introduction to mechnics

1. Engineering Mechanics
MEC107
Unit 1 - Introduction to Mechanics

2. Free body diagram (FBD)
A body which is free from all contact surfaces and
is shown with all the forces on it is called the free
body diagram.
Forces: all applied and non-applied forces.

3. Free body diagram (FBD)
❑ The string can have only tension in it, but
cannot have compression in it.
❑ The wall reaction is push, but not a pull on the
body.
❑ The line of reaction should be determined
accurately, but the direction can be assumed.
❑ Positive or Negative direction depends upon
the reaction value.

4. How to draw a free-body diagram?
You can draw a free-body diagram of an object following these 3 steps:
1.Sketch what is happening
2.Determine the forces that act on the object
3.Draw the object in isolation with the forces that act on it

5. Reaction Force

9. A sphere weighing 100 N is tied to a smooth wall by a string as shown in
fig. Find the reaction T and reaction R in the wall.

10. Resolution of Force T in vertical and horizontal direction
Vertical Component - T Cos (15)
Horizontal Component – T Sin (15)
T Cos (15)
T Sin (15) R
100 N
T cos (15) = 100
T = 100/cos(15)
T = 103.52 N
Tsin(15) = R
R = 26.79 N

11. Determine the horizontal force F to be applied to the block weighing
1500N to hold it in position on a smooth inclined plane AB which makes
an angle 30 deg. With the horizontal.

19. The moment of a force
Also known as the turning effect of a
force.
The moment of a force about any pointis
defined as: force x perpendicular
distance from the turning point to the
line of action of the force
moment = F x d
Unit: Newton-metre (Nm)
Moments can be either CLOCKWISE or
ANTICLOCKWISE
Force F exerting an
ANTICLOCKWISE
moment through the
spanner on the nut
.

20. Moment of a Force

23. Varignon’s Principle of Moments
or Law of moments
• If a number of coplanar forces are acting simultaneously on a particle, the
algebraic sum of the moments of all the forces about any point is equal to
the moment of their resultant force about the same point

24. Calculate moment …..?
Taking moment about point O
Mo = 15 x 0.8 (Clockwise)
Mo = 12 N-m (CW)
Taking moment about point O
Mo = 15cos(60) x 0 + 15sin(60) x 0.8
Mo = 12.99 x 0.8
Mo = 10.392 N-m (CW)
15 Sin(60)
15cos(60)

25. Determine the moment of 100 N force acting at B about moment centre A as shown
First step is to resolve the force in x and y direction
i.e. Fx = 100 cos(60)
Fy = 100 sin(60)
Assume Clockwise Moment as “–ve”
Taking Moment about point A
100 cos(60)
100 sin(60)
If, F = 100 N

27. Couple
Two parallel forces equal in magnitude and opposite
in direction and separated by a definite distance are
said to form a couple.

28. Properties of couple
• A couple consists of a pair of the equal and opposite parallel
forces which are separated by definite distance.
• The translator effect of a couple on the body is zero.
• The rotational effect of a couple about any point is constant.
• The couple is rotated through any angle
• The couple is shifted to any other position
• The couple is replaced by another pair of the forces whose
rotational effect is the same.

29. Comparison of Torque and Moment
Sr.No. Torque Moment
01 Torque is a movement force. Moment is a static force.
02 Torque is often presented as
Nm/revolution
Moment is typically presented
as Nm.
03 Tends to be used when there
is an axle or pivot to be turned
around
Tends to be used in
essentially non-rotational
situations, such as analysis of
forces on a beam.
04 Twisting effect produced by
force
Turning effect produced by
force
05

30. Resolution of forces into the force and couple

31. X- and y- Intercepts of resultant

32. 60

34. Types of Supports and its Reaction Force

35. Types of Beams

36. (f) Cantilever, Simply Supported
(Propped)
Types of Beams

38. Types of load on beam
• Concentrated load
• Uniform distributed load
• Uniformly varying load
• Applied moment

43. END
OF
UNIT - 1