engineering mechanics - equilibrium

11
Engineering Mechanics Lecture 2 5/10, Friday 3-4am

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Engineering Mechanics - Equilibrium

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Page 1: Engineering Mechanics - Equilibrium

Engineering MechanicsLecture 25/10, Friday

3-4am

Page 2: Engineering Mechanics - Equilibrium

2-Particle Interactions

1

2

f21

f12

Sun : earth

Atom : atom

12 21= −f f

12 21+ =f f 0

Page 3: Engineering Mechanics - Equilibrium

3-Particle Interactions

1

2f21

f13

ij ji+ =f f 03

f12

f31

f23

f32

F3

F1

F2, 1, 2,3i j =

Page 4: Engineering Mechanics - Equilibrium

Many-Particle Interactions

1

2f21

f13

3

f12

f31

f23

f32

F3

F1

F2

i ijj i≠

+ =∑F f 0 1,2,i =

i iji i j i≠

+ =∑ ∑∑F f 0

Summation of the above equation gives

Newton’s 3rd law

iji j i≠

=∑∑ f 0

Therefore:

ii

=∑F 0

For an N particles system, in equilibrium

Newton’s 1st law

Page 5: Engineering Mechanics - Equilibrium

Free-Body-Diagram

Isolate the body in question with the representation of all external forces acting on it.

ii

=∑F 0

Page 6: Engineering Mechanics - Equilibrium

Problem 1

Page 7: Engineering Mechanics - Equilibrium

Remove Spike from Timber

It is desired to remove the spike from the timber by applying force along its horizontal axis. An obstruction A prevents direct access, so that two forces, one is 1.6kN and the other P, are applied by cables as shown. Compute the magnitude of P necessary to ensure axial tension T along the spike. Also find T.

Page 8: Engineering Mechanics - Equilibrium

Remove Spike from Timber

T

sin 1.6sinP a β=

αβ

Horizontal:

Vertical:

cos 1.6cosT P a β= +

Page 9: Engineering Mechanics - Equilibrium

Problem 2

Page 10: Engineering Mechanics - Equilibrium

Knot in Static Equilibrium?

Mass: m

α β

F1 F2

1 2cos cosF Fα β=

1 2sin sinF F mgα β+ =

Horizontal:

Vertical:

Page 11: Engineering Mechanics - Equilibrium

When strings are almost straight horizontally

1

2

cossin cos cos sin

cossin cos cos sin

F mg

F mg

βα β α β

αα β α β

=+

=+

when00

αβ→→

1

2

FF→∞→∞

Mass: m

α β

F1 F2