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_14_EGR1301_Fall2014_Statics_141006.pptx
Newton’s Laws
Calibri 20
I. Every object in a state of uniform motion tends to remain in
that state of motion unless an external force is applied to it.
Arial 18
I. Every object in a state of uniform motion tends to remain in
that state of motion unless an external force is applied to it.
Sans Serif 18 I. Every object in a state of uniform motion tends to remain in that
state of motion unless an external force is applied to it.
Times New I. Every object in a state of uniform motion tends to remain in
Roman 20
that state of motion unless an external force is applied to it.
Calibri 20
II. The relationship between an object's mass m, its acceleration
a, and the applied force F is F = ma. Acceleration and force are
vectors (as indicated by their symbols being displayed in slant
bold font); in this law the direction of the force vector is the
same as the direction of the acceleration vector.
Calibri 20
III. For every action there is an equal and opposite reaction.
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_14_EGR1301_Fall2014_Statics_141006.pptx
Force Terminology
I. Forces acting along the same line of action are called collinear
forces
II. Forces whose lines of action pass through the same point in
space are called concurrent forces
III. Forces that lie in the same plane are called coplanar forces
Moments at a Point
If only one force can cause rotation at a point, the moment is the product of that force
and the perpendicular distance from the line of action of the force to the point. If more
than once force is involved, then superimpose their individual moments on the point.
Counterclockwise is the reference (positive) direction of rotation.
Equilibrium
A body is in equilibrium when the sum of all external forces and moments acting on the
body is zero. This requires the body to be at rest or moving with a constant velocity.
𝑭𝒙 = 𝟎
π‘­π’š = 𝟎
𝑴 = 𝟎 for any point that can rotate. Use counterclockwise as positive.
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_14_EGR1301_Fall2014_Statics_141006.pptx
Forces Possible from Supports, Surfaces, and Connectors
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_14_EGR1301_Fall2014_Statics_141006.pptx
Draw the Free-Body Diagram
Cut the body free from all others, all supports, and all connectors. They are
replaced with external forces acting upon the body under consideration.
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_14_EGR1301_Fall2014_Statics_141006.pptx
Write the Equations
4.0 m
6.5 m
8.5 m
Unknowns 𝐴π‘₯ , 𝐴𝑦 , 𝐡𝑦
1 0 0 𝐴π‘₯
βˆ’47
0 1 1 𝐴𝑦 = 81.41
0 0 6.5 𝐡𝑦
880.0
47
𝐹π‘₯ = 𝐴π‘₯ + 94 cos 60° = 0
81.41
𝐹𝑦 = 𝐴𝑦 + 𝐡𝑦 βˆ’ 94 sin 60° = 0
All points are free to rotate. Usually the only points not free are a beam
imbedded into a wall, or a clamp. A good rule is to chose a point that has the
most unknowns. In this case, point A is the best choice.
βˆ’188.0
βˆ’188
-692.0
𝑀𝐴 = βˆ’94 cos 60° βˆ™ 4.0 + βˆ’94 sin 60° βˆ™ 8.5 + 𝐡𝑦 βˆ™ 6.5 = 0
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_14_EGR1301_Fall2014_Statics_141006.pptx
Draw the Free-Body Diagram
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_14_EGR1301_Fall2014_Statics_141006.pptx
Some trig, then write the equations
tπ‘Žπ‘› πœƒ =
6
,
8
P
πœƒ = 36.9°
Pang=90°+∝ +πœƒ
Py
P
Py
Px
Px
𝑃π‘₯ = π‘ƒπ‘π‘œπ‘ (π‘ƒπ‘Žπ‘›π‘” )
𝑃𝑦 = 𝑃𝑠𝑖𝑛(π‘ƒπ‘Žπ‘›π‘” )
= 196.14 N
Slope of GH is given as 1: 1, π‘ π‘œ ∝ +πœƒ = 45°, ∝ = 45° βˆ’ 36.9° = 8.1°
π‘ƒπ‘Žπ‘›π‘” = 90° + 45° = 135°, 𝑃π‘₯ = 𝑃 π‘π‘œπ‘  (135°), 𝑃𝑦 = 𝑃 𝑠𝑖𝑛 (135°)
𝐹π‘₯ = 𝐺π‘₯ + 𝑃π‘₯ = 0, 𝐺π‘₯ βˆ’0.7071𝑃 = 0
Unknowns 𝐺π‘₯ , 𝐺𝑦 , 𝑃
𝐹𝑦 = 𝐺𝑦 + 𝑃𝑦 βˆ’ 196.14, 𝐺𝑦 + 0.7071𝑃 βˆ’ 196.14 = 0
𝑀𝐺 = 𝑃𝑦 βˆ™ 10π‘π‘œπ‘ 8.1° βˆ’ 𝑃π‘₯ βˆ™ 10𝑠𝑖𝑛8.1° βˆ’ 196.14 βˆ™ 12π‘π‘œπ‘ 8.1° = 0,
𝑃 𝑠𝑖𝑛(135°) βˆ™ 10π‘π‘œπ‘ 8.1° βˆ’ 𝑃 π‘π‘œπ‘ (135°) βˆ™ 10𝑠𝑖𝑛8.1° βˆ’ 196.14 βˆ™ 12π‘π‘œπ‘ 8.1° = 0
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_14_EGR1301_Fall2014_Statics_141006.pptx
𝐹π‘₯ = 𝐺π‘₯ + 𝑃π‘₯ = 0, 𝐺π‘₯ βˆ’0.7071𝑃 = 0
Unknowns 𝐺π‘₯ , 𝐺𝑦 , 𝑃
𝐹𝑦 = 𝐺𝑦 + 𝑃𝑦 βˆ’ 196.14, 𝐺𝑦 + 0.7071𝑃 βˆ’ 196.14 = 0
𝑀𝐺 = 𝑃𝑦 βˆ™ 10π‘π‘œπ‘ 8.1° βˆ’ 𝑃π‘₯ βˆ™ 10𝑠𝑖𝑛8.1° βˆ’ 196.14 βˆ™ 12π‘π‘œπ‘ 8.1° = 0,
𝑃 𝑠𝑖𝑛(135°) βˆ™ 10π‘π‘œπ‘ 8.1° βˆ’ 𝑃 π‘π‘œπ‘ (135°) βˆ™ 10𝑠𝑖𝑛8.1° βˆ’ 196.14 βˆ™ 12π‘π‘œπ‘ 8.1° = 0,
𝑃 βˆ™ 0.7071 βˆ™ 10 βˆ™ 0.9900 βˆ’ 𝑃 βˆ™ (βˆ’0.7071) βˆ™ 10 βˆ™ 0.1409 βˆ’ 196.14 βˆ™ 12 βˆ™ 0.9900 = 0,
𝑃 βˆ™ 7.000 βˆ’ 𝑃 βˆ™ (βˆ’0.9963) βˆ’ 2330 = 0,
1 0 βˆ’0.7071 𝐺π‘₯
0
0 1 0.7071 𝐺𝑦 = 196.14
0 0
7.996
2330
𝑃
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_14_EGR1301_Fall2014_Statics_141006.pptx
Draw the Free-Body Diagram
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_14_EGR1301_Fall2014_Statics_141006.pptx
Write the Equations
𝐴π‘₯
𝐴𝑦
𝐹π‘₯ = 𝐴π‘₯ βˆ’ 270 = 0
𝐹𝑦 = 𝐴𝑦 βˆ’ 250 βˆ’ 180 + 𝐡𝑦 = 0
𝐡𝑦
𝑀𝐴 = βˆ’250 βˆ™ 1.3 βˆ’ 180 βˆ™ 3.0 βˆ’ 270 βˆ™ 1.0 + 𝐡𝑦 βˆ™ (1.3 + 1.7 + 1.0 βˆ’ 1.5) = 0,
βˆ’1135 + 2.5𝐡𝑦 = 0,
1
0
0
0 0 𝐴π‘₯
270
1 1 𝐴𝑦 = 430
0 2.5 𝐡𝑦
1135
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_14_EGR1301_Fall2014_Statics_141006.pptx
Draw the Free-Body Diagram
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_14_EGR1301_Fall2014_Statics_141006.pptx
Some Trig to Determine the Cable Angle
Tang
E
42⁰
D
π΄π΅π‘π‘œπ‘ 42° = 3.1β€² , 𝐴𝐡 = 4.171β€²
𝐷𝐡 = 𝐴𝐡𝑠𝑖𝑛42° = 2.791β€²
𝐢𝐸 = 𝐢𝐴 βˆ’ 𝐷𝐡 = 3.9 βˆ’ 2.791 = 1.109β€²
𝐢𝐸
1.109
π‘‡π‘Žπ‘›π‘” = 180° βˆ’ π‘Žπ‘‘π‘Žπ‘›
= 180° βˆ’ π‘Žπ‘‘π‘Žπ‘›
= 180° βˆ’ 19.7° = 160.3°
𝐸𝐡
3.1
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_14_EGR1301_Fall2014_Statics_141006.pptx
Freebody Diagram and Equations
Tang = πŸπŸ”πŸŽ. πŸ‘°
E
T
Ty
Tx
𝟐. πŸ•πŸ—πŸβ€²
𝟐. πŸ•πŸ—πŸβ€²
42⁰
Ax
D
Ay
𝐹π‘₯ = 𝐴π‘₯ + 𝑇π‘₯ = 0, 𝐴π‘₯ + π‘‡π‘π‘œπ‘  160.3° = 0
1 0
0 1
0 0
0
βˆ’0.9415 𝐴π‘₯
0.3371 𝐴𝑦 = 1200
7680
3.673
𝑇
𝐹𝑦 = 𝐴𝑦 + 𝑇𝑦 βˆ’ 1200 = 0, 𝐴𝑦 + 𝑇𝑠𝑖𝑛 160.3° βˆ’ 1200 = 0
𝑀𝐴 = 𝑇𝑦 βˆ™ 3.1 βˆ’ 𝑇π‘₯ βˆ™ 2.791 βˆ’ 1200 βˆ™ 6.4 = 0
= 𝑇𝑠𝑖𝑛 160.3° βˆ™ 3.1 βˆ’ π‘‡π‘π‘œπ‘  160.3° βˆ™ 2.791 βˆ’ 7680 = 0
1.045
-2.628
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_14_EGR1301_Fall2014_Statics_141006.pptx
Draw the Free-Body Diagram
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_14_EGR1301_Fall2014_Statics_141006.pptx
Draw the Free-Body Diagram
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_14_EGR1301_Fall2014_Statics_141006.pptx
Some trig, then write the equations
8.0 βˆ’ 6.0 π‘ π‘–π‘›πœƒ
𝐢
π‘ͺ
Angle of π‘ͺ
∝
𝐢𝑦
𝐢π‘₯
∝
6.0 π‘π‘œπ‘ πœƒ
If πœƒ is known (0° to 75° in increments of 1°),
β€’ Then angle ∝ of the top triangle can be found,
𝐴π‘₯
β€’ then the angle of C is (180 βˆ’ ∝),
β€’ then 𝐢π‘₯ = π‘π‘œπ‘  of C, and 𝐢𝑦 = 𝑠𝑖𝑛 of C
𝐹π‘₯ = 𝐴π‘₯ + 𝐢π‘₯ = 0, 𝐴π‘₯ + πΆπ‘π‘œπ‘ (π‘π‘Žπ‘›π‘”) = 0
𝐴𝑦
𝐹𝑦 = 𝐴𝑦 + 𝐢𝑦 βˆ’ 1200 = 0, 𝐴𝑦 + 𝐢𝑠𝑖𝑛 π‘π‘Žπ‘›π‘” βˆ’ 1200 = 0
𝑀𝐢 = βˆ’πΆπ‘₯ βˆ™ 6.0π‘ π‘–π‘›πœƒ + 𝐢𝑦 βˆ™ 6.0 π‘π‘œπ‘ πœƒ βˆ’ 1200 βˆ™ 8.0π‘π‘œπ‘ πœƒ = 0,
βˆ’πΆπ‘π‘œπ‘ (π‘π‘Žπ‘›π‘”) βˆ™ 6.0π‘ π‘–π‘›πœƒ + 𝐢𝑠𝑖𝑛 π‘π‘Žπ‘›π‘” βˆ™ 6.0 π‘π‘œπ‘ πœƒ βˆ’ 1200 βˆ™ 8.0π‘π‘œπ‘ πœƒ = 0
1
0
0
0
π‘π‘œπ‘ (π‘π‘Žπ‘›π‘”)
1
𝑠𝑖𝑛(π‘π‘Žπ‘›π‘”)
0 βˆ’6 cos π‘π‘Žπ‘›π‘” βˆ™ π‘ π‘–π‘›πœƒ + 6 𝑠𝑖𝑛 π‘π‘Žπ‘›π‘” βˆ™ π‘π‘œπ‘ πœƒ
𝐴π‘₯
0
𝐴𝑦 =
1200
9600π‘π‘œπ‘ πœƒ
𝐢
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_14_EGR1301_Fall2014_Statics_141006.pptx
Two-Dimensional Problem: Loosening a Bolt with a Wrench,
When the Force is Not Perpendicular to the Wrench
y
Bolt Head
𝑫
P
πœƒπ·
πœƒπΉ
x
𝑭
Perpendicular distance 𝑃 = 𝐷 𝑠𝑖𝑛(πœƒπ· βˆ’ πœƒπΉ ),
𝑭 = F πœƒπΉ = F cos πœƒπΉ π‘Žx + sin πœƒπΉ π‘Žy = 𝐹π‘₯ π‘Žx + 𝐹𝑦 π‘Žy
𝑫 = D πœƒπ· = D cos πœƒπ· π‘Žx + sin πœƒπ· π‘Žy = 𝐷π‘₯ π‘Žx + 𝐷𝑦 π‘Ž y
Moment 𝑀 = 𝐹 𝑃
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_14_EGR1301_Fall2014_Statics_141006.pptx
Useful trig identities for this problem are
sin 𝐴 ± 𝐡 = sin 𝐴 cos 𝐡 ± cos 𝐴 sin 𝐡
cos 𝐴 ± 𝐡 = cos 𝐴 cos 𝐡 βˆ“ sin 𝐴 sin 𝐡.
Thus, P can be expressed as
𝑃 = 𝐷 𝑠𝑖𝑛(πœƒπ· βˆ’ πœƒπΉ ) = 𝐷 sin πœƒπ· cos πœƒπΉ βˆ’ cos πœƒπ· sin πœƒπΉ .
Now, substitute P into the M equation,
Moment 𝑀 = 𝐹 𝑃 = 𝐹 𝐷 sin πœƒπ· cos πœƒπΉ βˆ’ cos πœƒπ· sin πœƒπΉ .
In determinant form, Moment 𝑀 = 𝐹 𝐷
cos πœƒπΉ
cos πœƒπ·
sin πœƒπΉ
sin πœƒπ·
𝑭 = F cos πœƒπΉ π‘Žx + sin πœƒπΉ π‘Žy = 𝐹π‘₯ π‘Žx + 𝐹𝑦 π‘Žy
𝑫 = D cos πœƒπ· π‘Žx + sin πœƒπ· π‘Žy = 𝐷π‘₯ π‘Žx + 𝐷𝑦 π‘Žy
π‘Žx π‘Žy
𝑴 = 𝑭 × π‘« = 𝐹π‘₯ 𝐹𝑦
𝐷π‘₯ 𝐷𝑦
Same
π‘Žz 0
𝐹𝑧 0
𝐷𝑧
= 𝐹π‘₯ 𝐷𝑦 βˆ’ 𝐹𝑦 𝐷π‘₯ π‘Žz
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_14_EGR1301_Fall2014_Statics_141006.pptx
Chapter 13. Strength of Materials
Stress 𝜎 is force per unit area. Unit is Pascal ( 𝜎 ) Newtons per square
meter.
The conversion to U.S. Customary units is
1 MPa = 145.05 psi (pounds force per square inch)
Example, 310 π‘€π‘ƒπ‘Ž
𝑝𝑠𝑖
145.05 π‘€π‘ƒπ‘Ž = 44,966 𝑝𝑠𝑖 = 44.966 π‘˜π‘π‘ π‘–
In reverse,
1 kpsi = 6.894 MPa
Example, 44.966 π‘˜π‘π‘ π‘–
π‘€π‘ƒπ‘Ž
6.894 π‘˜π‘π‘ π‘– = 45,966 𝑝𝑠𝑖 = 310 π‘€π‘ƒπ‘Ž
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_14_EGR1301_Fall2014_Statics_141006.pptx
Three Types of Stress ( 𝝈 )
β€’ Tensile. Axial load such as a cable. Tends to pull the atoms
apart.
β€’ Compressive. Axial load such as driving a fence post into the
ground. Tends to push atoms together
β€’ Shear. Tends to slide atoms across each other. Such as sliding
the top half of a stack of plywood across the bottom half,
without lifting.
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_14_EGR1301_Fall2014_Statics_141006.pptx
Strain ( 𝜺 )
β€’ Dimensionless ratio, change in
length to the original length
Stress ( 𝝈 )
βˆ†π‘³
𝜺=
𝑳
β€’ Proportional Limit – linear
region
β€’ Elastic Limit – Material will still
return to original length when
load is removed
Strain ( 𝜺 )
β€’ Yield Strength – permanently
deformed
β€’ Ultimate Strength – Fails
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_14_EGR1301_Fall2014_Statics_141006.pptx
Some trig, then write the equations
acd
T
cda
8.0 m
dac
ead
Tang
sin(π‘Žπ‘π‘‘)
9.2
=
sin(50°)
,
8.0
so π‘Žπ‘π‘‘ = 61.8°.
π‘‘π‘Žπ‘ = 180° βˆ’ π‘π‘‘π‘Ž βˆ’ π‘Žπ‘π‘‘, so π‘‘π‘Žπ‘ = 68.2°.
π‘’π‘Žπ‘‘ = 90° βˆ’ π‘‘π‘Žπ‘ = 21.8°.
π‘‘π‘’π‘Ž = 90° βˆ’ π‘Žπ‘π‘‘ = 28.2°.
dea
E
Tang = 180° βˆ’ π‘‘π‘’π‘Ž = 151.8°.
Material
Yield Strength MPa
Iron
210
Structural Steel
240
Stainless Steel
210
Aluminum
85
Copper, hard drawn
415
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_14_EGR1301_Fall2014_Statics_141006.pptx
Draw the Free-Body Diagram
Beam weight is modeled as a force at the center of mass
𝑻𝒙 = 𝑻𝒄𝒐𝒔(π‘»π’‚π’π’ˆ )
12.4 m
π‘»π’š = π‘»π’”π’Šπ’(π‘»π’‚π’π’ˆ)
6.2 m
8.2 kg/m β€’ 12.4 m = 101.7
kg, corresponding to 997 N
6.2 m
21.8⁰
π‘»π’š
𝑻𝒙
Ay
21.8⁰
𝐹𝑦 = 𝐴𝑦 + 𝑇𝑦 βˆ’ 1471 βˆ’ 997 = 0
𝑀𝐴 = βˆ’997 (6.2cos 21.8°) βˆ’ 𝑻𝒙 (9.2sin 21.8°)
+ π‘»π’š (9.2cos 21.8°) βˆ’ 1471(12.4cos 21.8°) = 0
Write the Equations
𝑻
𝐹π‘₯ = 𝐴π‘₯ + 𝑇π‘₯ = 0
150 kg β€’
9.807 =
1471 N
Beam 101.7 kg,
997 N
Tang = 151.8°
Ax
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_14_EGR1301_Fall2014_Statics_141006.pptx
𝑇π‘₯ = π‘‡π‘π‘œπ‘ (π‘‡π‘Žπ‘›π‘” ), 𝑇𝑦 = 𝑇𝑠𝑖𝑛(π‘‡π‘Žπ‘›π‘” ), π‘‡π‘Žπ‘›π‘” = 151.8°
-0.8813
𝐹π‘₯ = 𝐴π‘₯ + 𝑇π‘₯ = 0, 𝐴π‘₯ + π‘‡π‘π‘œπ‘ (151.8° ) = 0
0.4726
𝐹𝑦 = 𝐴𝑦 + 𝑇𝑦 βˆ’ 1471 βˆ’ 997 = 0, 𝐴𝑦 + 𝑇𝑠𝑖𝑛(151.8° ) βˆ’ 2468 = 0
5739
-0.8813
3.417
0.4726
8.542
𝑀𝐴 = βˆ’997 (6.2cos 21.8°) βˆ’ π‘‡π‘π‘œπ‘ (151.8° ) βˆ™ (9.2sin 21.8°) + 𝑇𝑠𝑖𝑛(151.8° ) βˆ™ (9.2cos 21.8°)
βˆ’ 1471(12.4cos 21.8°) = 0
16936
𝐴π‘₯
βˆ’0.8813
0
𝐴𝑦 =
0.4726
2468
βˆ’3.011 + 4.037 𝑇
5739 + 16936
1
0
0
0
1
0
1
0
0
0 βˆ’0.8813 𝐴π‘₯
0
1 0.4726 𝐴𝑦 = 2468
0
1.026
22675
𝑇
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_14_EGR1301_Fall2014_Statics_141006.pptx
25
_14_EGR1301_Fall2014_Statics_141006.pptx
Cubic Meter of Air, Moving at V m/s
Air density
π‘˜π‘”
𝜌 = 1.225𝐾𝑇 𝐾𝐴 3
π‘š
where 𝐾𝑇 and 𝐾𝐴 are correction
factors for altitude and pressure,
and equal 1 at standard 15⁰C, 1
atmosphere (sea level).
Air mass passing through A square meters of surface normal to V is
π‘˜π‘”
π‘š
π‘˜π‘”
2
𝜌 3 π΄π‘š
𝑉
= πœŒπ΄π‘‰
π‘š
𝑠
𝑠
Kinetic energy of the mass passing through surface A per second (i.e., power) is
1
π‘˜π‘”
πœŒπ΄π‘‰
2
𝑠
𝑉2
π‘š2
𝑠2
1
π‘˜π‘” βˆ™ π‘š2 1 1
π½π‘œπ‘’π‘™π‘’π‘ 
1
3
3
= πœŒπ΄π‘‰
βˆ™ = πœŒπ΄π‘‰
= πœŒπ΄π‘‰ 3 π‘Šπ‘Žπ‘‘π‘‘π‘ 
2
2
𝑠
𝑠 2
𝑠
2
Joules/sec
Force and pressure of the air are
1
π‘˜π‘” βˆ™ π‘š2 1
π‘š 1
1
3
2 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ ,
2 π‘ƒπ‘Žπ‘ π‘π‘Žπ‘™s
𝐹 = πœŒπ΄π‘‰
βˆ™
÷
𝑉
=
πœŒπ΄π‘‰
𝑃
=
πœŒπ‘‰
2
2
2
𝑠
𝑠
𝑠
2
Joules/sec ÷ meters/sec
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_14_EGR1301_Fall2014_Statics_141006.pptx
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Cubic meter of air, moving at V m/s
Air density
π‘˜π‘”
𝜌 = 1.225𝐾𝑇 𝐾𝐴 3
π‘š
where 𝐾𝑇 and 𝐾𝐴 are correction
factors for altitude and pressure,
and equal 1 at standard 15⁰C, 1
atmosphere (sea level).
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_14_EGR1301_Fall2014_Statics_141006.pptx
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Prob. 1. A playground see-saw has a 10 kg child on one end, and an 80 kg adult on
the other end. Write the equations to find the distance x for the center of mass of
100 kg to achieve balance.
80 kg
m
100 kg
m
10 kg
m
x
2.5 m
Pivot
2.5 m
𝐹π‘₯ = 0 (there are no forces in the x direction)
𝐹𝑦 = βˆ’10 βˆ’ 100 βˆ’ 80 9.807 + πΉπ‘π‘–π‘£π‘œπ‘‘ = 0, πΉπ‘π‘–π‘£π‘œπ‘‘ = 190 βˆ™ 9.807 𝑁
π‘€π‘π‘–π‘£π‘œπ‘‘ = 10 9.807 βˆ™ 2.5 + 100(9.807)π‘₯ βˆ’ 80(9.807) βˆ™ 2.5 = 0
In this problem, only the
π‘€π‘π‘–π‘£π‘œπ‘‘ = 9.807(10 βˆ™ 2.5 + 100π‘₯ βˆ’ 80 βˆ™ 2.5) = 0 moment equation is needed
to answer the question.
175
25 + 100π‘₯ βˆ’ 200 = 0,
π‘₯=
= 1.75 π‘š
100
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EGR1301, Test 2, Oct. 08, 2014. Show all work on these sheets. Name. __________________________
y-axis
3βˆ’3π’„π’π’”πŸ‘πŸŽ°
𝛂
Prob. 2. A 100 kg mass is suspended by a cable, twothirds the way up on a 3 meter long, weightless pipe.
The pipe is cabled to the wall.
Tension T
3π’„π’π’”πŸ”πŸŽ°
3π’„π’π’”πŸ‘πŸŽ°
ο‚·
1m
wall
ο‚·
3m
2m
3π’”π’Šπ’πŸ”πŸŽ°
2π’”π’Šπ’πŸ”πŸŽ°
πŸ‘πŸŽ°
Fy
100
kg
Write the two force equations, and the moment
equation. Use the joint at bottom of the beam for
the moment equation.
Put the equations into standard matrix form,
combining terms where possible. Do not solve.
𝛼 = π‘Žπ‘‘π‘Žπ‘›
3 βˆ’ 3π‘π‘œπ‘ 30°
3π‘π‘œπ‘ 60°
60⁰
Fx
x-axis
πΉπ‘œπ‘Ÿπ‘π‘’π‘ π‘₯ = 𝑇cos(180° βˆ’ 𝛼) + 𝐹π‘₯ = 0
2π’„π’π’”πŸ”πŸŽ°
3π’„π’π’”πŸ”πŸŽ°
πΉπ‘œπ‘Ÿπ‘π‘’π‘ π‘¦ = 𝑇 sin 180° βˆ’ 𝛼 + 𝐹𝑦 βˆ’ 100(9.807) = 0
𝑀 = βˆ’π‘‡π‘π‘œπ‘ (180° βˆ’ 𝛼) βˆ™ 3π‘π‘œπ‘ 30° + 𝑇 𝑠𝑖𝑛 180° βˆ’ 𝛼 βˆ™ 3π‘π‘œπ‘ 60° βˆ’ 100(9.807) βˆ™ 2π‘π‘œπ‘ 60° = 0
1
0
0
0
1
0
cos(180° βˆ’ 𝛼)
sin(180° βˆ’ 𝛼)
βˆ’π‘‡π‘π‘œπ‘ (180° βˆ’ 𝛼) βˆ™ 3π‘π‘œπ‘ 30° + 𝑇 𝑠𝑖𝑛 180° βˆ’ 𝛼 βˆ™ 3π‘π‘œπ‘ 60°
𝐹π‘₯
0
𝐹𝑦 =
980.7
980.7 βˆ™ 2π‘π‘œπ‘ 60°
𝑇
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Prob. 3. The Sky Cam at the football stadium is 60 kg and supported by four cables.
In this problem, you will consider a two-dimensional version of the Sky Cam.
Assume that the Sky Cam assembly is a weightless rigid frame, and the entire 60 kg
mass is in the large ball at the bottom.
ο‚· Write the two force equations, and the moment equation. Use the left-hand
ring for the moment equation.
ο‚· Put the equations into standard matrix form, combining terms where possible.
Do not solve.
y-axis
Tension TL
120⁰
Tension TR
1m
1m
𝐹π‘₯ = 𝑇𝐿 π‘π‘œπ‘ 120° + 𝑇𝑅 π‘π‘œπ‘ 30° = 0
30⁰
x-axis
2m
𝐹𝑦 = 𝑇𝐿 𝑠𝑖𝑛120° + 𝑇𝑅 𝑠𝑖𝑛30° βˆ’ 60(9.807) = 0
60 kg
𝑀 = 𝑇𝑅 𝑠𝑖𝑛30° βˆ™ 2π‘š βˆ’ 60(9.807) βˆ™ 1π‘š = 0
π‘π‘œπ‘ 120°
𝑠𝑖𝑛120°
0
π‘π‘œπ‘ 30°
𝑠𝑖𝑛30°
2𝑠𝑖𝑛30°
0
0
0
0
𝑇𝐿
𝑇𝑅 = 60(9.807)
60(9.807)
0
In this problem, the first two
equations were sufficient.
The third equation can be
used as a check.
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