statics (met 2214) prof. simin nasseri statics of particles met 2214

Statics (MET 2214)Prof. Simin Nasseri

Statics of Particles

MET 2214

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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.



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



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



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.



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.



Principal of Transmissibility

The 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.



Mechanical components

String or cable: A mechanical device that can only transmit a tensile force along itself.




Linear 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.



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.




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.



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



Free Body Diagram (FBD)How to draw a Free Body Diagram:

Draw outlined shape - Imagine the particle isolated or cut “free” from its surroundings

Show 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.



FBD



FBDF.B.D of the ring A:



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 the knot at C.knot at C.



Sphere

There 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.



FBD of sphere

This is the way we show the FBD of the sphere:

FCE

58.9 N



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.



FBD of the cord CE

FCE

FEC

C

E



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.



FBD of the knot at C

FCE

FCBA

FCD

C

60o



Example 2

Draw the FBD diagram of the ring A:

W= 2.452 KN



FBD of the ring A

Is this the FBD of A?

No! this is not the free body diagram of A!



FBD of the ring A



Example 3

Draw the free body diagrams of C and E and the cable CE:



FBD of E



FBD of C



FBD of cable EC



Example 4

Draw the FBD of ring A.

W=78.5 N



FBD of A



Part 2

Applying the Equilibrium Equations



FBD

Draw 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



FBD

0

0

x

y

F

F

W

N

W

N

f

300

0

x

y

F

F

x

y

x

y



yFBD

W

N

N

f

N = WN - W.cos 30 = 0f - W.sin 30 = 0

W.cos30

W.sin30

x

x

y



Example 2:

Determine the tension in cables AB and AD for equilibrium of the 250 kg engine.

FBD of the ring A



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 2

According 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


statics (met 2214) prof. simin nasseri statics of particles met 2214

Documents

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