214 wikarta kuliah ii

8/10/2019 214 Wikarta Kuliah II

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KULIAH II

MEKANIKA TEKNIK TI

KESEIMBANGAN

PARTIKEL (2D)

OLEH:

ALIEF WIKARTA, ST

JURUSAN TEKNIK MESIN,

FTI

ITS SURABAYA, 2007


2/15

Equilibrium of a Particle (2-D)

Todays Objectives:Students will be able to :

a) Draw a free body diagram

(FBD), and,

b) Apply equations of

equilibrium to solve a 2-D

problem.

For a given cable

strength, what is the

maximum weight

that can be lifted ?


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Free Body Diagram (FBD) (2-D)


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Equations of Equilibrium (2-D)

Since particle A is in equilibrium,the net force at A is zero.

So FAB + FAD + FAC = 0

or F = 0

Or, written in a scalar form,

These are two scalar equations of equilibrium (EofE). They

can be used to solve for up to twounknowns.

Fx= 0 and

Fy= 0

I n general, for a particle in equi li br ium, F = 0 orFxi + Fyj = 0 = 0 i + 0j (A vector equation)


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EXAMPLE

Write the scalar EofE:

+ Fx= TBcos 30 TD = 0

+ Fy = TBsin 30

2.452 kN = 0Solving the second equation gives: TB= 4.90 kN

From the first equation, we get: TD= 4.25 kN

Note : Engine mass = 250 Kg FBD at A


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Problem Solving (2-D)

Given: The car is towed at constant

speed by the 600 N force

and the angle is 25.

Find: The forces in the ropes AB

and AC.

Plan:

1. Draw a FBD for point A.

2. Apply the EofE to solve for the forces

in ropes AB and AC.

F= 600 N

= 25o


7/15

Problem Solving (2-D)

F= 600 N

= 25o3025

600 N

FABFAC

AFBD at point A

Applying the scalar EofE at A, we get;

+ Fx= FACcos 30FABcos 25= 0

+ Fy= -FACsin 30FABsin 25+ 600 = 0

Solving the above equations, we get;

FAB= 634 N

FAC= 664 N


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In a ship-unloading operation, a

15.6 kN automobile is supported bya cable. A rope is tied to the cable

and pulled to center the automobile

over its intended position. What is

the tension in the rope?

SOLUTION:

Construct a free-body diagram for the

particle at the junction of the rope and

cable.

Determine the unknown force

magnitudes.

Example


9/15

EXAMPLE

Given:Sack A weighs 20 N.

and geometry is asshown.

Find: Forces in the cables and

weight of sack B.

Plan:

1. Draw a FBD for Point E.

2. Apply EofE at Point E to solvefor the unknowns (TEG& TEC).

3. Repeat this process at C.


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EXAMPLE (continued)

The scalar EofE are:

+ Fx= TEG sin 30 TECcos 45 = 0

+ Fy= TEGcos 30 TECsin 45 20 N = 0

Solving these two simultaneous equations for the

two unknowns yields:

TEC= 38.6 N

TEG= 54.6 N

A FBD at E should look like the oneto the left. Note the assumed

directions for the two cable tensions.


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EXAMPLE(continued)

Fx= 38.64 cos 45 (4/5) TCD = 0

Fy = (3/5) TCD+ 38.64 sin 45

WB = 0

Solving the first equation and then the second yields

TCD= 34.2 N and WB= 47.8 N .

The scalar EofE are:

Now move on to ring C.

A FBD for C should look

like the one to the left.


12/15

READING QUIZ

1) When a particle is in equilibrium, the sum of forces acting

on it equals ___ . (Choose the most appropriate answer)

A) a constant B) a positive number C) zero

D) a negative number E) an integer.

2) For a frictionless pulley and cable, tensions in the cable(T1and T2) are related as _____ .

A) T1> T2

B) T1= T2C) T1< T2

D) T1= T2sin


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ATTENTION QUIZ

F2

20 N

F1

C

50

3. Using this FBD of Point C, the sum of

forces in the x-direction (FX) is ___ .Use a sign convention of + .

A) F2sin 50 20 = 0

B) F2cos 50 20 = 0

C) F2sin 50 F1 = 0

D) F2cos 50 + 20 = 0


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40 kg

80 kg

5 m

10 m

ab

A

B

C

P

Jika b = 4 m,

tentukan harga P dan jarak a

SOAL TANTANGAN


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TERIMA KASIH

214 wikarta kuliah ii

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