Engineering Mechanics I: Concurrent Forces - Determining Resultants and Components, High school final essays of Mathematics

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1

Engineering Mechanics I

Statics

SSP

Ch1: Concurrent Coplanar Forces 1

Parallelogram and triangle force

Examples (1)

2

Example 2.1√√

The screw eye is

subjected to two forces

𝑭𝟏 and 𝑭𝟐. Determine

the magnitude and

direction of the

resultant force.

Solution:

4

Cosine Law

Sine Law

Direction 𝝓 of 𝑭𝑅 measured from the horizontal

Use Triangle of Forces

212. 6 N

100 150 2 ( 100 )( 150 )cos 115

2 2



FR   

sin 115

212. 6

sin

150

sin sin 115

(^2)     

F F R

o sin 115 0. 639 39. 8

212. 8

150 sin      

o   39. 8  15  54. 8

St. Point

F 1  100 N o 15

F 2  150 N

65 o

End Point

FR

115 o 

5

Example 2.2√√

Resolve the horizontal

𝟔𝟎𝟎𝐍 force into

components acting along

𝒖 and 𝒗 axes and

determine the

magnitudes of these

components.

7

Example 2.

Determine the

magnitude of the

component force 𝑭 and

the magnitude of the

resultant force 𝑭𝑹 if 𝑭𝑹

is directed along the

positive 𝒚 -axis.

200 N

FR

Sine Law

8

60 o

45 o

St. Point

200 N 30 o

45 o

F FR

End Point

30 o

Triangle Force

245 N

sin 45

sin 60

  F 

F

273 N

sin 45

sin 75

  R 

R F

F

Solution

200 N

FR F

FR

10

Since the magnitude of

𝑭𝟏 is known but the

magnitudes of 𝑭𝑅 and 𝑭 2

are unknown, then 𝑭 2 can

be any vector that is drawn

from the head of 𝑭𝟏and

intersect the line of action

of F R as shown (End Point).

Solution:

F 1  800 N

F 2

FR

Triangle Force



St. Point

F 1  800 N

o 60

F 2

End Point?

11

 The magnitude of 𝑭𝟐𝐦𝐢𝐧 is the

shortest length when its line

of action is perpendicular to

the line of action of 𝑭𝑹 , that

is, 𝜽 = 𝟗𝟎

𝒐

 Then the magnitudes of 𝑭𝑹 and 𝑭𝟐 can

be obtained from the triangle 𝑨𝑩𝑪.

o   90

Triangle Force

 FR ( 800 )cos 60  400 N

 F 2 ( 800 )sin 60  693 N

Solution: F^1 ^800 N

F 2

FR

o 60 x

F 2 min

A

B

C

F 1  800 N

St. Point

o 60

End Point?

FR

F 2

F 2

60 o

6 kN

105  45  60

o

End point

FR

Cosine law

2 6 2 ( 2 )( 6 )cos 105 6. 8 kN

2 2 FR    

Sine law

sin

6

sin 105

R  F

o sin 105 0. 852 58. 46

6. 8

6 sin       

o

13

The triangular of forces is shown.

Solution

45 o

45 o

2 kN

St. point

 

14

F. Problem 2.

Determine the magnitude

of the resultant force and

its direction measured

counterclockwise from

the positive 𝒙 -axis.

o 30

800 N

600 N

y

x

F. Problem 2.6 √√

If force 𝑭 is to have a

component along the 𝒖

axis of 𝑭𝒖 = 𝟔 𝐤𝐍,

determine the magnitude

of 𝑭 and the magnitude

of its component 𝑭𝒗 along

the 𝒗 −axis.

F

u



o 105

o 45

16

End point

The triangular of forces is shown.

Sine law

sin 105

sin 30

F

( 6 ) 3. 11 kN sin 105

sin 30 F  

Sine law

sin 105

6

sin 45



F 

( 6 ) 4. 39 kN sin 105

sin 45 F   17

Solution

F

u

105 o

45 o

St. point

Fu  6 kN F 

60 o

30 o

F

o 45

o 45

105 o

Solution

Cosine law

Sine law

The triangular of forces is shown.

19

500 700 2 ( 500 )( 700 )cos 105 959. 8 N

2 2 F    

sin 105

sin( 90 )

 o sin 105 0. 704 ( 90 ) 44. 25

959. 8

sin( 90  )      

o    90  44. 75  45. 25



90 

o 105

St. point o 700 N 15

o 15

500 N

End point

F

20

Problem 2.6 √√

1- Resolve the force 𝑭𝟐 into

components acting along

the 𝒖 and 𝒗 axes and

determine the magnitudes

of the components.

2- If the resultant force 𝑹 of

𝑭𝟏 and 𝑭𝟐 directed in 𝒖

axis, and 𝑭𝟏 is minimum,

find 𝑹, 𝑭𝟏𝒎𝒊𝒏, and 𝜽.

F 1

F 2  500 N



o 45

o 70 u

