1. Turning Moments
We know its easier to turn a bolt with a long handled spanner and its easier to close a door pushing at the edge rather than near the hinges.
This is due to the force causing the rotation of the body. This force can be defined as:
The moment of a force about a point is found by multiplying the magnitude of the force by the perpendicular distance from the point to the line of action of the force.
The moment of force P about the point X is F x distance to pivot. (Nm) X P F A d
The moment of force has both a magnitude and direction.

2. A rod AB 5m long has a force of 10N applied to one end
A rod AB 5m long has a force of 10N applied to one end. The rod has a pivot point C 2m from A. A B C
10N 2m 3m
Turning moment about A = F x d
PA = 10 x 2
PA = 20 Nm anticlockwise
Turning moment about B = F x d
PB = 10 x 3
PA = 30 Nm anticlockwise

3. A rod AB 5m long has the following forces applied
A rod AB 5m long has the following forces applied. What is the overall turning moment around B. A B
10N 2m 8N C D
16N
Turning moment about B = (-10 x 5) + (16 x 3) + (-8 x 2) =
= - 18 Nm
There is an anticlockwise turning moment about B of 18Nm

4. Turning moment about A = F x d PA = 20 x 6 cos 40o
The distance has to be the perpendicular distance to the force:
20N
A rod AB 6m long a 20N force applied at B. The rod is inclined at 40o. What is the overall turning moment around A the pivot point. B 6m
40o A
6 cos 40o
Turning moment about A = F x d
PA = 20 x 6 cos 40o
PA = 91.9 Nm clockwise

5. Turning moment about A = F x d PA = 30 sin 50o x 8
You can also calculate the perpendicular force:
30N
A rod AB 8m long has a 30N force applied at an angle of 50o at B. What is the overall turning moment around A the pivot point.
30 sin 50o
50o A 8m B
Turning moment about A = F x d
PA = 30 sin 50o x 8
PA = Nm anticlockwise

6. Horizontal Turning moment PS = 6 x 3 PS = 18 Nm
If positions and Force are given in i and j notation we find the force acting in each direction and add them.
A point S has a position vector of (-3i + 4j)m, a force (6i + 5j)N acting at position T (2i + j)m is applied to it. Find the moment about S.
Horizontal Turning moment S
PS = 6 x 3
Plot this: 5N
PS = 18 Nm 3 6N
Vertical Turning moment 5 T
PS = 5 x 5
PS = 25 Nm
Total PS =
Total PS = 43Nm anticlockwise

7. Turning Moment CouplesF1
A coin can be spun using the forefingers of both hands. With practice the coin can be made to spin on one spot d F2
The weight of the coin acts through its axis and therefore has no effect on the moments. If there is no sideways movements there must be no resultant force acting on the coin.
Resolving Forces:
F1 + F2 = 0
F1 = F2
Take Moment about O the centre of the coin:
O = F x ½d + F x ½d
= F x d
This turning moment will have the same value no matter where it is taken.

8. A turning moment couple has a resultant force of zero and an overall turning moment which is not zero.
2) Does this system reduce to a couple? If so find the moment of the couple 3N 4N 1N O 6m 2m 8m
1) Calculate the moment of the couple shown.
10N
20cm A B
10N
hence no transitory movement
10 – 10 = 0
hence no transitory movement
4 – 3 – 1 = 0
A = 10 x 0.2
= 2Nm anticlockwise
O = (3x2)+(6x(-4))+(8x1)
= 6 –
= -10 Nm
= 10 Nm anticlockwise

9. 3) Two forces of (2i – 5j)N and (-2i + 5j)N act at the points with positions (4i + 3j)m and (2i)m respectively. Show that the forces reduce to a couple and state the moment of the couple which needs to be applies in order to produce equilibrium.
hence no transitory movement
(2i – 5j) + (-2i + 5j) = 0
A = (5x2)+(2x3)+(-5x0)+(-2x0) B 5N 2N = 2N 5N
= 16 Nm clockwise A

10. Replacing Parallel Forces by a Single Force
If the acting forces are parallel but do not form a couple, it is possible to replace the forces with a single parallel force.
It should be noted that if the parallel forces are in the same direction (like) the resultant force with be between the forces. If they are opposite (unlike) the resultant will lie outside the forces.
1) Two like parallel forces of 3N and 8N are 25m apart. Find the magnitude, direction and line of action of the resultant. 3N 8N R
25m
x m A A B B
R = = 11N
A = R x x
A = 8 x 25
200 = 11x
= 200 Nm anticlockwise
x = 18.2m

11. Parallel Forces in Equilibrium
A system of parallel forces is either equivalent to a couple or can be replaced by a single force. If the forces are in equilibrium they are not a couple as turning moment occurs.
For an equilibrium of parallel forces:
The resultant force in any direction must be zero
The sum of moments of the forces at any point must be zero
1) A horizontal beam of length 4m and mass 5kg has a mass of 6kg at one end and a mass of 3kg at the other. Find the position if the beam is balanced.
R = 6g + 5g + 3g = 14g N R
A =(6g x 0)+(5gx2)+(3gx4)
= 14g x x A B
x m
22g = 14g x x 5g 3g 6g
x = 1.57m
4 m

12. 2) A light horizontal beam of length 6m rests of smooth supports at the ends. The beam carries masses of 8kg and 3kg at distances 2m and 5m from A. Find the reaction at each support. P Q
1 m A B
2 m
3 m
A = (8g x 2) + (3gx5) = Q x 6 8g 3g
31g = 6Q
6 m
Q = 31g 6 N
As it is in equilibrium:
B = (3g x 1) + (8gx4) = P x 6
P + Q = 8g + 3g
35g = 6P
31g g 6 = (31+ 35)g 6
P = 35g 6 N
= 66g 6
= 11g as required

13. Non Parallel Forces in Equilibrium
A uniform beam AB of mass 50kg and length 2r m has its lower end resting on rough horizontal ground. It is held in equilibrium at an angle of 15o to the horizontal by a rope attached at B with an angle of 26o to AB. Find the tension in the rope.
Tcos26o
Tsin26o
40o
26o
15o r
50g T A B
Tcos26o and rcos15o can be ignored as they are along the line of the beam thus having no turning effect A
50g x rcos15o
= Tsin26o x 2r
T = 50g x rcos15o sin26o x 2r
rcos15o
T = 50g x cos15o sin26o x 2
15o
50g
T = 539.8N
rsin15o