1 / 11

AP Chapter 9

AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?. Your force is applied farther from the pivot point resulting in a larger lever arm. The lever arm combined with the force creates a new phenomena called torque

andie
Download Presentation

AP Chapter 9

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. AP Chapter 9

  2. Why is it easier to open the door from the outer edge instead of closer to the hinges?

  3. Your force is applied farther from the pivot point resulting in a larger lever arm. • The lever arm combined with the force creates a new phenomena called torque • τ = Fl τ is called tau and units are Nm

  4. What is the net torque on a 4 m teeter totter if you, 58 kg, and your friend, 45 kg, are on opposite ends with the fulcrum (pivot) in the center?

  5. 2m 2m 0 m • τ =(58kg)(9.8m/s2)(2m) –(45kg)(9.8m/s2)(2m) • τ = 255 Nm 58 kg 45 kg

  6. In equilibrium • ΣFx = 0 Σ Fy = 0 Σ τ = 0

  7. Steps to solve • 1) Free-body diagram • 2) Choose convenient axes • 3) Apply the force equations • 4) Choose convenient pivot point • 5) Apply torque equation • 6) Solve

  8. If you, 55 kg, climb a 6.0 m, 25 kg ladder and stand on the rung that is 5.6 m up and the ladder is 40o from the ground. How much force does the wall apply to the ladder assuming no friction with the wall? How about the ground?

  9. 6.0m Fw 5.6m 3.0m Fyou Fgy Flad 40o Fgx

  10. ΣFx = Fgx – Fw = 0 Fgx = Fw • ΣFy = Fgy – Flad – Fyou = 0 • Fgy = Flad + Fyou = • Fgy = (25kg)(9.8m/s2) + (55kg)(9.8m/s2) • Fgy = 784 N

  11. Στ = Fw(6.0m)sin40o – Fyou(5.6m)sin50o – Flad(3.0m)sin50o = 0 • Fw(6.0m)sin40o= (55kg)(9.8m/s2)(5.6m) sin50o + (25kg)(9.8m/s2)(3.0m) sin50o • Fw = 750 N • Fgx = Fw = 750 N

More Related