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Calculating Total Magnification Under Low and High Power Objectives

Understand total magnification under different objectives. Reminder for UT Quest and Test Paper submission. Time-sensitive task!

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Calculating Total Magnification Under Low and High Power Objectives

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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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