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Lecture 03 BY Engr Muhammad Usman Mechanical Engineering Department CECOS University

Engineering Mechanice Lecture 05

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Page 1: Engineering Mechanice Lecture 05

Lecture 03

BY

Engr Muhammad Usman

Mechanical Engineering

Department

CECOS University

Page 2: Engineering Mechanice Lecture 05

COURSE OUTLINE

• Fundamental concept and principle of

mechanics, important vector quantity

• Force system i.e concurrent

• Non concurrent and parallel force system

• Resultant of forces

• Moment and couple

• Equilibrium of forces (law & type) concept of

free body diagram

Page 3: Engineering Mechanice Lecture 05

Resultants

Page 4: Engineering Mechanice Lecture 05

Vectors in opposite directions:

6 m s-1 10 m s-1 = 4 m s-1

6 N 10 N = 4 N

Resultant of Two Vectors

Vectors in the same direction:

6 N 4 N = 10 N

6 m

= 10 m

4 m

The resultant is the sum or the combined effect of two vector quantities

Page 5: Engineering Mechanice Lecture 05

Graphical Vector Addition

2. Parallelogram method

8 N

4 N

3 N

3 forces act on an object at the same

time. Fnet is not 15 N because these

forces aren’t working together. But

they’re not completely opposing each

either. So how do find Fnet ? The

answer is to add the vectors ... not their

magnitudes, but the vectors themselves.

There are two basic ways to add

vectors: 1. Tip to tail method

Page 6: Engineering Mechanice Lecture 05

Head to Tail Methodin-line examples

Place the tail of one vector at the

tip of the other. The vector sum

(also called the resultant) is shown

in red. It starts where the black

vector began and goes to the tip of

the blue one. In these cases, the

vector sum represents the net force.

You can only add or subtract

magnitudes when the vectors are

in-line!

16 N

20 N

4 N

20 N

16 N

12 N9 N

9 N

12 N

21 N

Page 7: Engineering Mechanice Lecture 05

Head to Tail – 2 Vectors

5 m

2 m

To add the red and blue displacement vectors first note:

• Vectors can only be added if they are of the same

quantity—in this case, displacement.

• The magnitude of the resultant must be less than

7 m (5 + 2 = 7) and greater than 3 m (5 - 2 = 3).

Interpretation: Walking 5 m in the direction

of the blue vector and then 2 m in the

direction of the black one is equivalent to

walking in the direction of the red vector.

The distance walked this way is the red

vector’s magnitude.

Page 8: Engineering Mechanice Lecture 05

The Parallelogram Law When two vectors are joined

tail to tail

Complete the parallelogram

The resultant is found by drawing the diagonal

When two vectors are joined head to tail

Draw the resultant vector by completing the triangle

Page 9: Engineering Mechanice Lecture 05

Parallelogram Method1. Create parallelogram using

“copies” of the two vectors

2. Draw Resultant vector from

tail of first vector to tip of last

vector.

3. You cannot add more than 2

vectors at a time with this

method, so…how do you add 3

or 4 vectors with this

method???

Note: Opposite sides of a

parallelogram are congruent.

Page 10: Engineering Mechanice Lecture 05

Comparison of Methods

Tip to tail method

Parallelogram method

The resultant has

the same

magnitude and

direction

regardless of the

method used.

Page 11: Engineering Mechanice Lecture 05

Law of Cosines

a b

c

R

A B

Law of Cosines: R2 = A2 + B2 - 2 AB cos Θ

This side is always opposite this angle.

These two sides are repeated.

It matters not which side is called A, B, and R, so long as the two rules above are

followed. This law is like the Pythagorean theorem with a built in correction term of -2

AB cos Θ . This term allows us to work with non-right triangles. Note if Θ = 90, this

term drops out (cos 90 = 0), and we have the normal Pythagorean theorem.

Θ

Page 12: Engineering Mechanice Lecture 05

• Here we see a table being

pulled by a force of 50 N at

a 30º angle to the horizontal

• When resolved we see that

this is the same as pulling

the table up with a force of

25 N and pulling it

horizontally with a force of

43.3 N

Practical Applications

y=25 N

x=43.3 N30º

We can see that it would be more efficient to pull the table with a horizontal force of 50 N

Page 13: Engineering Mechanice Lecture 05

Determine the resultant of the three forces below.

25o

45o

350 N

800 N600 N

60o

y

x

Page 14: Engineering Mechanice Lecture 05

Solution F x = 350 cos 25o + 800 cos 70o - 600 cos 60o

= 317.2 + 273.6 - 300 = 290.8 N

F y = 350 sin 25o + 800 sin 70o + 600 sin 60o

= 147.9 + 751 + 519.6 = 1419.3 N

i.e. F = 290.8 N i + 1419.3 N j

Resultant, F

F N

290 8 1419 3 1449

1419 3

290 878 4

2 2

1 0

. .

tan.

..

F = 1449 N 78.4o 25o

45o

350 N

800 N600 N

60o

y

x

Page 15: Engineering Mechanice Lecture 05

Determine the resultant of the four forces and one

couple which act on the plate shown.

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