2.4.1 Moments of forces
Figure 2.10(a) shows how a spanner is used to turn a nut. The force (F) is not acting directly on the nut but a distance (d) from the axis of the nut. This produces a turning effectcalled the moment of the forceabout the axis of the nut. It is also called the turn- ing moment or couplebut all these terms mean the same thing. The distance (d) is called the moment armor the leverage distance, both terms being interchangeable. Expressed mathematically:
MFd where:
Mmoment of the force (Nm) Fthe applied force (N)
dthe moment arm (leverage distance) (m)
Note: It is essential that the moment arm (d) is measured perpendicularly to the line of action of the force. If the force is inclined as shown in Fig. 2.10(b) the effective length of the spanner is shortened. The equation becomes:
MFde
where:
Mmoment of the force (Nm) Fthe applied force (N)
dethe effective moment arm (leverage distance) (m)
2.4.2 Principle of moments (related terminology)
Fulcrum The pivot point or axis about which rotation takes place or tends to take place.
Moment arm The distance between the line of action of an applied force and the fulcrum point measured at right angles to the line of action of the force.
Clockwise moment Any moment of a force that rotates or tends to rotate a body about its fulcrum in a clockwise direction. Clockwise moments are considered to be positive () as shown in Fig. 2.11.
Anticlockwise moment Any moment of a force that rotates or tends to rotate a body about its fulcrum in an anticlockwise direction. Anticlockwise moments are considered to be negative () as shown in Fig. 2.11.
Resultant moment This is the difference in magnitude between the total clock- wise moments and the total anticlockwise moments.
(b)
Spanner F
90° Eff a
ectiv e length (
dE) Nut
(a)
Nut Spanner
90°
F
d
Length (d)
Figure 2.10 Turning moments.
Anticlockwise ()
Clockwise () Fulcrum
Figure 2.11 Clockwise and anticlockwise moments.
2.4.3 Principle of moments
The algebraic sum of the moments of a number of forces acting about any point is equal to the resultant moment of force about the same point.
Expressed mathematically:
MRM1M2(M3)…Mn where:
MRthe resultant moment of force
M1to Mnany number of moments acting about the same fulcrum point
Remember: The positive moments will be acting in a clockwise direction, whereas the negative moments (such asM3) will be acting in an anticlockwise direction.
2 m
(a)
(b) F1 25 N
FR 15 N
Fulcrum F2 40 N
F2 50 N
2 m
2 m 3 m
Figure 2.12 Application of principle of moments.
Applying the above principles to the lever in Fig. 2.12(a), calculate:
(a) The magnitude of the resultant force.
(b) The magnitude of the resultant moment.
(c) The position of the line of action of the resultant force.
(a) Since the lines of action of the applied forces acting on the lever shown in Fig. 2.12(a) are parallel to each other, the resultant force (FR) can be obtained by simple addition and subtraction:
●Forces F1and F2tend to move the lever upwards. They are acting in the same direc- tion, therefore they can be added together.
●Force F3acts in the opposite direction, therefore it is subtracted from the sum of F1and F2.
Therefore:
FRF1F2F3
25 N40 N50 N 15 N
(b) To find the magnitude of the resultant moment (MR) Clockwise moments ():
M1F1(2 m2 m)25 N4 m100 Nm M2F22 m40 N2 m80 Nm Anticlockwise moments ():
M3F33 m50 N3 m150 Nm Resultant moment:
MRM1M2M3100 Nm80 Nm150 Nm 30 Nm
(c) To combine the results of (a) and (b) in order to find the position of the line of action of the resultant force (FR), refer to Fig. 2.12(b).
MRFRd30 Nm Therefore:
30 Nm15 Nd d30 Nm/15 N2 m
Therefore the forces F1, F2and F3can be replaced by the single force FRacting at a point 2 m from the fulcrum. Since MRis positive the force FRwill move or tend to move the lever in a clockwise direction so the force FRwill act in an upwards direction as shown.
2.4.4 Equilibrium
In the previous section, the moments have not acted in equal and opposite directions.
This has resulted in a residual, resultant force, causing or tending to cause a rotation about a pivot point (fulcrum).
Figure 2.13 shows a simple balance scale in which the masses M1 and M2in the scale pans are of equal magnitude so that the scales balance and the beam is horizontal. Since the force of gravity acts equally on M1and M2, the forces F1and F2are also equal. The reac- tion force (R) equals the sum of the downward forces.
F1 F2
M2
M1
FR Figure 2.13 Simple balance scale.
The forces are equidistant from the fulcrum, so the turning moments are also equal. Since the turning moments act in opposite directions these moments are also in equilibrium.
They balance each other and there is no tendency for the scale-beam to rotate. This can be summarized as follows:
When a body is in a state of equilibrium (balance) under the action of a number of forces, the clockwise moments about any point is equal to the anticlockwise moments about the same point.
Put more elegantly:
The algebraic sum of the moments about any point is zero.
For a body to be in a state of equilibrium the following conditions must exist:
●The algebraic sum of the horizontal forces acting on the body must equal zero, FH0.
●The algebraic sum of the vertical forces acting on the body must equal zero, FV0.
●The algebraic sum of the moments acting on the body must equal zero, M0.
Figure 2.14 shows a lever pivoted at its centre. The line of action of a force F1is 2 m to the left of the pivot and tends to impart an anticlockwise moment. The line of action of force F2is 3 m to the right of the pivot and tends to impart a clockwise moment. If the magni- tude of F2is 20 N, calculate:
1. The magnitude of the force F1required to put the lever into a state of equilibrium.
2. The magnitude of the reaction force (FR) at the pivot.
2 m
F1 F2 20 N
FR 3 m
Figure 2.14 Forces in equilibrium.
For equilibrium:
Clockwise momentsanticlockwise moments (20 N3 m)(F12 m)
F1(20 N3 m)/2 m30 N
The reaction force FR, acting upwards, is equal and opposite to the sum of F1and F2act- ing downwards. Therefore:
RF1F220 N30 N50 N
Figure 2.15 shows a simple beam supported at each end. Since it is in equilibrium there is no tendency for it to rotate and the reaction forces RAand RBat the supports can be cal- culated using the principle of moments. This is achieved by taking moments about each support in turn.
600 N
RA RB
3 m 4 m 3 m
500 N
Figure 2.15 Calculation of reaction forces.
To determine the magnitude of RAtake moments aboutRBin order to eliminate it from the initial calculation so that only one unknown quantity has to be calculated as follows:
Clockwise momentsanticlockwise moments RA10 m(600 N7 m)(500 N3 m) RA10 m4200 Nm1500 Nm RA10 m5700 Nm
RA5700 Nm/10 m RA570 N
Similarly to determine the magnitude of RBtake moments aboutRAin order to eliminate it from the initial calculation so that only one unknown quantity has to be calculated as follows:
Clockwise momentsanticlockwise moments (600 N3 m)(500 N7 m)RB10 m
1800 Nm3500 NmRB10 m 5300 NmRB10 m
RB5300 Nm/10 m RB530 N
For a system of forces in equilibrium the downward forces must equal the upward forces.
Therefore:
RARB600 N500 N 570 N530 N600 N500 N
1100 N1100 N
which is true. Therefore the calculation for RAand RBare correct. Avoid taking the short- cut of calculating on one of the reaction forces and deducting it from the sum of the downward forces. Any error in the initial calculation will be carried forward and no check will be possible.
The above examples ignore the weight of the beam.
The weight of a uniform beam may be considered to be a concentrated force whose line of action can be assumed to act through the centre of the length of the beam.