MEKANIKA TEKNIK TI
KESEIMBANGAN PARTIKEL (2D)Equilibrium of a Particle (2-D)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 ?
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 two unknowns.
In general, for a particle in equilibrium, ? F = 0 or
?Fx i + ?Fy j = 0 = 0 i + 0 j (A vector equation)
Write the scalar EofE:
+ ? ? Fx = TB cos 30º – TD = 0
+ ? ?Fy = TB sin 30º – 2.452 kN = 0
Solving 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
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.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
? = 25o
FBD at point A
Applying the scalar EofE at A, we get;
+ ? ?Fx = FAC cos 30° – FAB cos 25° = 0
+ ? ?Fy = -FAC sin 30° – FAB sin 25° + 600 = 0
Solving the above equations, we get;
FAB = 634 N
FAC = 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?
Construct a free-body diagram for the particle at the junction of the rope and cable.
Determine the unknown force magnitudes.
Given: Sack A weighs 20 N. and geometry is as shown.Find: Forces in the cables and weight of sack B.1. Draw a FBD for Point E.
2. Apply EofE at Point E to solve for the unknowns (TEG & TEC).
3. Repeat this process at C.
The scalar EofE are:
+ ? ? Fx = TEG sin 30º – TEC cos 45º = 0
+ ? ? Fy = TEG cos 30º – TEC sin 45º – 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 one to the left. Note the assumed directions for the two cable tensions.
? ? ? 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.
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 (T1 and T2) are related as _____ .
A) T1 > T2
B) T1 = T2
C) T1 < T2
D) T1 = T2 sin ?
3. Using this FBD of Point C, the sum of forces in the x-direction (? FX) is ___ . Use a sign convention of + ? .
A) F2 sin 50° – 20 = 0
B) F2 cos 50° – 20 = 0
C) F2 sin 50° – F1 = 0
D) F2 cos 50° + 20 = 0