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

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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!

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August 26, 2010

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A free body diagram is the visual representation of force vectors W OE = — mg N OS = + mg F OA = maf OS = µN ∑ F y = mg - mg = 0 ∑ F x = ma - µN = ma net F net = ma net

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#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

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W OE = — mg N OS = + mg ∑ F y = N OS - W OE = 0 ∑ F y = mg - mg = 0

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W OE = — mg N OS ∑ F ynet = N OS - W OE = ma net

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W OE = — mg N OS = + mg F OA = ma ∑ F y = N OS - W OE = 0 = mg - mg = 0 ∑ F x = ma net

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W OE = — mg N OS = + mg F OA = maf OS = µN ∑ F y = mg - mg = 0 ∑ F x = ma - µN = ma net F net = ma net

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W OE = — mg N OS = + mg F OA = maf OS = µN ∑ F y = mg - mg = 0 ∑ F x = ma - µN = — ma net F net = — ma net

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T OR = + mg W OE = — mg

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T OR = + mg W OE = — mg F OA = ma

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T OR = + mg W OE = — mg Can we directly measure the Tension? Is it still equal to +mg? Why or why not? -x +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. F OA = ma

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Trigonometry: deals with angles and sides of triangles Sine (sin ) Opposite over Hypoteneuse C osine (cos ) Adjacent over Hypoteneuse T angent (tan ) Opposite over A djacent Y = Opp. X =Adj. Hyp.

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x adj y opp T S O HC A HT O A i p y o d y a p d n p p s j pn p j sin Ѳ = opp = y hyp T cos Ѳ = adj = x hyp T tan Ѳ = opp = y adj x Ѳ

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X T cos Ѳ =x adj Y T sin Ѳ = y opp T S O HC A HT O A i p y o d y a p d n p p s j pn p j T sin Ѳ = y T cos Ѳ = x Ѳ

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T OR = + mg W OE = — mg -x T cos Ѳ =x +y T sin Ѳ = +y ∑ F y = T sin Ѳ - mg = 0 ∑ F x = ma - T cos Ѳ = 0 F OA = ma

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W OE = — mg Ѳ Ѳ T OR2 T OR1 In order for this box to just be hanging, what does the upward y component of the tension HAVE to be??

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