
FR = 1.2kN
Fp
50mm
200mm
o
4.3 Equilibrium and Resultant of Parallel Force
Resultant of parallel force equal to zero when the system in equilibrium.
80N? N
40N 20N
x0.9m
0.6m
4.3 Equilibrium and Resultant of Parallel Force
F1 = 80NF4 = ? N
F2 = 40N F3 = 20N
x0.9m
0.6m
a dcb
4.4 General Principle of Moments
If a body at rest (equilibrium) under the action of several force, the total clockwise moment of the force about any axis is equal to the total anticlockwise moment of the forces about the same axis.
P R Q
x b
a
F
4.4General Principle of Moments
P R Q
x b
a
F
Equating the clockwise and anticlockwise moments at F, we have:
Q(b+x) – Rx – P(a-x) = 0Since;
R = P + Q
Qb + Qx – (P + Q)x – Pa + Px = 0 Qb + Qx – Px - Qx – Pa + Px = 0 Qb – Pa = 0 Qb = Pa
4.5 Reaction on a Horizontal Beam Supported at Two Points
To determine what is upward forces must act on a beam at its supports when the beam is loaded and supported at two points.
5kg
RA RB
2m 1m
4.5 Reaction on a Horizontal Beam Supported at Two Points
5kg
RA RB
2m 1m
4.5 Reaction on a Horizontal Beam Supported at Two Points
8kgRA RB
20kg
A uniform horizontal beam 6m long, rest on two supports A and B, 4m apart, A being at one end of the beam. The mass of the beam is 20kg. If a mass of 8kg be hung from the beam at distance of 1m from A, calculate the reaction of the supports.