1. MOMENT OF A FORCE ABOUT A POINT
The tendency of a body to rotate about an axis passing through a specific point O when acted upon by a force (sometimes called a torque).

2. APPLICATIONS
A torque or moment of 12 N · m is required to rotate the wheel. Which one of the two grips of the wheel above will require less force to rotate the wheel?

3. Couple Moment
The couple-moment is known as a free vector, meaning that it can be moved anywhere in space without changing its meaning.

4. 3.14 Equivalent Couples
100mm
150mm
135 N
90 N
Figure shows three couples which act successively on the same rectangular box. As seen in the
preceding section, the only motion a couple can impart to a rigid body is a rotation.
Since each of the three couples shown has the same moment M ( same direction and same magnitude
M = 135 N. m ), we can expect the three couples to have the same effect on the box.

5. MOMENT OF A FORCE ABOUT A POINT
Magnitude of a moment
Mo = F d N.m
Mo = Magnitude of the moment of F around point O
d = Perpendicular distance from O to the line of action
of F

7. DIRECTION OF MOMENT OF A FORCE
Moment produces a rotation.
Direction determined by using the Right-Hand Rule.
The thumb points along the moment axis and the other fingers are curled following the sense of rotation.
Could be ‘clockwise’ (CW) or ‘anti/counter clockwise’ (CCW).

8. Direction of Moment
Choose the convenient sense of rotation for each analysis.

9. Calculating moment
Scalar Analysis
Mo = F d
Mo = F (r sin θ)

10. Sample problem 3.1
A 450N vertical force is applied to the end of a lever
which is attached to a shaft at O.
Determine
the moment of the 450N force about O
(b) the horizontal force applied at A which creates the
same moment about O
(c) the smallest force applied at A which creates the
(d) how far from the shaft a 1100N vertical force must act
to create the same moment about O
(e) whether any one of the forces obtained in parts b,c and
d is equivalent to the original force.

11. Solution
Moment about O
The perpendicular distance from O to the line
of action of the 450N force is
d = (0.6m) cos 60˚ = 0.3 m
The magnitude of the moment about O of the
450N force is
Mo = Fd = (450 N) (0.3m) = 135 N.m
Since the force tends to rotate the lever clockwise
about O, the moment will be represented by a
vector Mo perpendicular to the plane of the figure and
pointing into the paper.We express this fact by writing.
Mo = 135 N.m

12. Solution
b. Horizontal Force
In this case, we have
d = (0.6m) sin 60˚ = 0.52 m
Since the moment about O must be 135N.m,
we write
Mo = Fd
135 N.m = F (0.52 m)
F = 260 N
F= 260 N

13. Solution Mo = Fd 135 N.m = F (0.6 m) F = 225 N F= 225 N 30˚
c. Smallest Force
Since Mo = Fd, the smallest value of F occurs
when d is maximum. We choose the force perpendicular
to OA and note that d = 0.6m, thus
Mo = Fd
135 N.m = F (0.6 m)
F = 225 N
F= 225 N
30˚

14. In this case Mo = Fd yields
Solution
d N Vertical Force
In this case Mo = Fd yields
135 N.m = (1100 N)d
d = 0.12 m
OB cos 60˚ = d
OB = d
cos 60˚
OB = 0.24 m
e. None of the forces considered in parts b,c and d is equivalent to
the original 450N force. Although they have the same moment about O,
they have different x and y components.
In other words, although each forces tends to rotate the shaft in the same
manner, each causes the lever to pull on the shaft in a different way.

15. Sample problem 3. 2 A force of 800 N acts on as bracket as shown
Sample problem 3.2 A force of 800 N acts on as bracket as shown. Determine the moment of a force about B.

16. MB = xFy + yFx = 0.2 (800 sin 60) + 0.16 (800 cos 60) = 138.564 + 64
Solution
0.2
0.16
Fx= 800 cos 60
Fy = 800 sin 60 B A
MB = xFy + yFx
= 0.2 (800 sin 60)
(800 cos 60) =
= Nm

17. A 135N force acts on the end of the 0.9 m lever as shown.
Sample problems 3.3
135N
A 135N force acts on the end of the 0.9 m
lever as shown.
Determine the moment of the force about O

18. Mo = -Q (0.9m) = -(46.2N)(0.9m) = -41.6N.m
Solution
135N
The force is replaced by two components, one component
P in the direction of OA and one component Q perpendicular
to OA.
Since O is on the line of action of P, the moment of P about O
is zero and the moment of the 135N force reduces to the moment
of Q, which is clockwise and, thus, is represented by a negative
scalar.
Q = (135N) sin 20˚ = 46.2 N
Mo = -Q (0.9m) = -(46.2N)(0.9m) = -41.6N.m
Since the value obtained for the scalar Mo is negative,
the moment Mo points into the paper. We write
Mo = 41.6 N.m

19. EXAMPLE 1
Given: A 400 N force is applied to the frame and  = 20°.
Find: The moment of the force at A.
1) Resolve the force along x and y axes.
2) Determine MA using scalar analysis.

20. + MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m = 1160 N·m
Solution
+  Fy = sin 20° N
+  Fx = cos 20° N
+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
= N·m

21. Problems 3.1
A 90-N Force is applied to the control rod AB as shown. Knowing that the length of the rod is 225mm,
determine the moment of the force about point B by resolving the force into components along AB and in a
direction perpendicular to AB.

22. Solution

23. Problems 3.2
A 90-N Force is applied to the control rod AB as shown. Knowing that the length of the rod is 225mm,
determine the moment of the force about point B by resolving the force into horizontal and
vertical components.

25. Solution 3.3
3.4 m
4.8 m
A 3N force P is applied to the lever which controls the auger of a snow blower. Determine the moment of P about
A when αis equal to 30º.

26. Equivalent System
Sliding VECTOR
MOVING A FORCE ITS LINE OF ACTION ON

27. MOVING A FORCE OF ITS LINE OF ACTION
Free VECTOR
MOVING A FORCE OF ITS LINE OF ACTION
OFF

28. Resultant of a Force and Couple System

29. If R and MRP are perpendicular to each other,
3.20 Further Reduction of a Force and Couple System
If R and MRP are perpendicular to each other,
the force-couple system at P can be further reduced to a
single resultant force.
Will be the case for system consisting either :
(a) concurrent force
(b) coplanar force
(c) parallel forces
(a) concurrent force

30. (b) Coplanar force systems
(c) Parallel force systems

31. FRd = F1d1 + F2d2 + F3d3 The moment of the resultant
force about the grip
the moment of all the forces
about the grip
FRd = F1d1 + F2d2 + F3d3

32. Sample problems 3.8
150N
600N
100N
250N
A 4.80m long beam is subjected to the forces shown.
Reduce the given system of forces to
an equivalent force couple system at A
b) single force or resultant

33. solution
use FRd = F1d1 + F2d2 + F3d3 ;

34. solution
150N
600N
100N
250N F3 F2 F1
FR = = N
= N
SMA = F1d F2d F3d3
= - 600(1.6m) + 100(2.8m) – 250(4.8)
= Nm

35. FR = 600 [N] , d = 3.13 [m] FRd = SMA (600) d = 1880 Nm d = 1880/600
solution A B d
FR = 600 [N]
FRd = SMA
(600) d = 1880 Nm
d = 1880/600
= 3.13 m
Thus, the single force equivalent to the given system;
FR = 600 [N] , d = 3.13 [m]

36. Solution For equivalence, F : - 16.4 – 16.4 – 14 = - R R = 46.8 N
0.25 m
0.85 m d
16.4 N
16.4 N
14 N L R = A B C D D
2.1 m
For equivalence,
F : – 16.4 – 14 = - R
R = 46.8 N
: - (0.25)(16.4) – (1.10)(16.4) – ( d)(14) = - L (46.8)
d = 46.8 L »Eq 1

37. Example 3 (cont)
For d = m, substitute the value of d in Eq 1,
(0.625) = 46.8 L
L = m
The resultant passes through a point m to the right of D
(b) For L = 1.05 m, substitute the value of L in Eq. 1,
d = 46.8 (1.05)
d = m

38. THE END