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Torque & Equilibrium AP Physics
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Torque (𝝉) Turning or twisting effects due to a force
𝑭 End Slide Torque (𝝉) Pivot Point 𝑭 𝒓 ⊥ 𝜽 Turning or twisting effects due to a force The product of force and perpendicular distance from pivot point (a.k.a. level arm) Can produce changes in rotation 𝒓 𝝉=𝑭∗ 𝒓 ⊥ 𝝉=𝑭∗𝒓 𝐬𝐢𝐧 𝜽
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End Slide Torque (𝝉) Examples A bolt on a car engine needs to be tightened with a torque of 𝟑𝟓 𝑵∙𝒎. You use a 25-cm long wrench and pull on the end of the wrench at an angle of 𝟔𝟎.𝟎° from the perpendicular. How long is the lever arm, and how much force do you have to exert? 𝒓 ⊥ =𝒓 𝐬𝐢𝐧 𝜽 =𝟐𝟓∗ 𝐬𝐢𝐧 𝟔𝟎.𝟎° =𝟐𝟏.𝟕 𝒄𝒎 =𝟎.𝟐𝟏𝟕 𝒎 = 𝟑𝟓 𝟎.𝟐𝟏𝟕 ⇒𝑭= 𝝉 𝒓 ⊥ 𝝉=𝑭∗ 𝒓 ⊥ =𝟏𝟔𝟐 𝑵
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Torque (𝝉) Examples = 𝟓𝟓.𝟎 𝟏𝟑𝟓 ⇒ 𝒓 ⊥ = 𝝉 𝑭 𝝉=𝑭∗ 𝒓 ⊥ =𝟎.𝟒𝟎𝟕 𝒎 =𝟒𝟎.𝟕 𝒄𝒎
If a torque of 𝟓𝟓.𝟎 𝑵∙𝒎 is required and the largest force that can be exerted by you is 𝟏𝟑𝟓 𝑵, what is the length of the lever arm that must be used? You have a 23.4-cm long wrench. A job requires a torque of 𝟑𝟐.𝟒 𝑵∙𝒎, and you can exert a force of 𝟐𝟑𝟐 𝑵. What is the smallest angle, with respect to the wrench, at which the force can be exerted? = 𝟓𝟓.𝟎 𝟏𝟑𝟓 ⇒ 𝒓 ⊥ = 𝝉 𝑭 𝝉=𝑭∗ 𝒓 ⊥ =𝟎.𝟒𝟎𝟕 𝒎 =𝟒𝟎.𝟕 𝒄𝒎 = 𝟑𝟐.𝟒 𝟐𝟑𝟐∗𝟎.𝟐𝟑𝟒 ⇒ 𝐬𝐢𝐧 𝜽 > 𝝉 𝑭∗𝒓 𝝉=𝑭∗ 𝒓 ⊥ =𝑭∗𝒓 𝐬𝐢𝐧 𝜽 𝐬𝐢𝐧 𝜽 >𝟎.𝟓𝟗𝟕 ⇒𝜽> 𝐬𝐢𝐧 −𝟏 𝟎.𝟓𝟗𝟕 =𝟑𝟔.𝟔° You stand on the pedal of a bicycle. If you have a mass of 𝟔𝟓 𝒌𝒈, the pedal makes an angle of 𝟑𝟓° above the horizontal, and the pedal is 𝟏𝟖 𝒄𝒎 from the center of the chain ring, how much torque would you exert at this moment?
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Torque (𝝉) Examples 𝑭 𝒈 𝝉=𝑭∗ 𝒓 ⊥ =𝒎𝒈∗ 𝒓 ⊥ =𝟔𝟓∗𝟗.𝟖∗𝟎.𝟏𝟒𝟕 =𝟗𝟑.𝟗 𝑵∙𝒎
End Slide Torque (𝝉) Examples You stand on the pedal of a bicycle. If you have a mass of 𝟔𝟓 𝒌𝒈, the pedal makes an angle of 𝟑𝟓° above the horizontal, and the pedal is 𝟏𝟖 𝒄𝒎 from the center of the chain ring, how much torque would you exert at this moment? 𝑭 𝒈 𝝉=𝑭∗ 𝒓 ⊥ =𝒎𝒈∗ 𝒓 ⊥ =𝟔𝟓∗𝟗.𝟖∗𝟎.𝟏𝟒𝟕 𝒓 =𝟗𝟑.𝟗 𝑵∙𝒎 𝟑𝟓° 𝒓 ⊥ =𝒓 𝐜𝐨𝐬 𝟑𝟓° =𝟎.𝟏𝟖 𝐜𝐨𝐬 𝟑𝟓° =𝟎.𝟏𝟒𝟕 𝒎
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Equilibrium Conditions of Equilibrium:
End Slide Equilibrium 𝑭 𝟏 = 𝑭 𝒈𝟏 + 𝑭 𝒈𝟐 Conditions of Equilibrium: 𝚺𝑭=𝟎 𝚺𝝉=𝟎 Counter-Clockwise = Positive Clockwise = Negative In the diagram to the right, with how much force is the fulcrum pushing up? What is 𝒓 𝟐 ? 𝑭 𝟏 =𝟎.𝟓∗𝟗.𝟖+𝟎.𝟔∗𝟗.𝟖 ⇒ 𝑭 𝒖𝒑 = 𝑭 𝒅𝒐𝒘𝒏 ⇒ 𝑭 𝒍𝒆𝒇𝒕 = 𝑭 𝒓𝒊𝒈𝒉𝒕 𝑭 𝟏 =𝟏𝟎.𝟕𝟖 𝑵 ⇒ 𝝉 𝑪𝑪𝑾 = 𝝉 𝑪𝑾 𝒓 𝟏 =𝟓𝟏 𝒄𝒎 𝒓 𝟐 =? 𝒎 𝟏 = 𝟓𝟎𝟎 𝒈 𝒎 𝟐 = 𝟔𝟎𝟎 𝒈 𝑭𝒖𝒍𝒄𝒓𝒖𝒎 (𝑷𝒊𝒗𝒐𝒕 𝑷𝒐𝒊𝒏𝒕) 𝟎.𝟓∗𝟗.𝟖∗𝟎.𝟓𝟏=𝟎.𝟔∗𝟗.𝟖∗ 𝒓 𝟐 𝟐.𝟒𝟗𝟗=𝟓.𝟖𝟖∗ 𝒓 𝟐 𝑭 𝒈𝟏 ∗ 𝒓 𝟏 = 𝑭 𝒈𝟐 ∗ 𝒓 𝟐 𝒓 𝟐 =𝟎.𝟒𝟐𝟓 𝒎 =𝟒𝟐.𝟓 𝒄𝒎
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Equilibrium Can I choose a different pivot point?
End Slide Equilibrium Can I choose a different pivot point? I don’t want to system to rotate about ANY point, right? What if I choose here, can I still determine 𝒓 𝟐 ? 𝑭 𝟏 =𝟏𝟎.𝟕𝟖 𝑵 𝒓 𝟏 =𝟓𝟏 𝒄𝒎 𝒓 𝟐 =? 𝒎 𝟏 = 𝟓𝟎𝟎 𝒈 𝒎 𝟐 = 𝟔𝟎𝟎 𝒈 𝝉 𝑪𝑪𝑾 = 𝝉 𝑪𝑾 𝑭𝒖𝒍𝒄𝒓𝒖𝒎 (𝑷𝒊𝒗𝒐𝒕 𝑷𝒐𝒊𝒏𝒕) 𝑭 𝟏 ∗ 𝒓 𝟏 = 𝑭 𝒈𝟐 ∗ 𝒓 𝟏 + 𝒓 𝟐 𝟏𝟎.𝟕𝟖∗𝟎.𝟓𝟏=𝟎.𝟔∗𝟗.𝟖∗ 𝟎.𝟓𝟏+ 𝒓 𝟐 𝒓 𝟐 =𝟎.𝟒𝟐𝟓 𝒎 𝟓.𝟓𝟎=𝟓.𝟖𝟖∗ 𝟎.𝟓𝟏+ 𝒓 𝟐 𝟎.𝟗𝟑𝟓=𝟎.𝟓𝟏+ 𝒓 𝟐 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒓 𝟐 =𝟎.𝟒𝟐𝟓
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End Slide Equilibrium Examples Carl, whose mass is 43 kg, sits 1.8 m from the center of a seesaw. Steve, whose mass is 52 kg, wants to balance Carl. How far from the center of the seesaw should Steve sit? 𝝉 𝑪𝑪𝑾 = 𝝉 𝑪𝑾 𝑭 𝒈𝟏 ∗ 𝒓 𝟏 = 𝑭 𝒈𝟐 ∗ 𝒓 𝟐 𝟒𝟑∗𝟗.𝟖∗𝟏.𝟖=𝟓𝟐∗𝟗.𝟖∗ 𝒓 𝟐 𝟕𝟓𝟖.𝟓𝟐=𝟓𝟎𝟗.𝟔∗ 𝒓 𝟐 𝒓 𝟐 =𝟏.𝟒𝟗 𝒎
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End Slide Car on a Bridge A bridge with an evenly distributed mass of 1200 kg spans a gap of 86 meters. A car of mass 720 kg is 25 m onto the bridge from the right. How much force is holding up each end of the bridge? 𝑭𝒊𝒓𝒔𝒕 𝑪𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝑺𝒆𝒄𝒐𝒏𝒅 𝑪𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝚺𝑭=𝟎 ⇒ 𝚺 𝑭 𝒖𝒑 =𝚺 𝑭 𝒅𝒐𝒘𝒏 𝚺𝝉=𝟎 ⇒ 𝚺 𝝉 𝑪𝑾 =𝚺 𝝉 𝑪𝑪𝑾 𝑭 𝟏 + 𝑭 𝟐 = 𝑭 𝒈𝑩 + 𝑭 𝒈𝑪 𝑭 𝒈𝑩 ∗ 𝒓 𝟏 + 𝑭 𝒈𝑪 ∗ 𝒓 𝟐 = 𝑭 𝟐 ∗ 𝒓 𝟑 𝑭 𝟏 + 𝑭 𝟐 =𝟏𝟐𝟎𝟎∗𝟗.𝟖+𝟕𝟐𝟎∗𝟗.𝟖 𝟏𝟐𝟎𝟎∗𝟗.𝟖∗𝟒𝟑+𝟕𝟐𝟎∗𝟗.𝟖∗𝟔𝟏= 𝑭 𝟐 ∗𝟖𝟔 𝑭 𝟏 =𝟏𝟖,𝟖𝟏𝟔− 𝑭 𝟐 𝟗𝟑𝟔,𝟎𝟗𝟔= 𝑭 𝟐 ∗𝟖𝟔 𝑭 𝟏 =𝟕,𝟗𝟑𝟏 𝑵 𝑭 𝟐 =𝟏𝟎,𝟖𝟖𝟒 𝑵 Pivot Point 𝒓 𝟏 =𝟒𝟑 𝒎 𝑭 𝟏 𝑭 𝟐 𝑭 𝒈𝑪 𝒓 𝟐 =𝟔𝟏 𝒎 𝑭 𝒈𝑩 𝒓 𝟑 =𝟖𝟔 𝒎
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End Slide Car on a Bridge The same bridge with mass of 1200 kg spanning an 86-meter gap has a 720-kg car that is now at 68 m from the right. How much force is holding up each end of the bridge this time? 𝑭𝒊𝒓𝒔𝒕 𝑪𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝑺𝒆𝒄𝒐𝒏𝒅 𝑪𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝚺𝑭=𝟎 ⇒ 𝚺 𝑭 𝒖𝒑 =𝚺 𝑭 𝒅𝒐𝒘𝒏 𝚺𝝉=𝟎 ⇒ 𝚺 𝝉 𝑪𝑾 =𝚺 𝝉 𝑪𝑪𝑾 𝑭 𝟏 + 𝑭 𝟐 = 𝑭 𝒈𝑩 + 𝑭 𝒈𝑪 𝑭 𝒈𝑩 ∗ 𝒓 𝟏 + 𝑭 𝒈𝑪 ∗ 𝒓 𝟐 = 𝑭 𝟐 ∗ 𝒓 𝟑 𝑭 𝟏 + 𝑭 𝟐 =𝟏𝟐𝟎𝟎∗𝟗.𝟖+𝟕𝟐𝟎∗𝟗.𝟖 𝟏𝟐𝟎𝟎∗𝟗.𝟖∗𝟒𝟑+𝟕𝟐𝟎∗𝟗.𝟖∗𝟏𝟖= 𝑭 𝟐 ∗𝟖𝟔 𝑭 𝟏 =𝟏𝟖,𝟖𝟏𝟔− 𝑭 𝟐 𝟔𝟑𝟐,𝟔𝟖𝟖= 𝑭 𝟐 ∗𝟖𝟔 𝑭 𝟏 =𝟏𝟏,𝟒𝟓𝟗 𝑵 𝑭 𝟐 =𝟕,𝟑𝟓𝟕 𝑵 𝑭 𝒈𝑪 Pivot Point 𝒓 𝟏 =𝟒𝟑 𝒎 𝑭 𝟏 𝑭 𝟐 𝒓 𝟐 =𝟏𝟖 𝒎 𝑭 𝒈𝑩 𝒓 𝟑 =𝟖𝟔 𝒎
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Tilt the Box 𝝉 𝑪𝑾 ≥ 𝝉 𝑪𝑪𝑾 𝑭∗ 𝒓 𝟏 ≥𝒎𝒈∗ 𝒓 𝟐 𝑭∗𝟏.𝟏≥𝟐𝟕∗𝟗.𝟖∗𝟎.𝟒
End Slide Tilt the Box Consider the rectangular block of mass 27 kg, height 1.1 m, and length 0.8 m. A force F is applied horizontally at the upper edge. A) What minimum force is required to start to tip the block? The acceleration of gravity is 9.8 m/s2. Answer in units of N. 𝝉 𝑪𝑾 ≥ 𝝉 𝑪𝑪𝑾 Normal Force and Force of Friction are being applied at the pivot point; thus, no torque. Because the box will be tilted, all of the Normal Force will be applied at the corner rather than evenly distributed on the bottom. 𝑭∗ 𝒓 𝟏 ≥𝒎𝒈∗ 𝒓 𝟐 𝑭∗𝟏.𝟏≥𝟐𝟕∗𝟗.𝟖∗𝟎.𝟒 𝑭 𝑭 𝑭∗𝟏.𝟏≥𝟏𝟎𝟓.𝟖𝟒 𝑭≥𝟗𝟔.𝟐 𝑵 𝟏.𝟏 𝒎 𝑭 𝑵 Pivot Point 𝑭 𝑵 𝟎.𝟖 𝒎 𝑭 𝒇 𝑭 𝒈 𝑭 𝒇 𝑭 𝒈
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End Slide Tilt the Box Consider the rectangular block of mass 27 kg, height 1.1 m, and length 0.8 m. A force F is applied horizontally at the upper edge. B) What minimum coefficient of static friction is required for the block to tip with the application of a force of this magnitude? 𝑭 𝑵 Free-Body Diagram 𝑭 𝑵 = 𝑭 𝒈 𝑭< 𝑭 𝒇 𝑭 𝒇 𝑭 𝑭 𝑵 =𝒎𝒈 𝑭< 𝝁 𝒔 𝑭 𝑵 𝑭 𝒈 𝑭 𝑭 𝑵 =𝟐𝟕∗𝟗.𝟖 𝟗𝟔.𝟐< 𝝁 𝒔 𝟐𝟔𝟒.𝟔 𝑭 𝑵 =𝟐𝟔𝟒.𝟔 𝑵 𝝁 𝒔 >𝟎.𝟑𝟔𝟒 Pivot Point 𝑭 𝑵 𝑭 𝒇 𝑭 𝒈
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Jill Climbs a Ladder What’s the system? The ladder
Consider a uniform ladder of weight 250 N and length 10 m leaning against a smooth, vertical wall and is resting on a floor with friction. Jill weighs 600 N and is standing on the ladder 2 m from the bottom. The foot of the ladder is 0.8 m from the base of the wall. A) What is the Normal force on the ladder exert by the ground? 𝑭 𝑵𝟐 Free-Body Diagram 𝟏𝟎 𝒎 𝟐 𝒎 𝟎.𝟖 𝒎 What’s the system? 𝟏𝟎 𝒎 The ladder 𝑭 𝒈𝟏 𝑭 𝑵𝟏 𝟐 𝒎 𝑭 𝒈𝟐 𝑭 𝒇 𝟎.𝟖 𝒎
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Jill Climbs a Ladder 𝚺 𝑭 𝒖𝒑 =𝚺 𝑭 𝒅𝒐𝒘𝒏 𝚺 𝑭 𝒍𝒆𝒇𝒕 =𝚺 𝑭 𝒓𝒊𝒈𝒉𝒕
End Slide Jill Climbs a Ladder Consider a uniform ladder of weight 250 N and length 10 m leaning against a smooth, vertical wall and is resting on a floor with friction. Jill weighs 600 N and is standing on the ladder 2 m from the bottom. The foot of the ladder is 0.8 m from the base of the wall. A) What is the Normal force on the ladder exert by the ground? 𝑭 𝑵𝟏 Free-Body Diagram 𝑭𝒊𝒓𝒔𝒕 𝑪𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝚺𝑭=𝟎 𝚺 𝑭 𝒖𝒑 =𝚺 𝑭 𝒅𝒐𝒘𝒏 𝚺 𝑭 𝒍𝒆𝒇𝒕 =𝚺 𝑭 𝒓𝒊𝒈𝒉𝒕 𝑭 𝑵𝟐 𝑭 𝒇 𝑭 𝑵𝟏 = 𝑭 𝒈𝟏 + 𝑭 𝒈𝟐 𝑭 𝒇 = 𝑭 𝑵𝟐 𝑭 𝒈𝟐 𝑭 𝒈𝟏 𝑭 𝑵𝟏 =𝟐𝟓𝟎+𝟔𝟎𝟎 𝑭 𝑵𝟏 =𝟖𝟓𝟎 𝑵
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Jill Climbs a Ladder Consider a uniform ladder of weight 250 N and length 10 m leaning against a smooth, vertical wall and is resting on a floor with friction. Jill weighs 600 N and is standing on the ladder 2 m from the bottom. The foot of the ladder is 0.8 m from the base of the wall. B) What is the force exerted by the wall? 𝑭 𝑵𝟐 𝑭 𝑵𝟐 𝟏𝟎 𝒎 No Torque 𝑭 𝒈𝟏 𝑭 𝑵𝟏 𝟎.𝟖 𝒎 𝑭 𝒈𝟐 𝑭 𝒈𝟐 𝑭 𝒈𝟏 No Torque Pivot Point 𝑭 𝒇
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Jill Climbs a Ladder 𝝉 𝑪𝑪𝑾 = 𝝉 𝑪𝑾
End Slide Jill Climbs a Ladder Consider a uniform ladder of weight 250 N and length 10 m leaning against a smooth, vertical wall and is resting on a floor with friction. Jill weighs 600 N and is standing on the ladder 2 m from the bottom. The foot of the ladder is 0.8 m from the base of the wall. B) What is the force exerted by the wall? 𝑺𝒊𝒏𝒄𝒆 𝒕𝒉𝒆 𝒄𝒆𝒏𝒕𝒆𝒓 𝒐𝒇 𝒎𝒂𝒔𝒔 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒍𝒂𝒅𝒅𝒆𝒓 𝒊𝒔 𝒉𝒂𝒍𝒇𝒘𝒂𝒚 𝒖𝒑 𝒕𝒉𝒆 𝒍𝒂𝒅𝒅𝒆𝒓, 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒍𝒆𝒗𝒆𝒓 𝒊𝒔 𝒉𝒂𝒍𝒇𝒘𝒂𝒚 𝒂𝒍𝒐𝒏𝒈 𝒕𝒉𝒆 𝒃𝒐𝒕𝒕𝒐𝒎. 𝑭 𝑵𝟐 𝑳𝒆𝒗𝒆𝒓 𝑨𝒓𝒎𝒔 𝑻𝒉𝒆 𝒑𝒆𝒓𝒔𝒐𝒏 𝒊𝒔 𝟏 𝟓 𝒕𝒉 𝒕𝒉𝒆 𝒘𝒂𝒚 𝒖𝒑 𝒕𝒉𝒆 𝒍𝒂𝒅𝒅𝒆𝒓. 𝑻𝒉𝒖𝒔, 𝒕𝒉𝒆 𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎 𝒊𝒔 𝟏 𝟓 𝒕𝒉 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒂𝒍𝒐𝒏𝒈 𝒕𝒉𝒆 𝒃𝒐𝒕𝒕𝒐𝒎. 𝒓 𝑵𝟐 =𝟗.𝟗𝟕 𝒎 𝒓 𝒈𝟏 =𝟎.𝟒𝟎 𝒎 𝟏𝟎 𝟐 − 𝟎.𝟖 𝟐 𝟏𝟎 𝒎 𝒓 𝒈𝟐 =𝟎.𝟏𝟔 𝒎 =𝟗.𝟗𝟕 𝒎 𝝉 𝑪𝑪𝑾 = 𝝉 𝑪𝑾 𝟎.𝟖 𝒎 𝑭 𝑵𝟐 ∗ 𝒓 𝑵𝟐 = 𝑭 𝒈𝟏 ∗ 𝒓 𝒈𝟏 + 𝑭 𝒈𝟐 ∗ 𝒓 𝒈𝟐 𝑭 𝒈𝟐 𝑭 𝒈𝟏 𝑭 𝑵𝟐 ∗𝟗.𝟗𝟕=𝟐𝟓𝟎∗𝟎.𝟒𝟎+𝟔𝟎𝟎∗𝟎.𝟏𝟔 𝑭 𝑵𝟐 ∗𝟗.𝟗𝟕=𝟏𝟗𝟔 𝑭 𝑵𝟐 =𝟏𝟗.𝟔𝟔 𝑵
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Jill Climbs a Ladder 𝚺 𝑭 𝒍𝒆𝒇𝒕 =𝚺 𝑭 𝒓𝒊𝒈𝒉𝒕 𝑭 𝒇 ≤ 𝝁 𝒔 𝑭 𝑵𝟏 𝑭 𝒇 = 𝑭 𝑵𝟐
End Slide Jill Climbs a Ladder Consider a uniform ladder of weight 250 N and length 10 m leaning against a smooth, vertical wall and is resting on a floor with friction. Jill weighs 600 N and is standing on the ladder 2 m from the bottom. The foot of the ladder is 0.8 m from the base of the wall. C) What is the minimum coefficient of static friction between the ladder and the ground? 𝚺 𝑭 𝒍𝒆𝒇𝒕 =𝚺 𝑭 𝒓𝒊𝒈𝒉𝒕 𝑭 𝒇 ≤ 𝝁 𝒔 𝑭 𝑵𝟏 𝑭 𝒇 = 𝑭 𝑵𝟐 𝟏𝟗.𝟔𝟔≤ 𝝁 𝒔 ∗𝟖𝟓𝟎 𝟏𝟎 𝒎 𝑭 𝒇 =𝟏𝟗.𝟔𝟔 𝑵 𝝁 𝒔 ≥𝟎.𝟎𝟐𝟑 𝟐 𝒎 𝟎.𝟖 𝒎
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