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Today’s 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 ?

Free Body Diagram (FBD) (2-D)

Equations of Equilibrium (2-D)

Since particle A is in equilibrium, the net force at A is zero.
So F_^AB + F_^AD + F_^AC = 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 two unknowns.

F_^x = 0 and  F_^y = 0

In general, for a particle in equilibrium,  F = 0   or
F_^x i + F_^y j   =   0   = 0 i + 0 j    (A vector equation)
EXAMPLE
Write the scalar EofE:

+   F_^x = T_^B cos 30º – T_^D =   0
+  F_^y = T_^B sin 30º – 2.452 kN = 0

Solving the second equation gives: T_^B = 4.90 kN
From the first equation, we get: T_^D = 4.25 kN
Note : Engine mass = 250 Kg

FBD at A

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

θ = 25^o

Problem Solving (2-D)

F = 600 N
θ = 25^o
30°
25°
600 N
F_^AB

F_^AC

A
FBD at point A

Applying the scalar EofE at A, we get;
+  F_^x = F_^AC cos 30° – F_^AB cos 25° = 0
+  F_^y = -F_^AC sin 30° – F_^AB sin 25° + 600 = 0
Solving the above equations, we get;
F_^AB = 634 N
F_^AC = 664 N
In a ship-unloading operation, a 15.6 kN automobile is supported by a 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

EXAMPLE

Given: Sack A weighs 20 N. and geometry is as shown.

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 solve for the unknowns (T_^EG & T_^EC).
3. Repeat this process at C.

EXAMPLE (continued)

The scalar EofE are:
+     F_^x = T_^EG sin 30º – T_^EC cos 45º = 0
+      F_^y = T_^EG cos 30º – T_^EC sin 45º – 20 N = 0

Solving these two simultaneous equations for the two unknowns yields:
T_^EC = 38.6 N
T_^EG = 54.6 N
A FBD at E should look like the one to the left. Note the assumed directions for the two cable tensions.

EXAMPLE (continued)

   F_^x = 38.64 cos 45 – (4/5) T_^CD = 0
   F_^y = (3/5) T_^CD + 38.64 sin 45 – W_^B = 0
Solving the first equation and then the second yields
T_^CD = 34.2 N     and    W_^B = 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.

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 (T_¹ and T_²) are related as _____ .
    A) T_¹ > T_²
    B) T_¹ = T_²
    C) T_¹ < T_²
    D) T_¹ = T_² sin 
ATTENTION QUIZ
F²

20 N
F¹

C
50°
3. Using this FBD of Point C, the sum of forces in the x-direction ( F_^X) is ___ .   Use a sign convention of +  .
    A) F_² sin 50° – 20 = 0
    B) F_² cos 50° – 20 = 0
    C) F_² sin 50° – F_¹   = 0
    D) F_² cos 50° + 20 = 0

Jika b = 4 m,

tentukan harga P dan jarak a

SOAL TANTANGAN

TERIMA KASIH

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Posted in Mekanika Teknik , Teknik Mesin |

Sabtu, 12 Desember 2009|

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