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12. Static Equilibrium

12. Static Equilibrium. Conditions for Equilibrium. A bridge is an example of a system in static equilibrium . The bridge undergoes neither linear nor rotational motion!. Conditions for Equilibrium. A system is in static equilibrium if:. the net external force is zero.

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12. Static Equilibrium

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  1. 12. Static Equilibrium

  2. Conditions for Equilibrium A bridge is an example of a system in static equilibrium. The bridge undergoes neither linear nor rotational motion!

  3. Conditions for Equilibrium A system is in static equilibrium if: the netexternal force is zero the netexternal torque is zero

  4. Problem Solving Guideline • All static equilibrium problems are solved the same • way: • Find all external forces • Choose a pivot • Find all external torques • Set net force to zero • Set net torque to zero • Solve for unknown quantities

  5. Problem Solving Guideline It is generally simpler to choose the pivot at the point of application of the force for which you have the least information.

  6. Example – A Drawbridge What is the tension in the supporting cable of a 14 m, 11,000 kg drawbridge? Forces:1. Force at pivot2. Tension in cable 3. Weight of bridge

  7. Example – A Drawbridge Pivot:A sensible choice is the hinge since we do not know the exact direction of the hinge force, nor do we care about it!

  8. Example – A Drawbridge Torques Due to the weightτg = –(L/2) mg sinθ1 (This torque is into the page. Why?) Due to the tension τT = LT sinθ2

  9. Examples of Static Equilibrium

  10. Example – A Leaning Ladder At what minimum angle can the ladder lean without slipping? The wall is frictionless and there is friction between the floor and the ladder.

  11. Example – A Leaning Ladder Forces: 1. Normal force at bottom of ladder 2. Friction force at bottom of ladder 3. Ladder’s weight 4. Normal force at top of ladder Pivot: Choose bottom of ladder Why?

  12. Example – A Leaning Ladder Torques: 1. Due to ladder’s weight 2. Due to normal force at top of ladder Solve: Force, x: μn1 – n2 = 0 Force, y: n1 – mg = 0 Torque: Ln2sinϕ – (L/2) mg cosϕ = 0

  13. Example – A Leaning Ladder From the force equations we getn2 = μmg. Therefore, μ sinϕ – (1/2)cosϕ = 0 and so, tanϕ = 1/(2μ)

  14. Example – Standing on a Plank

  15. Example – Standing on a Plank Net force: Net torque:

  16. Example – Standing on a Plank Force on right scale Force on left scale Do these make sense?

  17. Example – Force on Elbow What is the force on the elbow? m = 6 kg Assume biceps force acts 3.4 cm from pivot point O.

  18. Example – Force on Elbow Model forearm as a horizontal rod

  19. Example – Force on Elbow The force we know least about is the force on the elbow. So, let’s take the elbow (O) as the pivot. Net torque:

  20. Example – Force on Elbow Net force x: y:

  21. Summary • For a system to be in static equilibrium, both the net force and the net torque must be zero. • When solving static equilibrium problems, it often simplifies things to choose the pivot so that the torque from unknown forces is zero.

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