Statics free body diagram
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Transcript of Statics free body diagram
Statics (MET 2214)Prof. Simin Nasseri
Statics of ParticlesMET 2214
Statics (MET 2214)Prof. Simin Nasseri
Static Equilibrium for a Particle
Objective: To introduce the concept of the free-body diagram for a particle and to show how to solve particle equilibrium problems using the equations of equilibrium.
A particle: An object with inertia (mass) but of negligible dimensions.
A particle at rest: A particle is at rest if originally at rest or has a constant velocity if originally in motion.
Statics (MET 2214)Prof. Simin Nasseri
Equilibrium equations for a particle
A particle is in equilibrium if the resultant of ALL forces acting on the particle is equal to zero.
(Newton’s first law is that a body at rest is not subjected to any unbalanced forces).
Sum of all forces acting on a particle = 0F =∑
Statics (MET 2214)Prof. Simin Nasseri
Equilibrium equations in component form
In a rectangular coordinate system the equilibrium equations can be represented by three scalar equations:
0
0
0
x
y
z
F
F
F
=
=
=
∑∑∑
Statics (MET 2214)Prof. Simin Nasseri
Free-Body Diagram (FBD): To apply equilibrium equations we must account for all
known and unknown forces acting on the particle.
The best way to do this is to draw a free-body diagram of the particle.
FBD: A diagram showing the particle under consideration and all the forces and moments acting on this particle.
This is a sketch that shows the particle “free” from its
surroundings with all the forces acting on it.
Statics (MET 2214)Prof. Simin Nasseri
Parallelogram Law
Copyright of Ohio University
Two forces on a body can be replaced by a single force called the resultant by drawing the diagonal of the parallelogram with sides equivalent to the two forces.
Statics (MET 2214)Prof. Simin Nasseri
Principal of TransmissibilityThe conditions of equilibrium or motion of a body remain unchanged
if a force on the body is replaced by a force of the same magnitude and direction along the line of action of the original force.
Statics (MET 2214)Prof. Simin Nasseri
Mechanical components
String or cable: A mechanical device that can only transmit a tensile force along itself.
Statics (MET 2214)Prof. Simin Nasseri
Mechanical componentsLinear spring: A mechanical device which exerts a force along its
line of action and proportional to its extension (F = kX). K is constant of proportionality which is a measure of stiffness or
strength.
Statics (MET 2214)Prof. Simin Nasseri
Mechanical componentsCables:Cables are assumed to have negligible weight and they cannot stretch.
They can only support tension or pulling (you can’t push on a rope!).
Frictionless pulleys: Pulleys are assumed to be frictionless.
Statics (MET 2214)Prof. Simin Nasseri
Mechanical components
A continuous cable passing over a frictionless pulley must have tension force of a constant magnitude.
The tension force is always directed in the direction of the cable.
For a frictionless pulley in static equilibrium, the tension in the cable is the same on both sides of the pulley.
Statics (MET 2214)Prof. Simin Nasseri
Force typesForce types:Active Forces - tend to set the particle in motion.
Reactive Forces - result from constraints or supports and tend to prevent motion.
Active force
Reactive force Active force
Reactive force
Statics (MET 2214)Prof. Simin Nasseri
Free Body Diagram (FBD)How to draw a Free Body Diagram:
Draw outlined shape - Imagine the particle isolated or cut “free” from its surroundingsShow all forces and moments - Include “active forces” and “reactive forces”. Place each force and couple at the point that it is applied.Identify each force:
Known forces labeled with proper magnitude and direction. Letters used for unknown quantities.
Add any relevant dimensions onto your picture.
Statics (MET 2214)Prof. Simin Nasseri
FBDF.B.D of the ring A:
Statics (MET 2214)Prof. Simin Nasseri
Example 1The sphere has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere, cord CEsphere, cord CE, and the knot at C.the knot at C.
Statics (MET 2214)Prof. Simin Nasseri
SphereThere are two forces acting on the sphere. These are its weight
and the force of cord CE. The weight is: W = 6 kg (9.81 m/s2) = 58.9 N.
Statics (MET 2214)Prof. Simin Nasseri
FBD of sphereThis is the way we show the FBD of the sphere:
FCE
58.9 N
Statics (MET 2214)Prof. Simin Nasseri
Cord CEThere are two forces acting on the cord. These are the force of
the sphere, and the force of the knot. A cord is a tension only member. Newton’s third law applies.
Statics (MET 2214)Prof. Simin Nasseri
FBD of the cord CE
FCE
FEC
C
E
Statics (MET 2214)Prof. Simin Nasseri
Knot at CThere are three forces acting on the knot at C. These are the
force of the cord CBA, and the force of the cord CE, and the force of the spring CD.
Statics (MET 2214)Prof. Simin Nasseri
FBD of the knot at C
FCE
FCBA
FCD
C
60o
Statics (MET 2214)Prof. Simin Nasseri
Example 2Draw the FBD diagram of the ring A:
W= 2.452 KN
Statics (MET 2214)Prof. Simin Nasseri
FBD of the ring A
Is this the FBD of A?
No! this is not the free body diagram of A!
Statics (MET 2214)Prof. Simin Nasseri
FBD of the ring A
Statics (MET 2214)Prof. Simin Nasseri
Example 3Draw the free body diagrams of C and E and the cable CE:
Statics (MET 2214)Prof. Simin Nasseri
FBD of E
Statics (MET 2214)Prof. Simin Nasseri
FBD of C
Statics (MET 2214)Prof. Simin Nasseri
FBD of cable EC
Statics (MET 2214)Prof. Simin Nasseri
Example 4Draw the FBD of ring A.
W=78.5 N
Statics (MET 2214)Prof. Simin Nasseri
FBD of A
Statics (MET 2214)Prof. Simin Nasseri
Part 2
Applying the Equilibrium Equations
Statics (MET 2214)Prof. Simin Nasseri
FBDDraw the free body diagrams:
W
N
W
N
f
Normal force = The force you have when there is a contact between surfaces (the ball is in contact with the ground).Friction force = You have this when the surface in contact is not frictionless and the friction prevents the motion of the object.
30
Statics (MET 2214)Prof. Simin Nasseri
FBD
0
0
x
y
F
F
=
=∑∑
W
N
W
N
f
300
0
x
y
F
F
=
=∑∑
x
y
x
y
Statics (MET 2214)Prof. Simin Nasseri
yFBD
W
N
N
f
N = WN - W.cos 30 = 0f - W.sin 30 = 0
W.cos30
W.sin30
x
x
y
Statics (MET 2214)Prof. Simin Nasseri
Example 2:Determine the tension in cables AB and AD for equilibrium of the 250 kg engine.
FBD of the ring A
Statics (MET 2214)Prof. Simin Nasseri
B
0, cos30 0
0, sin 30 2.452 0
Solving for T :
sin 30 2.452 , 4.90
Subsituting into the first equation:
4.25
x B D
y B
B B
D
F T T
F T kN
T kN T kN
T kN
= − =
= − =
= =
=
∑∑
Solution of Example 2According to the free body diagram of the ring A, we have three forces acting on the ring. The forces TB and TD have unknown magnitudes but known directions. Cable AC exerts a downward force on A equal to:W = (250kg)(9.81m/s2) = 2452N = 2.245KN
TBcos30
TBsin30