Lecture 03BY

Engr Muhammad UsmanMechanical Engineering

DepartmentCECOS University

COURSE OUTLINE

Fundamental concept and principle of mechanics, important vector quantityForce system i.e concurrentNon concurrent and parallel force system

• Resultant of forces (Remaining)• Moment and couple (Start)• Equilibrium of forces (law & type) concept of

free body diagram

3

MOMENT

Moment of A Force The rotation or turning effect of a force is called

MOMENT. The moment M of a force F about a fixed point A

is defined as the product of the magnitude of force F and the perpendicular distance d from point A to the line of action of force F.

MA = F x d Where force F is in newtons, NAnd distance d is in meters, mThus moment MA is in newton-meter, Nm.

4

A

d F

Moment Sign Convention

5

Anti-clockwise : + VE(Counter-clockwise)

Clockwise : - VE

Principle of TransmissibilityIf we move (TRANSMIT) the Force P from point A to point B which lies on the Line of Action of Force P, the same effect would be expected.

6

Principle of transmissibility states that a force acting on a rigid body at different points along the force’s line of action will produce the same effect on the body.

Point of Application

Line of action

P

A

BP

Sample Problem # 01• Calculate the moment of the 500 N

force about the point A as shown in the diagram.

7

A1.5 m

500 N

Solution

8

A 1.5 m

500 N

Since the perpendicular distance from the force to the axis point A is 1.5 m, from

MA = F x d

  MA = - 500 x 1.5 = - 750 Nm

= 750 Nm

Sample Problem # 02 • Calculate the moment about point A

caused by the 500 N force as shown in the diagram.

9

A 1.5 m

60o

500 N

Solution:

10

MA = F x d = -500

x AB

= -500 x 1.5 sin 60

= -

649.5 Nm= 649.5 Nm

1.5 m

500 N

Line of action

600

600

d

B

A

A 1.5 m

60o

500 N

Sample Problem # 03 Determine the resulting moment about point A of the system of forces on bar ABC as shown in the diagram.

11

A B C

60 0

500 N 800 N

1.5 m 0.5m

Solution:

12

60 0

AD = (1.5 + 0.5) sin 60= 2.0 sin 60= 1.732 m

MA = (-500 x 1.5) + (-800 x 1.732)

= - 750 – 1385.6 = - 2135.6 Nm

= 2135.6 Nm

A B C

500 800 N1.5 m 0.5 m

A1.5 m 0.5 m

Line of Action

600

D

C

800 N

500 N

Varignon’s Theorem • Varignon’s Theorem states that “ the

moment of a force about any point is equal to the sum of the moments of its components about the same point”.

• To calculate the moment of any force with a

slope or at an angle to the x or y-axis, resolve the force into the Fx and the Fy components, and calculate the sum of the moment of these two force components about the same point.

13

Sample Problem # 04 Employing Varignon’s Theorem.

14

A B C

60 0

500 N 800 N

1.5 m 0.5m

Solution

15

A

MA

= (-500 x 1.5) + (- 800 sin 60 x 2)

= -2135.6 Nm = 2135.6 Nm

B C

60 0

500 N 800 N

1.5 m 0.5m

1.5 m 0.5m600

A

800 N

B C

Fx

Fy = 800 sin 600

500 N

600

Fx

Fy

= 800 cos 600

Sample Problem # 05Determine the resulting moment about point A for the system of forces acting on the plate ABCD as shown in the diagram.

15 cm

8cm

10.38 cm 45 N

20 N

35 N

80o

45o

30o

A B

C

D

16

Solution

17

45 N force: Fx = 45 cos 80 = 7.814 N Fy = 45 sin 80 = 44.32 N

15 cm

8cm

10.38 cm 45 N

20 N35 N

80o

45o

30o

A B

C

D

35 N force: Fx = 35 cos 60 = 17.5 N Fy = 35 sin 60 = 30.31 N

FxFy

45 N

800

Fx

Fy

35 N300

18

20 N force: Fx = 20 cos 45 = 14.14 N Fy = 20 sin 45 = 14.14 N

0.15 m

0.08m

0.1038 m 45 N

20 N

35 N80o

45o60o

A B

C

D

MA = (- 45 sin 80 x 0.1038) + (- 35 cos 60 x 0.08) + (20 cos 45 x 0.08) + (20 sin 45 x 0.15)

= - 4.591 – 1.4 + 1.131 + 2.121 = - 2.74 Nm = 2.74 Nm

FyFx

20 N

450

Problem 2/29

Problem 2/30

Problem 2/31