STATICS - Anasayfa

Equilibrium of a Particle3

STATICSAssist. Prof. Dr. Cenk Üstündağ

Condition for the Equilibrium of a Particle

• Particle at equilibrium if- At rest- Moving at constant a constant velocity

• Newton’s first law of motion∑F = 0

where ∑F is the vector sum of all the forces acting on the particle

Condition for the Equilibrium of a Particle

• Newton’s second law of motion∑F = ma

• When the force fulfill Newton's first law of motion, ma = 0

a = 0therefore, the particle is moving in constant velocity or at rest

The Free-Body Diagram

• Best representation of all the unknown forces (∑F) which acts on a body

• A sketch showing the particle “free” from the surroundings with all the forces acting on it

• Consider two common connections in this subject –– Spring– Cables and Pulleys


• Spring– Linear elastic spring: change in length is directly

proportional to the force acting on it– spring constant or stiffness k: defines the elasticity

of the spring– Magnitude of force when spring

is elongated or compressed F = ks


• Cables and Pulley– Cables (or cords) are assumed negligible weight and

cannot stretch– Tension always acts in the direction of the cable– Tension force must have a constant magnitude for

equilibrium– For any angle θ, the cable

is subjected to a constant tension T


Procedure for Drawing a FBD 1. Draw outlined shape2. Show all the forces

- Active forces: particle in motion- Reactive forces: constraints that prevent motion

3. Identify each forces- Known forces with proper magnitude and direction- Letters used to represent magnitude and directions

Example

The sphere has a mass of 6kg and is supported. Draw a free-body diagram of the sphere, the cord CE and the knot at C.

Solution

FBD at SphereTwo forces acting, weight and the force on cord CE. Weight of 6kg (9.81m/s2) = 58.9N

Cord CETwo forces acting: sphere and knotNewton’s 3rd Law: FCE is equal but oppositeFCE and FEC pull the cord in tensionFor equilibrium, FCE = FEC

Solution

FBD at Knot3 forces acting: cord CBA, cord CE and spring CD Important to know that the weight of the sphere does not act directly on the knot but subjected to by the cord CE

Coplanar Systems

• A particle is subjected to coplanar forces in the x-y plane

• For equilibrium ∑Fx = 0∑Fy = 0

• Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero

Coplanar Systems

• Procedure for Analysis1. Free-Body Diagram

- Establish the x, y axes- Label all the unknown and known forces

2. Equations of Equilibrium- Apply F = ks to find spring force - When negative result force is the reserve- Apply the equations of equilibrium

∑Fx = 0 ∑Fy = 0

Example

Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is l’AB = 0.4m, and the spring has a stiffness of kAB = 300N/m.

Solution

FBD at Point AThree forces acting, force by cable AC, force in spring AB and weight of the lamp.If force on cable AB is known, stretch of the spring is found by F = ks. +→ ∑Fx = 0; TAB – TAC cos30º = 0+↑ ∑Fy = 0; TABsin30º – 78.5N = 0Solving, TAC = 157.0kNTAB = 136.0kN

Solution

TAB = kABsAB; 136.0N = 300N/m(sAB)sAB = 0.453m

For stretched length, lAB = l’AB+ sAB

lAB = 0.4m + 0.453m= 0.853m

For horizontal distance BC, 2m = lACcos30° + 0.853mlAC = 1.32m

Force System Resultants4

STATICSAssist. Prof. Dr. Cenk Üstündağ

Moment of a Force – Scalar Formation

When a force is applied to a body it will produce a tendencyfor the body to rotate about a point that is not on the line ofaction of the force. This tendency to rotate is sometimescalled a torque, but most often it is called the moment of aforce or simply the moment.

If a force is applied to the handle ofthe wrench it will tend to turn thebolt about point O (or the z axis).The magnitude of the moment isdirectly proportional to themagnitude of F and theperpendicular distance or momentarm d.


As shown, d is the perpendicular distance from point O to the line of action of the force.

In 2-D, the direction of MO is either clockwise (CW) orcounter-clockwise (CCW), depending on the tendency forrotation.

In a 2-D case, the magnitude of the moment is MO = F d


Often it is easier to determineMO by using the components ofF as shown.Then MO = (FY a) – (FX b). Note the different signs on theterms! The typical sign convention for a moment in 2-D isthat counter-clockwise is considered positive. We candetermine the direction of rotation by imagining the bodypinned at O and deciding which way the body would rotatebecause of the force.

For example, MO = F d and the direction is counter-clockwise.

Fa

b

dO

abO

F

F x

F y


Resultant Moment • Resultant moment, MRO = moments of all the forces

MRO = ∑Fd

Example

For each case, determine the moment of the force about point O.

Solution

Line of action is extended as a dashed line to establish moment arm d.Tendency to rotate is indicated and the orbit is shown as a colored curl.

)(.200)2)(100()( CWmNmNMa o

Solution

)(.5.37)75.0)(50()( CWmNmNMb o

)(.229)30cos24)(40()( CWmNmmNMc o

Solution

)(.4.42)45sin1)(60()( CCWmNmNMd o

)(.0.21)14)(7()( CCWmkNmmkNMe o

Example

Force F acts at the end of the angle bracket in figure. Determine the moment of the force about point O .

Solution

Solution on whiteboard

Moment of a Couple

The moment of a couple is defined as

MO = Fd (using a scalar analysis)

A couple is defined as twoparallel forces with the samemagnitude but opposite indirection separated by aperpendicular distance “d.”

Moment of a Couple

The net external effect of a couple is thatthe net force equals zero and themagnitude of the net moment equals F·d.

Moment of a Couple

Moments due to couples can be addedtogether using the same rules as addingany vectors.

Since the moment of a couple dependsonly on the distance between the forces,the moment of a couple is a free vector.It can be moved anywhere on the bodyand have the same external effect on thebody.

Moment of a Couple

Equivalent Couples• 2 couples are equivalent if they produce the same

moment• Forces of equal couples lie on the same plane or plane

parallel to one another

Moment of a Couple

Resultant Couple Moment• Couple moments are free vectors and may be applied

to any point P and added vectorially• For resultant moment of two couples at point P,

MR = M1 + M2

Example

1) Add the two couples to find the resultant couple.

2) Equate the net moment to 1.5 kNm clockwise to find F.

Given: Two couples act on the beam with the geometry shown.

Find: The magnitude of F so that the resultant couple moment is 1.5 kNmclockwise.

Plan:

Solution

Solution:

The net moment is equal to:

+ M = – F (0.9) + (2) (0.3)

= – 0.9 F + 0.6

– 1.5 kNm = – 0.9 F + 0.6

Solving for the unknown force F, we get F = 2.33 kN

Example

Determine the magnitude and direction of the couple moment acting on the gear.

Solution

Solution on whiteboard

STATICS - Anasayfa

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