214 Wikarta Kuliah II
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Transcript of 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
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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
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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
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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.
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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