#### 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