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BELLRINGER

BELLRINGER. 6. Calculate the total magnification of an object viewed under a) the low power objective and b) the high power objective?. Housekeeping. Supply check Attendance Remember to sign up for UT Quest and submit your “Test Paper” to turnitin.com DUE TOMORROW

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BELLRINGER

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  1. BELLRINGER 6. Calculate the total magnification of an object viewed under a) the low power objective and b) the high power objective?

  2. Housekeeping • Supply check • Attendance • Remember to sign up for UT Quest and submit your “Test Paper” to turnitin.com • DUE TOMORROW • THIS IS THE EASIEST 100 YOU WILL EVER GET!

  3. August 26, 2010 Forces (cont’d)

  4. Free Body Diagram Fnet = manet NOS = + mg fOS = µN FOA = ma WOE = — mg ∑ Fy = mg - mg = 0 ∑ Fx = ma - µN = manet A free body diagram is the visual representation of force vectors

  5. Castro’s Guide to Forces • #1 & 2 together • 10 minutes to complete • #3 & 6 Groups 1, 4, 7 • #4 & 7 Groups 2, 5, 8 • #5 & 8 Groups 3, 6, 9

  6. #1 NOS = + mg WOE = — mg ∑ Fy = NOS - WOE = 0 ∑ Fy = mg - mg = 0

  7. #2 NOS WOE = — mg ∑ Fynet = NOS - WOE = manet

  8. #3 NOS = + mg FOA = ma ∑ Fy = NOS- WOE = 0 = mg - mg = 0 ∑ Fx = manet WOE = — mg

  9. #4 Fnet = manet NOS = + mg fOS = µN FOA = ma WOE = — mg ∑ Fy = mg - mg = 0 ∑ Fx = ma - µN = manet

  10. #5 Fnet = — manet NOS = + mg fOS = µN FOA = ma WOE = — mg ∑ Fy = mg - mg = 0 ∑ Fx = ma - µN = — manet

  11. #6 TOR = + mg WOE = — mg

  12. #6 with applied horizontal tension TOR = + mg FOA = ma WOE = — mg

  13. #6 with applied horizontal tension Can we directly measure the Tension? Is it still equal to +mg? Why or why not? TOR = + mg +y The x and y are just the components of the actual forces. This is why they’re drawn with dotted lines. You must ALWAYS draw components with dotted lines. FOA = ma -x WOE = — mg

  14. Trig Time! • Trigonometry: deals with angles and sides of triangles • S ine  (sin ) • O pposite over • H ypoteneuse • C osine  (cos ) • A djacent over • H ypoteneuse • T angent  (tan ) • O pposite over • A djacent Hyp. Y = Opp.  X =Adj.

  15. Trig sinѲ= opp = y hyp T cos Ѳ = adj = x hyp T tan Ѳ = opp = y adjx T y opp Ѳ x adj S O H C A H T O A i p y o d y a p d n p p s j p n p j

  16. Now we need to rearrange the formulas to solve for the x & y components T sinѲ= y T cos Ѳ = x T Y T sinѲ = y opp Ѳ X T cos Ѳ =x adj S O H C A H T O A i p y o d y a p d n p p s j p n p j

  17. #6 with applied horizontal tension TOR = + mg +y T sinѲ = +y FOA = ma -x T cos Ѳ =x ∑ Fy = T sinѲ - mg = 0 ∑ Fx = ma - T cosѲ = 0 WOE = — mg

  18. #7 TOR1 TOR2 Ѳ Ѳ In order for this box to just be hanging, what does the upward y component of the tension HAVE to be?? WOE = — mg

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