Transcript Static Equilibrium Examples

Static Equilibrium Examples
•
•
•
•
•
•
•
•
Evaluating individual torques
Strut with sign
Ladder
More complicated Ladder
Traffic light with diagonal strut
Pushing lamp across floor
Door on hinges
Pulling tire over curb
Torque Definition
𝜏 = 𝑟 𝐹 𝑠𝑖𝑛𝜃
• Force times distance to rotation
point, times sinϴ
• Sin(θ) gives “perpendicularness”
– Can be F┴ times d (Type “b”)
– Can be r┴ times F (Type “a”)
Torque Tips
• Choose any point of rotation
– 2-D plane problems, x and y and one axis rotation
– Choose to minimize unknown variables
– “Ridiculous” rotations OK
• Evaluate individual torques
–
–
–
–
F┴ . r (moment arms horz and vert) (strut/wire) “Type b”
F . r┴ (forces horz and vert) (diagonal crane) “Type a”
Diagonal moment arms mix x and y (wired gate) combination
All finite object weights at CM
• Add them all up
– Keep aware of rotation point
– CCW + , CW – by convention
Example 1 – Strut with sign
• Free body diagram
• Guess where forces must be
– Weight down by beam
– Weight down by sign
– Force up at hinge
– Force up by cable
– Force to left by cable
– Force to right at hinge
Example 1 – Strut with sign (2)
• Setup
• Type “b” problem, replace FT with vert. and horz. components!
• Equilibrium in x and y
𝐹𝑦 = 0
𝐹𝐻𝑦 + 𝐹𝑇𝑦 − 245 𝑁 − 274.4 = 0
𝐹𝐻𝑦 + 𝐹𝑇𝑦 = 519.4 𝑁
𝐹𝑥 = 0
𝐹𝐻𝑥 − 𝐹𝑇𝑥 = 0
𝐹𝐻𝑥 = 𝐹𝑇𝑥
• Torque equilibrium around hinge
𝜏 = − 245 𝑁
1.1𝑚 − 274.4𝑁 2.2𝑚 + (𝐹𝑇𝑦 )(2.2𝑚) = 0
(Note FTx sinθ = 0, no torque!)
Example 1 – Strut with sign (3)
• Solve
• Note by taking torques around hinge, unknown FH disappears!
𝜏 = − 245 𝑁
1.1𝑚 − 274.4𝑁 2.2𝑚 + (𝐹𝑇𝑦 )(2.2𝑚) = 0
(𝐹𝑇𝑦 ) 2.2𝑚 = 873.2𝑚𝑁
𝐹𝑇𝑦 = 397 𝑁
𝐹𝐻𝑦 = 519.4 − 397 = 122.4 𝑁
• Little trigonometry
𝐹𝑇 =
𝐹𝑇𝑦
𝑠𝑖𝑛30
= 794 𝑁
𝐹𝑇𝑥 = 794 𝑁 cos 30 = 686.6 𝑁
𝐹𝐻𝑥 = 𝐹𝑇𝑥 = 686.6 𝑁
• Could also have taken torques around FT (ridiculous rotation)
Example 2 - Ladder
• Free body diagram (x0= 3 m by Pythagorean)
• Guess where forces must be
– Weight down by ladder at CM
– Force up by floor
– Force to right at floor
– Force to left at wall
𝐹𝑦 = 0
𝐹𝐶𝑦 − 𝑚𝑔 = 0
𝐹𝑐𝑦 = 117.6 𝑁
𝐹𝑥 = 0
𝐹𝑐𝑥 − 𝐹𝑤 = 0
𝐹𝑐𝑥 = 𝐹𝑤
Example 2 – Ladder (2)
• Type “a” problem
• Torque equilibrium around floor
𝜏 = − 117.6 𝑁
1.5 𝑚 + 𝐹𝑤 4𝑚 = 0
• r┴ moment arms
– mg - half distance along floor to wall
– Fw – height along wall to top of ladder
𝐹𝑤 =
117.6 𝑁 1.5 𝑚
4𝑚
= 44.1𝑁
𝐹𝐶𝑥 = 𝐹𝑊 = 44.1 𝑁
Note how torque “connects” x and y forces!
Example 3 – More complicated ladder
• Guess where forces must be
– Weight down by ladder at CM
– Weight down by painter
– Force up by floor
– Force to right at floor
– Force to left at wall
𝐹𝑦 = 0
𝐹𝐶𝑦 − 𝑚𝑙 𝑔 − 𝑚𝑝 𝑔 = 0
𝐹𝑐𝑦 = 656.6 𝑁
𝐹𝑥 = 0
𝐹𝑐𝑥 − 𝐹𝑤 = 0
𝐹𝑐𝑥 = 𝐹𝑤
Example 3 – More complicated ladder (2)
• Type “a” problem
• Torque equilibrium around floor
𝜏 = − 117.6 𝑁
1.5 𝑚 − (539 𝑁)(.7 ∙ 3 𝑚) + 𝐹𝑤 4𝑚 = 0
• r┴ moment arms
– mlg - half distance along floor to wall
– mpg – 0.7 distance along floor to wall
– Fw – height along wall to top of ladder
𝐹𝑤 =
1308.3 𝑚 𝑁
4𝑚
= 327 𝑁
𝐹𝐶𝑥 = 𝐹𝑊 = 327 𝑁
𝜇𝑠 =
𝐹𝑓𝑟𝑖𝑐
𝐹𝑁
=
327 𝑁
656.6
= 0.5
Example 4 -Traffic light on diagonal strut
• Guess forces on pole
– Weight down by strut at CM
– Weight down by light
– Force up at point A
– Force to left at point D
– Force to right point A
𝐹𝑦 = 0
𝐹𝐴𝑦 − 𝑚𝑙 𝑔 − 𝑚𝑝 𝑔 = 0
𝐹𝐴𝑦 = 328.3 𝑁
𝐹𝑥 = 0
𝐹𝐴𝑥 − 𝐹𝐷 = 0
𝐹𝐴𝑥 = 𝐹𝐷
There is no vertical force by cable
Example 4 -Traffic light on diagonal strut (2)
• Type “a” problem
• Torque equilibrium around point A
𝜏 = −(117.6 𝑁)(3.75 𝑚 𝑐𝑜𝑠37) − 210.7 𝑁
• r┴ moment arms
– mpg – half pole length in horiz,
– mlg – entire pole length in horiz.
– FD – vertical distance A to cable
𝐹𝐷 =
1614.4 𝑚𝑁
3.8 𝑚
= 425 𝑁
𝐹𝐴𝑥 = 𝐹𝐷 = 425 𝑁
7.5 𝑚 𝑐𝑜𝑠37 + 𝐹𝐷 3.8𝑚 = 0
Example 5 – Pushing lamp across floor
• Guess forces on lamp
– Weight down by lamp at CM
– Normal force up by floor
– Pushing force to right
– Friction force to left
𝐹𝑦 = 0
𝐹𝑁 − 𝑚𝑔 = 0
𝐹𝑁 = 70.6 𝑁
𝐹𝑥 = 0
𝐹𝑃 − 𝐹𝑓𝑟𝑖𝑐 = 0
𝐹𝑃 = 𝐹𝑓𝑟𝑖𝑐 = 𝜇𝐹𝑁 = 14.1 𝑁
Example 5 – Pushing lamp across floor (2)
• Type “a” problem
• At tipping point it will be teetering on right edge of 12 cm base
• Torque equilibrium around here
𝜏 = − 𝐹𝑃 𝑥 + 70.6 𝑁
14.1 𝑁 𝑥 − 70.6 𝑁
𝑥 = 0.6 𝑚
.12 𝑚 = 0
.12 𝑚 = 0
Example 6 - Door on hinges
Example 6 - Door on hinges (2)
• Guess forces on door
– Weight down by door at CM
– Force up by upper hinge
– Force up by lower hinge
– Force to right by upper hinge (assumed)
– Force to right lower hinge (assumed)
𝐹𝑦 = 0
𝐹𝑈𝑦 + 𝐹𝐿𝑦 − 𝑚𝑔 = 0
𝐹𝑈𝑦 + 𝐹𝐿𝑦 = 127.4 𝑁
𝐹𝑥 = 0
𝐹𝑈𝑥 + 𝐹𝐿𝑥 = 0
𝐹𝐿𝑥 = −𝐹𝑈𝑥
Example 6 - Door on hinges (3)
• Type “a” problem
• Torque equilibrium around lower hinge A
𝜏 = (127.4 𝑁)(0.65 𝑚) − 𝐹𝑈𝑥
𝐹𝑈𝑥 =
127.4 𝑁 0.65 𝑚
1.5 𝑚
= 55.2 𝑁
𝐹𝐿𝑥 = −𝐹𝑈𝑥 = −55.2 𝑁
𝑡𝑜 𝑙𝑒𝑓𝑡
𝐹𝑈𝑦 + 𝐹𝐿𝑦 = 127.4 𝑁
•
Without more information you can’t
separate FUy and FLy further
1.5 𝑚 = 0
Example 7 – Wheel on curb
• Guess forces on wheel
– Weight down by wheel at CM
– Vertical force up by curb
– Pulling force to right
– Curb force to left
• No force under wheel for it to lift!
𝐹𝑦 = 0
𝐹𝐶𝑦 − 𝑚𝑔 = 0
𝐹𝐶𝑦 = 𝑚𝑔
𝐹𝑥 = 0
−𝐹𝐶𝑥 + 𝐹 = 0
𝐹𝐶𝑥 = 𝐹
Example 7 – Wheel on curb (2)
A little geometry
• Vertical distance from
center to curb
(𝑅 − ℎ)
• Horizontal distance from
center to curb
𝑅2 − 𝑅 − ℎ
2𝑅ℎ − ℎ2
ℎ(2𝑅 − ℎ)
2
Example 7 – Wheel on curb (3)
Back to torques
• Type “a” problem
• Around curb, for force at (b)
𝜏𝑐𝑢𝑟𝑏 = 0
−𝐹 𝑅 − ℎ + 𝑀𝑔 ℎ(2𝑅 − ℎ) = 0
𝐹=
𝑀𝑔 ℎ(2𝑅−ℎ)
𝑅−ℎ
• Around curb, for force at (a)
𝐹=
𝑀𝑔 ℎ(2𝑅−ℎ)
2𝑅−ℎ
Other problems
• Problem 23 – meter stick
• Problem 22 – gymnast