1 Equilibrium of Concurrent, Coplanar Force Systems EF 202 - Week 5.

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Equilibrium of Concurrent,

Coplanar Force SystemsEF 202 - Week 5

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Equilibrium•Newton’s First Law - If, and only if, an object’s “mass center” has zero acceleration, then the sum of ALL of the forces acting on the object (body) is zero.

•Where is a body’s mass center? We’ll study that in Module 4. But in this class, the entire body is at rest, so we know that the mass center, wherever it is, has zero acceleration.

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Problem Solving Strategy

1.Identify ALL forces acting on the body by making a “free-body diagram” (FBD).

2.Invoke Newton’s First Law. (Add all the force vectors on the FBD and set the result equal to zero.)

3.Solve.

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Free-Body Diagram

1.Define the “body.”

2.“Free” the body from the rest of the world by cutting through EVERYTHING that connects the body to anything else.

3.Sketch the freed body.

4.To the sketch, add EVERY force that is (or may be) acting on the body.

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Concurrent, Coplanar Force

System•Coplanar system

•The lines of action of all forces lie in a common plane.

•Concurrent system

•The lines of action of all forces intersect at a common point.

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Independent Equations

•In a given problem, we can find only as many unknowns as we have independent equations.

•For a system of coplanar, concurrent forces, Newton’s First Law yields only two independent equations for a given FBD.

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1. Make FBD.2. Newton’s First Law3. Solve.

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rF ∑ = 0

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F cosθˆ i + sinθˆ j ( ) + −4.5 kN( )ˆ i +L

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L 7.5 kN −sin30oˆ i − cos30oˆ j ( ) +L

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L 2.25 kN cos60oˆ i − sin60oˆ j ( ) = 0

1. Figure is already an FBD.2.

Problem 3-4Find magnitude and direction of F.

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For the components:

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ˆ i

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F cosθ − 4.5 kN − 7.5 kNsin 30o + 2.25 kNcos60o = 0

For the components:

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ˆ j

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F sinθ − 7.5 kNcos30o − 2.25 kNsin60o = 0

COUNT EQUATIONS AND UNKNOWNS!!!!!!!

Now, what?

(1)

(2)

Two equations, (1) and (2), and two unknowns, F and .

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θ

3. Collect and equate components.

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From (1) and (2),

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F cosθ = 7.125 kN

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F sinθ = 8.444 kN

(3)

(4)

Now, what?

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tanθ =1.185⇒ θ = 49.8o

For the magnitude,

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F = Fx2 + Fy

2 = F cosθ( )2

+ F sinθ( )2

F =11.0 kN

For the angle, divide (4)

by (3) to get

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Ideal Cable

•Neglect weight (massless).

•Neglect bending stiffness.

•Force parallel to cable.

•Force only tensile.

•Neglect stretching (inextensible).

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1. Make FBD.2. Newton’s First Law3. Solve.

Problem 3-8Find the forces in cables AB and AC.

12 kg

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COUNT EQUATIONS AND UNKNOWNS!!!!!!!

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Problem 3-40Find the forces in all five cables.

30kg

Since there are 5 unknowns, we’ll need at least 3 FBD’s.

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Linear Spring - I

•Like an ideal cable, but extensible and resists either stretching or shortening.

•Force is either tensile or compressive.

•When the force is zero, the spring is undeformed and its length is called its free length or natural length.

•When the force is not zero, it is proportional to the spring’s deformation (stretch or contraction).

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Linear Spring - II

•The constant of proportionality between the force in a spring and its deformation is called the spring’s stiffness or its spring rate.

•Stiffness is usually represented by the letter k.

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Equation for Spring

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l0 = natural/free/undeformed length

(length when F = 0)

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F = k l − l0( )stretch

1 2 3

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F = k l0 − l( )contraction1 2 3

Tension

Compression

OR

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Assumed Tension

•Directions of loads are reversed on objects to which springs are attached (Newton III).

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F = k l − l0( )

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Assumed Compression

•Direction assumed for spring load cannot be wrong!

•Equation used for spring load can be wrong!

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F = k l0 − l( )

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Problem 3-14Spring AB is stretched 3m. What is the mass of the suspended block?

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